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Irregular triangle read by rows: the group of 2n + 1 integers starting at A014105(n).
+20
5
0, 3, 4, 5, 10, 11, 12, 13, 14, 21, 22, 23, 24, 25, 26, 27, 36, 37, 38, 39, 40, 41, 42, 43, 44, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 136, 137
COMMENTS
The squares of numbers in each row can be gathered in an equation with the first n terms on one side, the next n+1 terms on the other. The third row, for example, could be rendered as 10^2 + 11^2 + 12^2 = 13^2 + 14^2.
This sequence contains all nonnegative integers that are within a distance of n from 2n^2 + 2n where n is any nonnegative integer. The nonnegative integers that are not in this sequence are of the form 2n^2 + k where n is any positive integer and -n <= k <= n-1. Also, when n is the product of two consecutive integers, a(n) = 2n; for example, a(20) = 40. See explicit formulas for the sequence in the formula section below. - Dennis P. Walsh, Aug 09 2013 Numbers k with the property that the largest Dyck path of the symmetric representation of sigma(k) has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
FORMULA
As a triangle, T(n,k) = 2n^2 + 2n + k where -n <= k <= n and n = 0,1,... - Dennis P. Walsh, Aug 09 2013
EXAMPLE
Triangle begins:
0;
3, 4, 5;
10, 11, 12, 13, 14;
21, 22, 23, 24, 25, 26, 27;
36, 37, 38, 39, 40, 41, 42, 43, 44;
55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65;
78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90;
...
MAPLE
seq(n+floor(sqrt(n))*(floor(sqrt(n))+1), n=0..100); # Dennis P. Walsh, Aug 09 2013
CROSSREFS
Row sums give the even-indexed terms of A027480.
EXTENSIONS
Definition clarified, 8th row terms corrected by R. J. Mathar, Jul 19 2009
Irregular triangle read by rows in which row n lists the row A014105(n) of A237591, n >= 1.
+20
5
2, 1, 6, 2, 1, 1, 11, 4, 3, 1, 1, 1, 19, 6, 4, 2, 2, 1, 1, 1, 28, 10, 5, 3, 3, 2, 1, 1, 1, 1, 40, 13, 7, 5, 3, 2, 2, 2, 1, 1, 1, 1, 53, 18, 10, 5, 4, 3, 3, 2, 1, 2, 1, 1, 1, 1, 69, 23, 12, 7, 5, 4, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 86, 29, 15, 9, 6, 5, 4, 2, 3, 2, 2, 1, 2, 1, 1, 1, 1, 1
COMMENTS
The characteristic shape of the symmetric representation of sigma( A014105(n)) consists in that in the main diagonal of the diagram the smallest Dyck path has a peak and the largest Dyck path has a valley. So knowing this characteristic shape we can know if a number is a second hexagonal number (or not) just by looking at the diagram, even ignoring the concept of second hexagonal number.
Therefore we can see a geometric pattern of the distribution of the second hexagonal numbers in the stepped pyramid described in A245092. T(n,k) is also the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma( A014105(n)), from the border to the center, hence the sum of the n-th row of triangle is equal to A014105(n). T(n,k) is also the difference between the total number of partitions of all positive integers <= n-th second hexagonal number into exactly k consecutive parts, and the total number of partitions of all positive integers <= n-th second hexagonal number into exactly k + 1 consecutive parts.
EXAMPLE
Triangle begins:
2, 1;
6, 2, 1, 1;
11, 4, 3, 1, 1, 1;
19, 6, 4, 2, 2, 1, 1, 1;
28, 10, 5, 3, 3, 2, 1, 1, 1, 1;
40, 13, 7, 5, 3, 2, 2, 2, 1, 1, 1, 1;
53, 18, 10, 5, 4, 3, 3, 2, 1, 2, 1, 1, 1, 1;
69, 23, 12, 7, 5, 4, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1;
86, 29, 15, 9, 6, 5, 4, 2, 3, 2, 2, 1, 2, 1, 1, 1, 1, 1;
...
Illustration of initial terms:
Column h gives the n-th second hexagonal number ( A014105). Column S gives the sum of the divisors of the second hexagonal numbers which equals the area (and the number of cells) of the associated diagram.
----------------------------------------------------------------------------
n h S Diagram
----------------------------------------------------------------------------
_ _ _
| | | | | |
_ _|_| | | | |
1 3 4 |_ _|1 | | | |
2 | | | |
_ _| | | |
| _ _| | |
_ _|_| | |
| _|1 | |
_ _ _ _ _| | 1 | |
2 10 18 |_ _ _ _ _ _|2 | |
6 _ _ _ _|_|
| |
_| |
| _|
_ _|_|
_ _| _|1
|_ _ _|1 1
| 3
|4
_ _ _ _ _ _ _ _ _ _ _| \
3 21 32 |_ _ _ _ _ _ _ _ _ _ _| \
11 |\
_| \
| \
_ _| _\
_ _| _| \
| _|1 \
_ _ _| _ _|1 1
| | 2
| _ _ _ _|2
| | 4
| |
| |6
| |
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
4 36 91 |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
19
.
The symmetric representation of sigma(36) is partially illustrated because it is too big to include totally here.
1, 8, 1024, 2097152, 68719476736, 36028797018963968, 302231454903657293676544, 40564819207303340847894502572032, 87112285931760246646623899502532662132736, 2993155353253689176481146537402947624255349848014848
COMMENTS
a(n) is the number of simple labeled graphs on 2(n+1) nodes such that every vertex has odd degree. The complements of these graphs are precisely the Eulerian graphs on 2(n+1) nodes. a(1) = 8 because we have: K_4; K_1,3; and K_2 + K_2 with 1,4, and 3 labelings respectively: 1 + 4 + 3 = 8. Cf. A006125. - Geoffrey Critzer, Feb 16 2020
FORMULA
a(n) = (-1)^floor(n/2)/Product_{i=1..2*n} cos(i*Pi/(2*n+1))^i.
MATHEMATICA
Table[2^(Binomial[n, 2] - (n - 1)), {n, 2, 20, 2}] (* Geoffrey Critzer, Feb 16 2020 *)
PROG
(PARI) a(n)=2^(n*(2*n+1))
1, 2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86
MATHEMATICA
a=2; b=1;
F[n_]:=a*n^2+b*n;
R[n_]:=(n/a+((b-1)/(2a))^2)^(1/2);
G[n_]:=n-1+Ceiling[R[n]-(b-1)/(2a)];
Table[G[n], {n, 100}]
PROG
(Python)
from math import isqrt
Convolution of the even-indexed triangular numbers ( A014105) and the squares ( A000290).
+20
2
0, 0, 3, 22, 88, 258, 623, 1316, 2520, 4476, 7491, 11946, 18304, 27118, 39039, 54824, 75344, 101592, 134691, 175902, 226632, 288442, 363055, 452364, 558440, 683540, 830115, 1000818, 1198512, 1426278, 1687423, 1985488, 2324256, 2707760, 3140291, 3626406
COMMENTS
This sequence was originally proposed in a comment on A071245 by J. M. Bergot as giving the first differences. Therefore, a(n) gives the partial sums of A071245.
FORMULA
O.g.f.: x^2*(1 + x)*(3 + x)/(1 - x)^6 = (x*(3 + x)/(1 - x)^3)*(x*(1 + x)/(1 - x)^3).
a(n) = binomial(n+1, 3)*(4*n^2 + 5*n + 4)/10 = (n - 1)*n*(n + 1)*(4*n^2 + 5*n + 4)/60.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5. - Colin Barker, Oct 21 2016
MATHEMATICA
Table[(n - 1) n (n + 1) (4 n^2 + 5 n + 4)/60, {n, 0, 40}] (* Bruno Berselli, Oct 21 2016 *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 0, 3, 22, 88, 258}, 40] (* Harvey P. Dale, Jun 04 2023 *)
PROG
(PARI) concat(vector(2), Vec(x^2*(1+x)*(3+x)/(1-x)^6 + O(x^50))) \\ Colin Barker, Oct 21 2016 (Magma) [Binomial(n+1, 3)*(4*n^2 +5*n +4)/10: n in [0..40]]; // G. C. Greubel, Oct 22 2018
Numbers that are product of a second hexagonal number ( A014105) and a square pyramidal numbers ( A000330) in at least two ways.
+20
1
0, 105, 300, 855, 1155, 2940, 13860, 14700, 17850, 20790, 22230, 27300, 33930, 70125, 73920, 87780, 114400, 116025, 135135, 145530, 157080, 195000, 213150, 235290, 304590, 347655, 381150, 431340, 451044, 471975, 566580, 632700, 764400, 796950, 942480, 950040
COMMENTS
We have A000330(n) = 1 + 2^2 + ... + n^2 and A014105(m) = 0^2 - 1^2 + 2^2 -+ ... + (2m)^2, so the terms of this sequence are the numbers that are a product, in at least two ways, of a partial sum of squares times a (positive) partial sum of squares with alternating signs (with + for even terms; cf. A306121 for the opposite convention). The initial a(1) = 0 is added for completeness.
