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Smallest n such that A001542(a(n)) == 0 (mod n), i.e., x= A001541(a(n)) and y= A001542(a(n)) is the fundamental solution of the Pell equation x^2 - 2*(n*y)^2 = 1.
+20
5
1, 1, 2, 2, 3, 2, 3, 4, 6, 3, 6, 2, 7, 3, 6, 8, 4, 6, 10, 6, 6, 6, 11, 4, 15, 7, 18, 6, 5, 6, 15, 16, 6, 4, 3, 6, 19, 10, 14, 12, 5, 6, 22, 6, 6, 11, 23, 8, 21, 15, 4, 14, 27, 18, 6, 12, 10, 5, 10, 6, 31, 15, 6, 32, 21, 6, 34, 4, 22, 3, 35, 12, 18, 19, 30
COMMENTS
The fundamental solution of the Pell equation x^2 - 2*(n*y)^2 = 1, is the smallest solution of x^2 - 2*y^2 = 1 satisfying y == 0 (mod n).
If n is prime (i.e., n in A000040) then a(n) divides (n - Legendre symbol (n/2)); the Legendre symbol (n/2), or more general Kronecker symbol (n/2) is A091337(n). - A.H.M. Smeets, Jan 23 2018 Stronger, but conjectured:
If n is prime (i.e., in A000040) and n in {2,3,5,7,11,13,19,23} (mod 24) then (n - Legendre symbol (n/2)) / a(n) == 2 (mod 4). If n is a safe prime (i.e., in A005385) and n in {7,23} (mod 24) then (n - Legendre symbol (n/2)) / a(n) = 2, i.e., a(n) is a Sophie Germain prime ( A005384). If n is prime (i.e., in A000040) and n in {1,17} (mod 24) then (n - Legendre symbol (n/2)) / a(n) == 0 (mod 4). (End)
REFERENCES
Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.
FORMULA
if n | m then a(n) | a(m).
a(2^(m+1)) = 2^m for m>=0.
MATHEMATICA
b[n_] := b[n] = Switch[n, 0, 0, 1, 2, _, 6 b[n - 1] - b[n - 2]];
a[n_] := For[k = 1, True, k++, If[Mod[b[k], n] == 0, Return[k]]];
PROG
(Python)
xf, yf = 3, 2
x, n = 2*xf, 0
while n < 20000:
n = n+1
y1, y0, i = 0, yf, 1
while y0%n != 0:
y1, y0, i = y0, x*y0-y1, i+1
print(n, i)
1, 105, 20213, 3998709, 791704585, 156753394977, 31036379835581, 6145046450172525, 1216688160731724433, 240898110778299543129, 47696609245941810082565, 9443687732585695622131557
FORMULA
G.f.: (1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
MATHEMATICA
CoefficientList[Series[(1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)), {x, 0, 30}], x] (* G. C. Greubel, Jul 15 2018 *)
PROG
(PARI) x='x+O('x^30); Vec((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1))) \\ G. C. Greubel, Jul 15 2018 (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)))); // G. C. Greubel, Jul 15 2018
3, 83, 2811, 95483, 3243603, 110187011, 3743114763, 127155714923, 4319551192611, 146737584833843, 4984758333158043, 169335045742539611, 5752406796913188723, 195412496049305876963
FORMULA
a(n) = A038761(n)^2 + 2, e.g., 95483 = 309^2 + 2. a(n) = A001652(2*n+1) - A001109(n+1)^2 - Sum_{k=1..n-1} A038723(2*n), e.g., 95483 = 137903 - 204^2 - (23 + 781). For n