Below 10^8, the number 17850 is the only one to have four representations of the given form, and 6347250 is the only one to have exactly three.
PROG
(PARI) {my(L=10^6, A14105(a)=a*(2*a+1), A330(b)=(b+1)*b*(2*b+1)/6, A=S=[]); for(b=1, sqrtnint(L\A14105(1)\3, 3), for(a=1, oo, if( setsearch(S, t=A14105(a)*A330(b)), A=setunion(A, [t]), t>L&&next(2); S=setunion(S, [t])))); A}
Triangular numbers: a(n) = binomial(n+1,2) = n*(n+1)/2 = 0 + 1 + 2 + ... + n.
(Formerly M2535 N1002)
+10
4676
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431
COMMENTS
Also referred to as T(n) or C(n+1, 2) or binomial(n+1, 2) (preferred).
Also generalized hexagonal numbers: n*(2*n-1), n=0, +-1, +-2, +-3, ... Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. In this case k = 6. - Omar E. Pol, Sep 13 2011 and Aug 04 2012 Number of edges in complete graph of order n+1, K_{n+1}.
Number of legal ways to insert a pair of parentheses in a string of n letters. E.g., there are 6 ways for three letters: (a)bc, (ab)c, (abc), a(b)c, a(bc), ab(c). Proof: there are C(n+2,2) ways to choose where the parentheses might go, but n + 1 of them are illegal because the parentheses are adjacent. Cf. A002415. For n >= 1, a(n) is also the genus of a nonsingular curve of degree n+2, such as the Fermat curve x^(n+2) + y^(n+2) = 1. - Ahmed Fares (ahmedfares(AT)my_deja.com), Feb 21 2001
From Harnack's theorem (1876), the number of branches of a nonsingular curve of order n is bounded by a(n-1)+1, and the bound can be achieved. See also A152947. - Benoit Cloitre, Aug 29 2002. Corrected by Robert McLachlan, Aug 19 2024 Number of tiles in the set of double-n dominoes. - Scott A. Brown, Sep 24 2002 Number of ways a chain of n non-identical links can be broken up. This is based on a similar problem in the field of proteomics: the number of ways a peptide of n amino acid residues can be broken up in a mass spectrometer. In general, each amino acid has a different mass, so AB and BC would have different masses. - James A. Raymond, Apr 08 2003 Triangular numbers - odd numbers = shifted triangular numbers; 1, 3, 6, 10, 15, 21, ... - 1, 3, 5, 7, 9, 11, ... = 0, 0, 1, 3, 6, 10, ... - Xavier Acloque, Oct 31 2003 [Corrected by Derek Orr, May 05 2015] Centered polygonal numbers are the result of [number of sides * A000217 + 1]. E.g., centered pentagonal numbers (1,6,16,31,...) = 5 * (0,1,3,6,...) + 1. Centered heptagonal numbers (1,8,22,43,...) = 7 * (0,1,3,6,...) + 1. - Xavier Acloque, Oct 31 2003 Maximum number of lines formed by the intersection of n+1 planes. - Ron R. King, Mar 29 2004 Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - Mike Zabrocki, Aug 26 2004 Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012 Number of ways two different numbers can be selected from the set {0,1,2,...,n} without repetition, or, number of ways two different numbers can be selected from the set {1,2,...,n} with repetition.
Conjecturally, 1, 6, 120 are the only numbers that are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
Binomial transform is {0, 1, 5, 18, 56, 160, 432, ...}, A001793 with one leading zero. - Philippe Deléham, Aug 02 2005 Each pair of neighboring terms adds to a perfect square. - Zak Seidov, Mar 21 2006 Number of transpositions in the symmetric group of n+1 letters, i.e., the number of permutations that leave all but two elements fixed. - Geoffrey Critzer, Jun 23 2006 With rho(n):=exp(i*2*Pi/n) (an n-th root of 1) one has, for n >= 1, rho(n)^a(n) = (-1)^(n+1). Just use the triviality a(2*k+1) == 0 (mod (2*k+1)) and a(2*k) == k (mod (2*k)).
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3)^(n-1). - Sergio Falcon, Feb 12 2007 a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 2 variables. - Richard Barnes, Sep 06 2017 Equal to the rank (minimal cardinality of a generating set) of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, May 03 2007 a(n) gives the total number of triangles found when cevians are drawn from a single vertex on a triangle to the side opposite that vertex, where n = the number of cevians drawn+1. For instance, with 1 cevian drawn, n = 1+1 = 2 and a(n)= 2*(2+1)/2 = 3 so there is a total of 3 triangles in the figure. If 2 cevians are drawn from one point to the opposite side, then n = 1+2 = 3 and a(n) = 3*(3+1)/2 = 6 so there is a total of 6 triangles in the figure. - Noah Priluck (npriluck(AT)gmail.com), Apr 30 2007
For n >= 1, a(n) is the number of ways in which n-1 can be written as a sum of three nonnegative integers if representations differing in the order of the terms are considered to be different. In other words, for n >= 1, a(n) is the number of nonnegative integral solutions of the equation x + y + z = n-1. - Amarnath Murthy, Apr 22 2001 (edited by Robert A. Beeler) a(n) is the number of levels with energy n + 3/2 (in units of h*f0, with Planck's constant h and the oscillator frequency f0) of the three-dimensional isotropic harmonic quantum oscillator. See the comment by A. Murthy above: n = n1 + n2 + n3 with positive integers and ordered. Proof from the o.g.f. See the A. Messiah reference. - Wolfdieter Lang, Jun 29 2007 Numbers m >= 0 such that round(sqrt(2m+1)) - round(sqrt(2m)) = 1.
Numbers m >= 0 such that ceiling(2*sqrt(2m+1)) - 1 = 1 + floor(2*sqrt(2m)).
Numbers m >= 0 such that fract(sqrt(2m+1)) > 1/2 and fract(sqrt(2m)) < 1/2, where fract(x) is the fractional part of x (i.e., x - floor(x), x >= 0). (End)
If Y and Z are 3-blocks of an n-set X, then, for n >= 6, a(n-1) is the number of (n-2)-subsets of X intersecting both Y and Z. - Milan Janjic, Nov 09 2007 a(n) is also a perfect number A000396 if n is a Mersenne prime A000668, assuming there are no odd perfect numbers. - Omar E. Pol, Sep 05 2008 The number of matches played in a round robin tournament: n*(n-1)/2 gives the number of matches needed for n players. Everyone plays against everyone else exactly once. - Georg Wrede (georg(AT)iki.fi), Dec 18 2008
-a(n+1) = E(2)*binomial(n+2,2) (n >= 0) where E(n) are the Euler numbers in the enumeration A122045. Viewed this way, a(n) is the special case k=2 in the sequence of diagonals in the triangle A153641. - Peter Luschny, Jan 06 2009 Equivalent to the first differences of successive tetrahedral numbers. See A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009 The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-3)(P(2,1)-(-1)^k P(2,2k+1))|. - Peter Luschny, Jul 12 2009 a(n) is the smallest number > a(n-1) such that gcd(n,a(n)) = gcd(n,a(n-1)). If n is odd this gcd is n; if n is even it is n/2. - Franklin T. Adams-Watters, Aug 06 2009 The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - Paul Muljadi, Jan 25 2010 More generally, a(2k+1) == j*(2j-1) (mod 2k+2j+1) and
a(2k) == [-k + 2j*(j-1)] (mod 2k+2j).
Column sums of:
1 3 5 7 9 ...
1 3 5 ...
1 ...
...............
---------------
1 3 6 10 15 ...
Sum_{n>=1} 1/a(n)^2 = 4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).
(End)
1/a(n+1), n >= 0, has e.g.f. -2*(1+x-exp(x))/x^2, and o.g.f. 2*(x+(1-x)*log(1-x))/x^2 (see the Stephen Crowley formula line). -1/(2*a(n+1)) is the z-sequence for the Sheffer triangle of the coefficients of the Bernoulli polynomials A196838/ A196839. - Wolfdieter Lang, Oct 26 2011 (End)
Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - J. M. Bergot, May 04 2012 The sum of four consecutive triangular numbers, beginning with a(n)=n*(n+1)/2, minus 2 is 2*(n+2)^2. a(n)*a(n+2)/2 = a(a(n+1)-1). - J. M. Bergot, May 17 2012 (a(n)*a(n+3) - a(n+1)*a(n+2))*(a(n+1)*a(n+4) - a(n+2)*a(n+3))/8 = a((n^2+5*n+4)/2). - J. M. Bergot, May 18 2012 a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - J. M. Bergot, May 22 2012 In general, a(n)*a(n+1) + a(n+k)*a(n+k+1) + a(k-1)*a(k) = a(n^2 + (k+2)*n + k*(k+1)). - Charlie Marion, Sep 11 2012 a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - J. M. Bergot, May 22 2012 In general, a(n)*a(n+k) + a(n+1)*a(n+k-1) = a(n^2 + (k+1)*n + k-1). - Charlie Marion, Sep 11 2012 a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - J. M. Bergot, May 22 2012 Three points (a(n),a(n+1)), (a(n+1),a(n)) and (a(n+2),a(n+3)) form a triangle with area 4*a(n+1). - J. M. Bergot, May 23 2012 a(n) + a(n+k) = (n+k)^2 - (k^2 + (2n-1)*k -2n)/2. For k=1 we obtain a(n) + a(n+1) = (n+1)^2 (see below). - Charlie Marion, Oct 02 2012 In n-space we can define a(n-1) nontrivial orthogonal projections. For example, in 3-space there are a(2)=3 (namely point onto line, point onto plane, line onto plane). - Douglas Latimer, Dec 17 2012 For n >= 1, a(n) is equal to the rank (minimal cardinality of a generating set) and idempotent rank (minimal cardinality of an idempotent generating set) of the semigroup P_n\S_n, where P_n and S_n denote the partition monoid and symmetric group on [n].
For n >= 3, a(n-1) is equal to the rank and idempotent rank of the semigroup T_n\S_n, where T_n and S_n denote the full transformation semigroup and symmetric group on [n].
(End)
For n >= 3, a(n) is equal to the rank and idempotent rank of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, Jan 15 2013 The formula, a(n)*a(n+4k+2)/2 + a(k) = a(a(n+2k+1) - (k^2+(k+1)^2)), is a generalization of the formula a(n)*a(n+2)/2 = a(a(n+1)-1) in Bergot's comment dated May 17 2012. - Charlie Marion, Mar 28 2013 For odd m = 2k+1, we have the recurrence a(m*n + k) = m^2*a(n) + a(k). Corollary: If number T is in the sequence then so is 9*T+1. - Lekraj Beedassy, May 29 2013 Euler, in Section 87 of the Opera Postuma, shows that whenever T is a triangular number then 9*T + 1, 25*T + 3, 49*T + 6 and 81*T + 10 are also triangular numbers. In general, if T is a triangular number then (2*k + 1)^2*T + k*(k + 1)/2 is also a triangular number. - Peter Bala, Jan 05 2015 Using 1/b and 1/(b+2) will give a Pythagorean triangle with sides 2*b + 2, b^2 + 2*b, and b^2 + 2*b + 2. Set b=n-1 to give a triangle with sides of lengths 2*n,n^2-1, and n^2 + 1. One-fourth the perimeter = a(n) for n > 1. - J. M. Bergot, Jul 24 2013 a(n) = A028896(n)/6, where A028896(n) = s(n) - s(n-1) are the first differences of s(n) = n^3 + 3*n^2 + 2*n - 8. s(n) can be interpreted as the sum of the 12 edge lengths plus the sum of the 6 face areas plus the volume of an n X (n-1) X (n-2) rectangular prism. - J. M. Bergot, Aug 13 2013 Number of positive roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013 A formula for the r-th successive summation of k, for k = 1 to n, is binomial(n+r,r+1) [H. W. Gould]. - Gary Detlefs, Jan 02 2014 For n >= 3, a(n-2) is the number of permutations of 1,2,...,n with the distribution of up (1) - down (0) elements 0...011 (n-3 zeros), or, the same, a(n-2) is up-down coefficient {n,3} (see comment in A060351). - Vladimir Shevelev, Feb 14 2014 a(n) is the dimension of the vector space of symmetric n X n matrices. - Derek Orr, Mar 29 2014 Non-vanishing subdiagonal of A132440^2/2, aside from the initial zero. First subdiagonal of unsigned A238363. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of complete graphs. - Tom Copeland, Apr 05 2014 The number of Sidon subsets of {1,...,n+1} of size 2. - Carl Najafi, Apr 27 2014 Number of factors in the definition of the Vandermonde determinant V(x_1,x_2,...,x_n) = Product_{1 <= i < k <= n} x_i - x_k. - Tom Copeland, Apr 27 2014 Number of weak compositions of n into three parts. - Robert A. Beeler, May 20 2014 Suppose a bag contains a(n) red marbles and a(n+1) blue marbles, where a(n), a(n+1) are consecutive triangular numbers. Then, for n > 0, the probability of choosing two marbles at random and getting two red or two blue is 1/2. In general, for k > 2, let b(0) = 0, b(1) = 1 and, for n > 1, b(n) = (k-1)*b(n-1) - b(n-2) + 1. Suppose, for n > 0, a bag contains b(n) red marbles and b(n+1) blue marbles. Then the probability of choosing two marbles at random and getting two red or two blue is (k-1)/(k+1). See also A027941, A061278, A089817, A053142, A092521. - Charlie Marion, Nov 03 2014 Let O(n) be the oblong number n(n+1) = A002378 and S(n) the square number n^2 = A000290(n). Then a(4n) = O(3n) - O(n), a(4n+1) = S(3n+1) - S(n), a(4n+2) = S(3n+2) - S(n+1) and a(4n+3) = O(3n+2) - O(n). - Charlie Marion, Feb 21 2015 Consider the partition of the natural numbers into parts from the set S=(1,2,3,...,n). The length (order) of the signature of the resulting sequence is given by the triangular numbers. E.g., for n=10, the signature length is 55. - David Neil McGrath, May 05 2015 a(n) counts the partitions of (n-1) unlabeled objects into three (3) parts (labeled a,b,c), e.g., a(5)=15 for (n-1)=4. These are (aaaa),(bbbb),(cccc),(aaab),(aaac),(aabb),(aacc),(aabc),(abbc),(abcc),(abbb),(accc ),(bbcc),(bccc),(bbbc). - David Neil McGrath, May 21 2015Conjecture: the sequence is the genus/deficiency of the sinusoidal spirals of index n which are algebraic curves. The value 0 corresponds to the case of the Bernoulli Lemniscate n=2. So the formula conjectured is (n-1)(n-2)/2. - Wolfgang Tintemann, Aug 02 2015 Conjecture: Let m be any positive integer. Then, for each n = 1,2,3,... the set {Sum_{k=s..t} 1/k^m: 1 <= s <= t <= n} has cardinality a(n) = n*(n+1)/2; in other words, all the sums Sum_{k=s..t} 1/k^m with 1 <= s <= t are pairwise distinct. (I have checked this conjecture via a computer and found no counterexample.) - Zhi-Wei Sun, Sep 09 2015 The Pisano period lengths of reading the sequence modulo m seem to be A022998(m). - R. J. Mathar, Nov 29 2015 For n >= 1, a(n) is the number of compositions of n+4 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016 In this sequence only 3 is prime. - Fabian Kopp, Jan 09 2016 Suppose you are playing Bulgarian Solitaire (see A242424 and Chamberland's and Gardner's books) and, for n > 0, you are starting with a single pile of a(n) cards. Then the number of operations needed to reach the fixed state {n, n-1,...,1} is a(n-1). For example, {6}->{5,1}->{4,2}->{3,2,1}. - Charlie Marion, Jan 14 2016 Every perfect cube is the difference of the squares of two consecutive triangular numbers. 1^2-0^2 = 1^3, 3^2-1^2 = 2^3, 6^2-3^2 = 3^3. - Miquel Cerda, Jun 26 2016 For n > 1, a(n) = tau_n(k*) where tau_n(k) is the number of ordered n-factorizations of k and k* is the square of a prime. For example, tau_3(4) = tau_3(9) = tau_3(25) = tau_3(49) = 6 (see A007425) since the number of divisors of 4, 9, 25, and 49's divisors is 6, and a(3) = 6. - Melvin Peralta, Aug 29 2016 In an (n+1)-dimensional hypercube, number of two-dimensional faces congruent with a vertex (see also A001788). - Stanislav Sykora, Oct 23 2016 Generalizations of the familiar formulas, a(n) + a(n+1) = (n+1)^2 (Feb 19 2004) and a(n)^2 + a(n+1)^2 = a((n+1)^2) (Nov 22 2006), follow: a(n) + a(n+2k-1) + 4a(k-1) = (n+k)^2 + 6a(k-1) and a(n)^2 + a(n+2k-1)^2 + (4a(k-1))^2 + 3a(k-1) = a((n+k)^2 + 6a(k-1)). - Charlie Marion, Nov 27 2016 a(n) is also the greatest possible number of diagonals in a polyhedron with n+4 vertices. - Vladimir Letsko, Dec 19 2016 For n > 0, 2^5 * (binomial(n+1,2))^2 represents the first integer in a sum of 2*(2*n + 1)^2 consecutive integers that equals (2*n + 1)^6. - Patrick J. McNab, Dec 25 2016 Does not satisfy Benford's law (cf. Ross, 2012). - N. J. A. Sloane, Feb 12 2017 Number of ordered triples (a,b,c) of positive integers not larger than n such that a+b+c = 2n+1. - Aviel Livay, Feb 13 2017 Number of inequivalent tetrahedral face colorings using at most n colors so that no color appears only once. - David Nacin, Feb 22 2017 Also the Wiener index of the complete graph K_{n+1}. - Eric W. Weisstein, Sep 07 2017 Number of intersections between the Bernstein polynomials of degree n. - Eric Desbiaux, Apr 01 2018 a(n) is the area of a triangle with vertices at (1,1), (n+1,n+2), and ((n+1)^2, (n+2)^2). - Art Baker, Dec 06 2018 For n > 0, a(n) is the smallest k > 0 such that n divides numerator of (1/a(1) + 1/a(2) + ... + 1/a(n-1) + 1/k). It should be noted that 1/1 + 1/3 + 1/6 + ... + 2/(n(n+1)) = 2n/(n+1). - Thomas Ordowski, Aug 04 2019 Upper bound of the number of lines in an n-homogeneous supersolvable line arrangement (see Theorem 1.1 in Dimca). - Stefano Spezia, Oct 04 2019 For n > 0, a(n+1) is the number of lattice points on a triangular grid with side length n. - Wesley Ivan Hurt, Aug 12 2020 Maximum number of distinct nonempty substrings of a string of length n.
Maximum cardinality of the sumset A+A, where A is a set of n numbers. (End)
a(n) is the number of parking functions of size n avoiding the patterns 123, 132, and 312. - Lara Pudwell, Apr 10 2023 Suppose two rows, each consisting of n evenly spaced dots, are drawn in parallel. Suppose we bijectively draw lines between the dots of the two rows. For n >= 1, a(n - 1) is the maximal possible number of intersections between the lines. Equivalently, the maximal number of inversions in a permutation of [n]. - Sela Fried, Apr 18 2023 The following equation complements the generalization in Bala's Comment (Jan 05 2015). (2k + 1)^2*a(n) + a(k) = a((2k + 1)*n + k). - Charlie Marion, Aug 28 2023 a(n) + a(n+k) + a(k-1) + (k-1)*n = (n+k)^2. For k = 1, we have a(n) + a(n+1) = (n+1)^2. - Charlie Marion, Nov 17 2023
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See Chapter 1.
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 109ff.
Marc Chamberland, Single Digits: In Praise of Small Numbers, Chapter 3, The Number Three, p. 72, Princeton University Press, 2015.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 309 pp 46-196, Ellipses, Paris, 2004
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.
Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.
James Gleick, The Information: A History, A Theory, A Flood, Pantheon, 2011. [On page 82 mentions a table of the first 19999 triangular numbers published by E. de Joncort in 1762.]
Cay S. Horstmann, Scala for the Impatient. Upper Saddle River, New Jersey: Addison-Wesley (2012): 171.
Labos E.: On the number of RGB-colors we can distinguish. Partition Spectra. Lecture at 7th Hungarian Conference on Biometry and Biomathematics. Budapest. Jul 06 2005.
A. Messiah, Quantum Mechanics, Vol.1, North Holland, Amsterdam, 1965, p. 457.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
T. Trotter, Some Identities for the Triangular Numbers, Journal of Recreational Mathematics, Spring 1973, 6(2).
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 91-93 Penguin Books 1987.
LINKS
C. Hamberg, Triangular Numbers Are Everywhere, llinois Mathematics and Science Academy, IMSA Math Journal: a Resource Notebook for High School Mathematics (1992), pp. 7-10. Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)] Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc). Eric Weisstein's World of Mathematics, Absolute Value, Binomial Coefficient, Composition, Distance, Golomb Ruler, Line Line Picking, Polygonal Number, Triangular Number, Trinomial Coefficient, and Wiener Index
FORMULA
E.g.f.: exp(x)*(x+x^2/2).
a(n) = a(-1-n).
a(n+1) = ((n+2)/n)*a(n), Sum_{n>=1} 1/a(n) = 2. - Jon Perry, Jul 13 2003 For n > 0, a(n) = A001109(n) - Sum_{k=0..n-1} (2*k+1)* A001652(n-1-k); e.g., 10 = 204 - (1*119 + 3*20 + 5*3 + 7*0). - Charlie Marion, Jul 18 2003 With interpolated zeros, this is n*(n+2)*(1+(-1)^n)/16. - Benoit Cloitre, Aug 19 2003 a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j) = binomial(i+j+1, i). - Benoit Cloitre, Aug 19 2003 a(n) = ((n+1)^3 - n^3 - 1)/6. - Xavier Acloque, Oct 24 2003
a(n) = a(n-1) + (1 + sqrt(1 + 8*a(n-1)))/2. This recursive relation is inverted when taking the negative branch of the square root, i.e., a(n) is transformed into a(n-1) rather than a(n+1). - Carl R. White, Nov 04 2003 a(n) = a(n-2) + 2*n - 1. - Paul Barry, Jul 17 2004 a(n) = sqrt(sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)^3)) = (Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i*j*k)^3)^(1/6). - Alexander Adamchuk, Oct 26 2004 a(n) == 1 (mod n+2) if n is odd and a(n) == n/2+2 (mod n+2) if n is even. - Jon Perry, Dec 16 2004 a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 1. - Miklos Kristof, Mar 09 2005 a(n) = floor((2*n+1)^2/8). - Paul Barry, May 29 2006 a(n)^2 + a(n+1)^2 = a((n+1)^2) [R B Nelsen, Math Mag 70 (2) (1997), p. 130]. - R. J. Mathar, Nov 22 2006 a(n)*a(n+k)+a(n+1)*a(n+1+k) = a((n+1)*(n+1+k)). Generalizes previous formula dated Nov 22 2006 [and comments by J. M. Bergot dated May 22 2012]. - Charlie Marion, Feb 04 2011 (sqrt(8*a(n)+1)-1)/2 = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
A general formula for polygonal numbers is P(k,n) = (k-2)*(n-1)n/2 + n = n + (k-2)* A000217(n-1), for n >= 1, k >= 3. - Omar E. Pol, Apr 28 2008 and Mar 31 2013 If we define f(n,i,a) = Sum_{j=0..k-1} (binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,1), for n >= 1. - Milan Janjic, Dec 20 2008 An exponential generating function for the inverse of this sequence is given by Sum_{m>=0} ((Pochhammer(1, m)*Pochhammer(1, m))*x^m/(Pochhammer(3, m)*factorial(m))) = ((2-2*x)*log(1-x)+2*x)/x^2, the n-th derivative of which has a closed form which must be evaluated by taking the limit as x->0. A000217(n+1) = (lim_{x->0} d^n/dx^n (((2-2*x)*log(1-x)+2*x)/x^2))^-1 = (lim_{x->0} (2*Gamma(n)*(-1/x)^n*(n*(x/(-1+x))^n*(-x+1+n)*LerchPhi(x/(-1+x), 1, n) + (-1+x)*(n+1)*(x/(-1+x))^n + n*(log(1-x)+log(-1/(-1+x)))*(-x+1+n))/x^2))^-1. - Stephen Crowley, Jun 28 2009 With offset 1, a(n) = floor(n^3/(n+1))/2. - Gary Detlefs, Feb 14 2010 a(n) = 4*a(floor(n/2)) + (-1)^(n+1)*floor((n+1)/2). - Bruno Berselli, May 23 2010 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=1. - Mark Dols, Aug 20 2010 a(n) + 2*a(n-1) + a(n-2) = n^2 + (n-1)^2; and
a(n) + 3*a(n-1) + 3*a(n-2) + a(n-3) = n^2 + 2*(n-1)^2 + (n-2)^2.
In general, for n >= m > 2, Sum_{k=0..m} binomial(m,m-k)*a(n-k) = Sum_{k=0..m-1} binomial(m-1,m-1-k)*(n-k)^2.
a(n) - 2*a(n-1) + a(n-2) = 1, a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 and a(n) - 4*a(n-1) + 6*a(n-2) - 4*(a-3) + a(n-4) = 0.
In general, for n >= m > 2, Sum_{k=0..m} (-1)^k*binomial(m,m-k)*a(n-k) = 0.
(End)
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} 4*(sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011 a(n) = -s(n+1,n), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012 a(n)*a(n+1) = a(Sum_{m=1..n} A005408(m))/2, for n >= 1. For example, if n=8, then a(8)*a(9) = a(80)/2 = 1620. - Ivan N. Ianakiev, May 27 2012 G.f.: x * (1 + 3x + 6x^2 + ...) = x * Product_{j>=0} (1+x^(2^j))^3 = x * A(x) * A(x^2) * A(x^4) * ..., where A(x) = (1 + 3x + 3x^2 + x^3). - Gary W. Adamson, Jun 26 2012 G.f.: G(0) where G(k) = 1 + (2*k+3)*x/(2*k+1 - x*(k+2)*(2*k+1)/(x*(k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 23 2012 G.f.: x + 3*x^2/(Q(0)-3*x) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013 a(n) + a(n+1) + ... + a(n+8) + 6*n = a(3*n+15). - Charlie Marion, Mar 18 2013 a(n) + a(n+1) + ... + a(n+20) + 2*n^2 + 57*n = a(5*n+55). - Charlie Marion, Mar 18 2013 In general, a(k*n) = (2*k-1)*a(n) + a((k-1)*n-1). - Charlie Marion, Apr 20 2015 Also, a(k*n) = a(k)*a(n) + a(k-1)*a(n-1). - Robert Israel, Apr 20 2015 a(n+1) = det(binomial(i+2,j+1), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013 a(n) = floor(n/2) + ceiling(n^2/2) = n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013 Sum_{n>=1} a(n)/n! = 3*exp(1)/2 by the e.g.f. Also see A067764 regarding ratios calculated this way for binomial coefficients in general. - Richard R. Forberg, Jul 15 2013 Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 2 = 0.7725887... . - Richard R. Forberg, Aug 11 2014 Let O(n) be the oblong number n(n+1) = A002378(n) and S(n) the square number n^2 = A000290(n). Then a(n) + a(n+2k) = O(n+k) + S(k) and a(n) + a(n+2k+1) = S(n+k+1) + O(k). - Charlie Marion, Jul 16 2015 A generalization of the Nov 22 2006 formula, a(n)^2 + a(n+1)^2 = a((n+1)^2), follows. Let T(k,n) = a(n) + k. Then for all k, T(k,n)^2 + T(k,n+1)^2 = T(k,(n+1)^2 + 2*k) - 2*k. - Charlie Marion, Dec 10 2015 a(n)^2 + a(n+1)^2 = a(a(n) + a(n+1)). Deducible from N. J. A. Sloane's a(n) + a(n+1) = (n+1)^2 and R. B. Nelson's a(n)^2 + a(n+1)^2 = a((n+1)^2). - Ben Paul Thurston, Dec 28 2015 a(n)^2 + a(n+3)^2 + 19 = a(n^2 + 4*n + 10). - Charlie Marion, Nov 23 2016 G.f.: x/(1-x)^3 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^3 = (1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6). - Gary W. Adamson, Dec 03 2016 a(n) = sum of the elements of inverse of matrix Q(n), where Q(n) has elements q_i,j = 1/(1-4*(i-j)^2). So if e = appropriately sized vector consisting of 1's, then a(n) = e'.Q(n)^-1.e. - Michael Yukish, Mar 20 2017 a(n) = Sum_{k=1..n} ((2*k-1)!!*(2*n-2*k-1)!!)/((2*k-2)!!*(2*n-2*k)!!). - Michael Yukish, Mar 20 2017 Sum_{i=0..k-1} a(n+i) = (3*k*n^2 + 3*n*k^2 + k^3 - k)/6. - Christopher Hohl, Feb 23 2019 a(n) == 0 (mod n) iff n is odd (see De Koninck reference). - Bernard Schott, Jan 10 2020 8*a(k)*a(n) + ((a(k)-1)*n + a(k))^2 = ((a(k)+1)*n + a(k))^2. This formula reduces to the well-known formula, 8*a(n) + 1 = (2*n+1)^2, when k = 1. - Charlie Marion, Jul 23 2020 a(k)*a(n) = Sum_{i = 0..k-1} (-1)^i*a((k-i)*(n-i)). - Charlie Marion, Dec 04 2020 Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)/(2*Pi).
Product_{n>=2} (1 - 1/a(n)) = 1/3. (End)
a(n) = Sum_{k=1..2*n-1} (-1)^(k+1)*a(k)*a(2*n-k). For example, for n = 4, 1*28 - 3*21 + 6*15 - 10*10 + 15*6 - 21*3 + 28*1 = 10. - Charlie Marion, Mar 23 2022 2*a(n) = A000384(n) - n^2 + 2*n. In general, if P(k,n) = the n-th k-gonal number, then (j+1)*a(n) = P(5 + j, n) - n^2 + (j+1)*n. More generally, (j+1)*P(k,n) = P(2*k + (k-2)*(j-1),n) - n^2 + (j+1)*n. - Charlie Marion, Mar 14 2023 a(n) = (1/6)* Sum_{k = 0..3*n} (-1)^(n+k+1) * k*(k + 1) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
EXAMPLE
G.f.: x + 3*x^2 + 6*x^3 + 10*x^4 + 15*x^5 + 21*x^6 + 28*x^7 + 36*x^8 + 45*x^9 + ...
When n=3, a(3) = 4*3/2 = 6.
Example(a(4)=10): ABCD where A, B, C and D are different links in a chain or different amino acids in a peptide possible fragments: A, B, C, D, AB, ABC, ABCD, BC, BCD, CD = 10.
a(2): hollyhock leaves on the Tokugawa Mon, a(4): points in Pythagorean tetractys, a(5): object balls in eight-ball billiards. - Bradley Klee, Aug 24 2015 The a(1) = 1 through a(5) = 15 ordered triples of positive integers summing to n + 2 [Beeler, McGrath above] are the following. These compositions are ranked by A014311. (111) (112) (113) (114) (115)
(121) (122) (123) (124)
(211) (131) (132) (133)
(212) (141) (142)
(221) (213) (151)
(311) (222) (214)
(231) (223)
(312) (232)
(321) (241)
(411) (313)
(322)
(331)
(412)
(421)
(511)
The unordered strict case is A001399(n-6), with Heinz numbers A007304. (End)
MAPLE
istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return true else return false; end if; end proc; # N. J. A. Sloane, May 25 2008 ZL := [S, {S=Prod(B, B, B), B=Set(Z, 1 <= card)}, unlabeled]:
seq(combstruct[count](ZL, size=n), n=2..55); # Zerinvary Lajos, Mar 24 2007 isA000217 := proc(n)
issqr(1+8*n) ;
end proc: # R. J. Mathar, Nov 29 2015 [This is the recipe Leonhard Euler proposes in chapter VII of his "Vollständige Anleitung zur Algebra", 1765. Peter Luschny, Sep 02 2022]
MATHEMATICA
CoefficientList[Series[x / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 30 2014 *) (* For Mathematica 10.4+ *) Table[PolygonalNumber[n], {n, 0, 53}] (* Arkadiusz Wesolowski, Aug 27 2016 *) (* The following Mathematica program, courtesy of Steven J. Miller, is useful for testing if a sequence is Benford. To test a different sequence only one line needs to be changed. This strongly suggests that the triangular numbers are not Benford, since the second and third columns of the output disagree. - N. J. A. Sloane, Feb 12 2017 *) fd[x_] := Floor[10^Mod[Log[10, x], 1]]
benfordtest[num_] := Module[{},
For[d = 1, d <= 9, d++, digit[d] = 0];
For[n = 1, n <= num, n++,
{
d = fd[n(n+1)/2];
If[d != 0, digit[d] = digit[d] + 1];
}];
For[d = 1, d <= 9, d++, digit[d] = 1.0 digit[d]/num];
For[d = 1, d <= 9, d++,
Print[d, " ", 100.0 digit[d], " ", 100.0 Log[10, (d + 1)/d]]];
];
benfordtest[20000]
Table[Length[Join@@Permutations/@IntegerPartitions[n, {3}]], {n, 0, 15}] (* Gus Wiseman, Oct 28 2020 *)
PROG
(PARI) A000217(n) = n * (n + 1) / 2; (PARI) list(lim)=my(v=List(), n, t); while((t=n*n++/2)<=lim, listput(v, t)); Vec(v) \\ Charles R Greathouse IV, Jun 18 2021 (Haskell)
a000217 n = a000217_list !! n
(Scala) (1 to 53).scanLeft(0)(_ + _) // Horstmann (2012), p. 171
(Python) for n in range(0, 60): print(n*(n+1)/2, end=', ') # Stefano Spezia, Dec 06 2018 (Python) # Intended to compute the initial segment of the sequence, not
# isolated terms. If in the iteration the line "x, y = x + y + 1, y + 1"
# is replaced by "x, y = x + y + k, y + k" then the figurate numbers are obtained,
# for k = 0 (natural A001477), k = 1 (triangular), k = 2 (squares), k = 3 (pentagonal), k = 4 (hexagonal), k = 5 (heptagonal), k = 6 (octagonal), etc. def aList():
x, y = 1, 1
yield 0
while True:
yield x
x, y = x + y + 1, y + 1
CROSSREFS
Cf. A000096, A000124, A000292, A000330, A000396, A000668, A001082, A001788, A002024, A002378, A002415, A003056 (inverse function), A004526, A006011, A007318, A008953, A008954, A010054 (characteristic function), A028347, A036666, A046092, A051942, A055998, A055999, A056000, A056115, A056119, A056121, A056126, A062717, A087475, A101859, A109613, A143320, A210569, A245031, A245300, A060544, A016754. Cf. A002817 (doubly triangular numbers), A075528 (solutions of a(n)=a(m)/2). Cf. A104712 (first column, starting with a(1)). Some generalized k-gonal numbers are A001318 (k=5), this sequence (k=6), A085787 (k=7), etc. A011782 counts compositions of any length.
A337461 counts pairwise coprime triples, with unordered version A307719.
The cubes: a(n) = n^3.
(Formerly M4499 N1905)
+10
1030
0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507
COMMENTS
a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002 Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004. See Propp and Propp-Gubin for a proof. Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004 Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005 Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007 Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009 Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009 One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010 Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010 The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011 The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011 Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013 For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014 Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014 One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015 We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
1
2 2
3 3 3
4 4 4 4
5 5 5
6 6
7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016 Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017 Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017 Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018 By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018
REFERENCES
Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019] J. Propp and A. Propp-Gubin, "Counting Triangles in Triangles", Pi Mu Epsilon Journal (to appear).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.
LINKS
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010). M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384. T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (8). Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
FORMULA
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007 Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007 a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019] a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013 a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016 a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016 a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017 Sum_{n>=1} 1/a(n) = zeta(3) ( A002117). Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 ( A197070). (End) Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
EXAMPLE
For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - Patrick J. McNab, Mar 28 2016
MAPLE
isA000578 := proc(r)
local p;
if r = 0 or r =1 then
true;
else
for p in ifactors(r)[2] do
if op(2, p) mod 3 <> 0 then
return false;
end if;
end do:
true ;
end if;
MATHEMATICA
CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *) Accumulate[Table[3n^2+3n+1, {n, 0, 20}]] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 8, 27, 64}, 20](* Harvey P. Dale, Aug 18 2018 *)
PROG
(Haskell)
a000578 = (^ 3)
a000578_list = 0 : 1 : 8 : zipWith (+)
(map (+ 6) a000578_list)
(map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
(Magma) I:=[0, 1, 8, 27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014 (Python)
A000578_list, m = [], [6, -6, 1, 0]
for _ in range(10**2):
for i in range(3):
CROSSREFS
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Oblong (or promic, pronic, or heteromecic) numbers: a(n) = n*(n+1).
(Formerly M1581 N0616)
+10
786
0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450, 2550
COMMENTS
4*a(n) + 1 are the odd squares A016754(n). The word "pronic" (used by Dickson) is incorrect. - Michael SomosAccording to the 2nd edition of Webster, the correct word is "promic". - R. K. Guya(n) is the number of minimal vectors in the root lattice A_n (see Conway and Sloane, p. 109).
Let M_n denote the n X n matrix M_n(i, j) = (i + j); then the characteristic polynomial of M_n is x^(n-2) * (x^2 - a(n)*x - A002415(n)). - Benoit Cloitre, Nov 09 2002 The greatest LCM of all pairs (j, k) for j < k <= n for n > 1. - Robert G. Wilson v, Jun 19 2004 First differences are a(n+1) - a(n) = 2*n + 2 = 2, 4, 6, ... (while first differences of the squares are (n+1)^2 - n^2 = 2*n + 1 = 1, 3, 5, ...). - Alexandre Wajnberg, Dec 29 2005 25 appended to these numbers corresponds to squares of numbers ending in 5 (i.e., to squares of A017329). - Lekraj Beedassy, Mar 24 2006 A rapid (mental) multiplication/factorization technique -- a generalization of Lekraj Beedassy's comment: For all bases b >= 2 and positive integers n, c, d, k with c + d = b^k, we have (n*b^k + c)*(n*b^k + d) = a(n)*b^(2*k) + c*d. Thus the last 2*k base-b digits of the product are exactly those of c*d -- including leading 0(s) as necessary -- with the preceding base-b digit(s) the same as a(n)'s. Examples: In decimal, 113*117 = 13221 (as n = 11, b = 10 = 3 + 7, k = 1, 3*7 = 21, and a(11) = 132); in octal, 61*67 = 5207 (52 is a(6) in octal). In particular, for even b = 2*m (m > 0) and c = d = m, such a product is a square of this type. Decimal factoring: 5609 is immediately seen to be 71*79. Likewise, 120099 = 301*399 (k = 2 here) and 99990000001996 = 9999002*9999998 (k = 3). - Rick L. Shepherd, Jul 24 2021 Number of circular binary words of length n + 1 having exactly one occurrence of 01. Example: a(2) = 6 because we have 001, 010, 011, 100, 101 and 110. Column 1 of A119462. - Emeric Deutsch, May 21 2006 The sequence of iterated square roots sqrt(N + sqrt(N + ...)) has for N = 1, 2, ... the limit (1 + sqrt(1 + 4*N))/2. For N = a(n) this limit is n + 1, n = 1, 2, .... For all other numbers N, N >= 1, this limit is not a natural number. Examples: n = 1, a(1) = 2: sqrt(2 + sqrt(2 + ...)) = 1 + 1 = 2; n = 2, a(2) = 6: sqrt(6 + sqrt(6 + ...)) = 1 + 2 = 3. - Wolfdieter Lang, May 05 2006 Nonsquare integers m divisible by ceiling(sqrt(m)), except for m = 0. - Max Alekseyev, Nov 27 2006 The number of off-diagonal elements of an (n + 1) X (n + 1) matrix. - Artur Jasinski, Jan 11 2007 a(n) is equal to the number of functions f:{1, 2} -> {1, 2, ..., n + 1} such that for a fixed x in {1, 2} and a fixed y in {1, 2, ..., n + 1} we have f(x) <> y. - Aleksandar M. Janjic and Milan Janjic, Mar 13 2007 Numbers m >= 0 such that round(sqrt(m+1)) - round(sqrt(m)) = 1. - Hieronymus Fischer, Aug 06 2007 Numbers m >= 0 such that ceiling(2*sqrt(m+1)) - 1 = 1 + floor(2*sqrt(m)). - Hieronymus Fischer, Aug 06 2007 Numbers m >= 0 such that fract(sqrt(m+1)) > 1/2 and fract(sqrt(m)) < 1/2 where fract(x) is the fractional part (fract(x) = x - floor(x), x >= 0). - Hieronymus Fischer, Aug 06 2007 X values of solutions to the equation 4*X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007 Nonvanishing diagonal of A132792, the infinitesimal Lah matrix, so "generalized factorials" composed of a(n) are given by the elements of the Lah matrix, unsigned A111596, e.g., a(1)*a(2)*a(3) / 3! = - A111596(4,1) = 24. - Tom Copeland, Nov 20 2007 If Y is a 2-subset of an n-set X then, for n >= 2, a(n-2) is the number of 2-subsets and 3-subsets of X having exactly one element in common with Y. - Milan Janjic, Dec 28 2007 a(n) coincides with the vertex of a parabola of even width in the Redheffer matrix, directed toward zero. An integer p is prime if and only if for all integer k, the parabola y = kx - x^2 has no integer solution with 1 < x < k when y = p; a(n) corresponds to odd k. - Reikku Kulon, Nov 30 2008 The third differences of certain values of the hypergeometric function 3F2 lead to the squares of the oblong numbers i.e., 3F2([1, n + 1, n + 1], [n + 2, n + 2], z = 1) - 3*3F2([1, n + 2, n + 2], [n + 3, n + 3], z = 1) + 3*3F2([1, n + 3, n + 3], [n + 4, n + 4], z = 1) - 3F2([1, n + 4, n + 4], [n + 5, n + 5], z = 1) = (1/((n+2)*(n+3)))^2 for n = -1, 0, 1, 2, ... . See also A162990. - Johannes W. Meijer, Jul 21 2009 Generalized factorials, [a.(n!)] = a(n)*a(n-1)*...*a(0) = A010790(n), with a(0) = 1 are related to A001263. - Tom Copeland, Sep 21 2011 For n > 1, a(n) is the number of functions f:{1, 2} -> {1, ..., n + 2} where f(1) > 1 and f(2) > 2. Note that there are n + 1 possible values for f(1) and n possible values for f(2). For example, a(3) = 12 since there are 12 functions f from {1, 2} to {1, 2, 3, 4, 5} with f(1) > 1 and f(2) > 2. - Dennis P. Walsh, Dec 24 2011 a(n) gives the number of (n + 1) X (n + 1) symmetric (0, 1)-matrices containing two ones (see [Cameron]). - L. Edson Jeffery, Feb 18 2012 a(n) is the number of positions of a domino in a rectangled triangular board with both legs equal to n + 1. - César Eliud Lozada, Sep 26 2012 a(n) is the number of ordered pairs (x, y) in [n+2] X [n+2] with |x-y| > 1. - Dennis P. Walsh, Nov 27 2012 a(n) is the number of injective functions from {1, 2} into {1, 2, ..., n + 1}. - Dennis P. Walsh, Nov 27 2012 a(n) is the sum of the positive differences of the partition parts of 2n + 2 into exactly two parts (see example). - Wesley Ivan Hurt, Jun 02 2013 Number of positive roots in the root system of type D_{n + 1} (for n > 2). - Tom Edgar, Nov 05 2013 Number of roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013 a(m), for m >= 1, are the only positive integer values t for which the Binet-de Moivre formula for the recurrence b(n) = b(n-1) + t*b(n-2) with b(0) = 0 and b(1) = 1 has a root of a square. PROOF (as suggested by Wolfdieter Lang, Mar 26 2014): The sqrt(1 + 4t) appearing in the zeros r1 and r2 of the characteristic equation is (a positive) integer for positive integer t precisely if 4t + 1 = (2m + 1)^2, that is t = a(m), m >= 1. Thus, the characteristic roots are integers: r1 = m + 1 and r2 = -m. Let m > 1 be an integer. If b(n) = b(n-1) + a(m)*b(n-2), n >= 2, b(0) = 0, b(1) = 1, then lim_{n->oo} b(n+1)/b(n) = m + 1. (End)
Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices (chromatic polynomial) of the complete graphs (here simply K_2). - Tom Copeland, Apr 05 2014 The set of integers k for which k + sqrt(k + sqrt(k + sqrt(k + sqrt(k + ...) ... is an integer. - Leslie Koller, Apr 11 2014 a(n-1) is the largest number k such that (n*k)/(n+k) is an integer. - Derek Orr, May 22 2014 Number of ways to place a domino and a singleton on a strip of length n - 2. - Ralf Stephan, Jun 09 2014 With offset 1, this appears to give the maximal number of crossings between n nonconcentric circles of equal radius. - Felix Fröhlich, Jul 14 2014 For n > 1, the harmonic mean of the n values a(1) to a(n) is n + 1. The lowest infinite sequence of increasing positive integers whose cumulative harmonic mean is integral. - Ian Duff, Feb 01 2015 a(n) is the maximum number of queens of one color that can coexist without attacking one queen of the opponent's color on an (n+2) X (n+2) chessboard. The lone queen can be placed in any position on the perimeter of the board. - Bob Selcoe, Feb 07 2015 With a(0) = 1, a(n-1) is the smallest positive number not in the sequence such that Sum_{i = 1..n} 1/a(i-1) has a denominator equal to n. - Derek Orr, Jun 17 2015 The positive members of this sequence are a proper subsequence of the so-called 1-happy couple products A007969. See the W. Lang link there, eq. (4), with Y_0 = 1, with a table at the end. - Wolfdieter Lang, Sep 19 2015 For n > 0, a(n) is the reciprocal of the area bounded above by y = x^(n-1) and below by y = x^n for x in the interval [0, 1]. Summing all such areas visually demonstrates the formula below giving Sum_{n >= 1} 1/a(n) = 1. - Rick L. Shepherd, Oct 26 2015 It appears that, except for a(0) = 0, this is the set of positive integers n such that x*floor(x) = n has no solution. (For example, to get 3, take x = -3/2.) - Melvin Peralta, Apr 14 2016 If two independent real random variables, x and y, are distributed according to the same exponential distribution: pdf(x) = lambda * exp(-lambda * x), lambda > 0, then the probability that n - 1 <= x/y < n is given by 1/a(n). - Andres Cicuttin, Dec 03 2016 a(n) is equal to the sum of all possible differences between n different pairs of consecutive odd numbers (see example). - Miquel Cerda, Dec 04 2016 a(n+1) is the dimension of the space of vector fields in the plane with polynomial coefficients up to order n. - Martin Licht, Dec 04 2016 It appears that a(n) + 3 is the area of the largest possible pond in a square ( A268311). - Craig Knecht, May 04 2017 Also the number of 3-cycles in the (n+3)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jul 27 2017 The left edge of a Floyd's triangle that consists of even numbers: 0; 2, 4; 6, 8, 10; 12, 14, 16, 18; 20, 22, 24, 26, 28; ... giving 0, 2, 6, 12, 20, ... The right edge generates A028552. - Waldemar Puszkarz, Feb 02 2018 a(n+1) is the order of rowmotion on a poset obtained by adjoining a unique minimal (or maximal) element to a disjoint union of at least two chains of n elements. - Nick Mayers, Jun 01 2018 For n > 0, 1/a(n) = n/(n+1) - (n-1)/n.
For example, 1/6 = 2/3 - 1/2; 1/12 = 3/4 - 2/3.
Corollary of this:
Take 1/2 pill.
Next day, take 1/6 pill. 1/2 + 1/6 = 2/3, so your daily average is 1/3.
Next day, take 1/12 pill. 2/3 + 1/12 = 3/4, so your daily average is 1/4.
And so on. (End)
For an oblong number m >= 6 there exists a Euclidean division m = d*q + r with q < r < d which are in geometric progression, in this order, with a common integer ratio b. For b >= 2 and q >= 1, the Euclidean division is m = qb*(qb+1) = qb^2 * q + qb where (q, qb, qb^2) are in geometric progression.
Some examples with distinct ratios and quotients:
6 | 4 30 | 25 42 | 18
----- ----- -----
2 | 1 , 5 | 1 , 6 | 2 ,
and also:
42 | 12 420 | 100
----- -----
6 | 3 , 20 | 4 .
Some oblong numbers also satisfy a Euclidean division m = d*q + r with q < r < d that are in geometric progression in this order but with a common noninteger ratio b > 1 (see A335064). (End) For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {2, 2n}]. For n=1, this collapses to [1; {2}]. - Magus K. Chu, Sep 09 2022 a(n-2) is the maximum irregularity over all trees with n vertices. The extremal graphs are stars. (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023 For n > 0, number of diagonals in a regular 2*(n+1)-gon that are not parallel to any edge (cf. A367204). - Paolo Xausa, Mar 30 2024 a(n-1) is the maximum Zagreb index over all trees with n vertices. The extremal graphs are stars. (The Zagreb index of a graph is the sum of the squares of the degrees over all vertices of the graph.) - Allan Bickle, Apr 11 2024 For n >= 1, a(n) is the determinant of the distance matrix of a cycle graph on 2*n + 1 vertices (if the length of the cycle is even such a determinant is zero). - Miquel A. Fiol, Aug 20 2024
REFERENCES
W. W. Berman and D. E. Smith, A Brief History of Mathematics, 1910, Open Court, page 67.
J. H. Conway and R. K. Guy, The Book of Numbers, 1996, p. 34.
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag.
L. E. Dickson, History of the Theory of Numbers, Vol. 1: Divisibility and Primality. New York: Chelsea, p. 357, 1952.
L. E. Dickson, History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, pp. 6, 232-233, 350 and 407, 1952.
H. Eves, An Introduction to the History of Mathematics, revised, Holt, Rinehart and Winston, 1964, page 72.
Nicomachus of Gerasa, Introduction to Arithmetic, translation by Martin Luther D'Ooge, Ann Arbor, University of Michigan Press, 1938, p. 254.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
C. S. Ogilvy and J. T. Anderson, Excursions in Number Theory, Oxford University Press, 1966, pp. 61-62.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
F. J. Swetz, From Five Fingers to Infinity, Open Court, 1994, p. 219.
LINKS
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
FORMULA
a(n) = a(n-1) + 2*n, a(0) = 0.
Sum_{n >= 1} a(n) = n*(n+1)*(n+2)/3 (cf. A007290, partial sums). Sum_{n >= 1} 1/a(n) = 1. (Cf. Tijdeman)
Sum_{n >= 1} (-1)^(n+1)/a(n) = log(4) - 1 = A016627 - 1 [Jolley eq (235)]. 1 = 1/2 + Sum_{n >= 1} 1/(2*a(n)) = 1/2 + 1/4 + 1/12 + 1/24 + 1/40 + 1/60 + ... with partial sums: 1/2, 3/4, 5/6, 7/8, 9/10, 11/12, 13/14, ... - Gary W. Adamson, Jun 16 2003 a(n)*a(n+1) = a(n*(n+2)); e.g., a(3)*a(4) = 12*20 = 240 = a(3*5). - Charlie Marion, Dec 29 2003 Log 2 = Sum_{n >= 0} 1/a(2n+1) = 1/2 + 1/12 + 1/30 + 1/56 + 1/90 + ... = (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + ... = Sum_{n >= 0} (-1)^n/(n+1) = A002162. - Gary W. Adamson, Jun 22 2003 (2, 6, 12, 20, 30, ...) = binomial transform of (2, 4, 2). - Gary W. Adamson, Nov 28 2007 a(0) = 0, a(n) = a(n-1) + 1 + floor(x), where x is the minimal positive solution to fract(sqrt(a(n-1) + 1 + x)) = 1/2. - Hieronymus Fischer, Dec 31 2008 E.g.f.: ((-x+1)*log(-x+1)+x)/x^2 also Integral_{x = 0..1} ((-x+1)*log(-x+1) + x)/x^2 = zeta(2) - 1. - Stephen Crowley, Jul 11 2009 a(n-1) = floor(n^5/(n^3 + n^2 + 1)). - Gary Detlefs, Feb 11 2010 For n > 0 a(n) = 1/(Integral_{x=0..Pi/2} 2*(sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011 Sum_{n >= 1} 1/(a(n))^(2s) = Sum_{t = 1..2*s} binomial(4*s - t - 1, 2*s - 1) * ( (1 + (-1)^t)*zeta(t) - 1). See Arxiv:1301.6293. - R. J. Mathar, Feb 03 2013 a(n) = floor(n^2 * e^(1/n)) and a(n-1) = floor(n^2 / e^(1/n)). - Richard R. Forberg, Jun 22 2013 Binomial transform of [0, 2, 2, 0, 0, 0, ...]. - Alois P. Heinz, Mar 10 2015 For n > 0, a(n) = 1/(Integral_{x=0..1} (x^(n-1) - x^n) dx). - Rick L. Shepherd, Oct 26 2015 For n > 0, a(n) = lim_{m -> oo} (1/m)*1/(Sum_{i=m*n..m*(n+1)} 1/i^2), with error of ~1/m. - Richard R. Forberg, Jul 27 2016 Dirichlet g.f.: zeta(s-2) + zeta(s-1).
Convolution of nonnegative integers ( A001477) and constant sequence ( A007395). Sum_{n >= 0} a(n)/n! = 3*exp(1). (End)
a(n)*a(n+2k-1) + (n+k)^2 = ((2n+1)*k + n^2)^2.
a(n)*a(n+2k) + k^2 = ((2n+1)*k + a(n))^2. (End)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi. - Amiram Eldar, Jan 20 2021 A generalization of the Dec 29 2003 formula, a(n)*a(n+1) = a(n*(n+2)), follows. a(n)*a(n+k) = a(n*(n+k+1)) + (k-1)*n*(n+k+1). - Charlie Marion, Jan 02 2023
EXAMPLE
a(3) = 12, since 2(3)+2 = 8 has 4 partitions with exactly two parts: (7,1), (6,2), (5,3), (4,4). Taking the positive differences of the parts in each partition and adding, we get: 6 + 4 + 2 + 0 = 12. - Wesley Ivan Hurt, Jun 02 2013 G.f. = 2*x + 6*x^2 + 12*x^3 + 20*x^4 + 30*x^5 + 42*x^6 + 56*x^7 + ... - Michael Somos, May 22 2014 a(1) = 2, since 45-43 = 2;
a(2) = 6, since 47-45 = 2 and 47-43 = 4, then 2+4 = 6;
a(3) = 12, since 49-47 = 2, 49-45 = 4, and 49-43 = 6, then 2+4+6 = 12. (End)
MATHEMATICA
oblongQ[n_] := IntegerQ @ Sqrt[4 n + 1]; Select[Range[0, 2600], oblongQ] (* Robert G. Wilson v, Sep 29 2011 *) LinearRecurrence[{3, -3, 1}, {2, 6, 12}, {0, 20}] (* Eric W. Weisstein, Jul 27 2017 *)
PROG
(PARI) {a(n) = n*(n+1)};
(PARI) concat(0, Vec(2*x/(1-x)^3 + O(x^100))) \\ Altug Alkan, Oct 26 2015 (Haskell)
a002378 n = n * (n + 1)
a002378_list = zipWith (*) [0..] [1..]
(Python)
def a(n): return n*(n+1)
CROSSREFS
Partial sums of A005843 (even numbers). Twice triangular numbers ( A000217). Cf. A035106, A087811, A119462, A127235, A049598, A124080, A033996, A028896, A046092, A000217, A005563, A046092, A001082, A059300, A059297, A059298, A166373, A002943 (bisection), A002939 (bisection), A078358 (complement). Cf. A045943 (4-cycles in triangular honeycomb acute knight graph), A028896 (5-cycles), A152773 (6-cycles).
Hexagonal numbers: a(n) = n*(2*n-1).
(Formerly M4108 N1705)
+10
442
0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560
COMMENTS
Number of edges in the join of two complete graphs, each of order n, K_n * K_n. - Roberto E. Martinez II, Jan 07 2002 The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1-x)log(1-x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero). - Tommaso Toffoli (tt(AT)bu.edu), May 06 2002
Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n-1. Such a triangle has consecutive longer sides, with short leg 2n-1, hypotenuse a(n) - (n-1) = A001844(n), and area (n-1)*a(n) = 6* A000330(n-1). - Lekraj Beedassy, Apr 23 2003 More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n-1) or equivalently of any member of A054753^(n-1). - Ant King, Aug 29 2011 Number of standard tableaux of shape (2n-1,1,1) (n>=1). - Emeric Deutsch, May 30 2004 It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n-1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2. - Charlie Marion, Dec 16 2006 Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Jan 09 2009 Let Hex(n)=hexagonal number, T(n)=triangular number, then Hex(n)=T(n)+3*T(n-1). - Vincenzo Librandi, Nov 10 2010 For n>=1, 1/a(n) = Sum_{k=0..2*n-1} ((-1)^(k+1)*binomial(2*n-1,k)*binomial(2*n-1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.
The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants. - Paul Cleary, Dec 21 2010 For n>0, a(n-1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(|w-x|,|x-y|) = |w-y|. - Clark Kimberling, Jun 12 2012 a(n) is the number of positions of one domino in an even pyramidal board with base 2n. - César Eliud Lozada, Sep 26 2012 Let a triangle have T(0,0) = 0 and T(r,c) = |r^2 - c^2|. The sum of the differences of the terms in row(n) and row(n-1) is a(n). - J. M. Bergot, Jun 17 2013 With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 11 2013 a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1-x^2)^3. - Geoffrey Critzer, Jan 02 2014 For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m) - n. - Derek Orr, Sep 04 2014 Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...). - Gary W. Adamson, Oct 05 2015 a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4. - Wolfdieter Lang, Oct 21 2015 For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4. - Milan Janjic, Jan 07 2016 Also the number of minimum dominating sets and maximal irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Jun 29 and Aug 17 2017 As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n-1) because (2*n-1) renders the triangle number. - Bruce J. Nicholson, Nov 05 2017 Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.) - Omar E. Pol, Aug 28 2018 Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values. - Ralf Steiner, Feb 25 2020 It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1. - Omar E. Pol, Aug 28 2021 The n-th hexagonal number equals the sum of the n consecutive integers with the same parity starting at 2*n-1; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the n-th 2k-gonal number is the sum of the n consecutive integers with the same parity starting at (k-2)*n - (k-3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively. - Charlie Marion, Mar 02 2022 Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that |A|=|B|=k and A+B={0,1,2,...,2*a(n)}} = 2*n. - Michael Chu, Mar 09 2022
REFERENCES
Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 122-123.
LINKS
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, The Ramanujan Journal, October 2011, 26:109. DOI: 10.1007/s11139-011-9325-y. Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
FORMULA
E.g.f.: exp(x)*(x+2x^2). - Paul Barry, Jun 09 2003 G.f.: x*(1+3*x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation, dropping the initial zero a(n) = right term of M^n * [1,0,0], where M = the 3 X 3 matrix [1,0,0; 1,1,0; 1,4,1]. Example: a(5) = 45 since M^5 *[1,0,0] = [1,5,45]. - Gary W. Adamson, Dec 24 2006 Starting with offset 1, = binomial transform of [1, 5, 4, 0, 0, 0, ...]. Also, A004736 * [1, 4, 4, 4, ...]. - Gary W. Adamson, Oct 25 2007 a(n)^2 + (a(n)+1)^2 + ... + (a(n)+n-1)^2 = (a(n)+n+1)^2 + ... + (a(n)+2n-1)^2 + n^2; e.g., 6^2 + 7^2 = 9^2 + 2^2; 28^2 + 29^2 + 30^2 + 31^2 = 33^2 + 34^2 + 35^2 + 4^2. - Charlie Marion, Nov 10 2007 a(n) = binomial(n+1,2) + 3*binomial(n,2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=6. - Jaume Oliver Lafont, Dec 02 2008 a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Aug 26 2011 a(n+1) = trinomial(2*n+1, 2) = trinomial(2*n+1, 4*n), for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = (n+1)*(2*n+1) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*n-2, x) with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 2, rewritten with x = 2*cos(phi)]. - Wolfdieter Lang, Apr 19 2018 Product_{n>=2} (1 - 1/a(n)) = 2/3. - Amiram Eldar, Jan 21 2021 a(n) = floor(Sum_{k=(n-1)^2..n^2} sqrt(k)), for n >= 1. - Amrit Awasthi, Jun 13 2021
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {0, 1, 6}, 50] (* Harvey P. Dale, Sep 10 2015 *) Join[{0}, Accumulate[Range[1, 312, 4]]] (* Harvey P. Dale, Mar 26 2016 *) (* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[6], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *) CoefficientList[Series[x*(1 + 3*x)/(1 - x)^3 , {x, 0, 100}], x] (* Stefano Spezia, Sep 02 2018 *)
PROG
(PARI) a(n)=n*(2*n-1)
(PARI) a(n) = binomial(2*n, 2) \\ Altug Alkan, Oct 06 2015 (Haskell)
a000384 n = n * (2 * n - 1)
a000384_list = scanl (+) 0 a016813_list
(Python) # Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
x, y = 1, 1
yield 0
while True:
yield x
x, y = x + y + 4, y + 4
CROSSREFS
a(n)= A093561(n+1, 2), (4, 1)-Pascal column. Cf. A002939 (twice a(n): sums of Pythagorean triples (X, Y, Z=Y+1)).
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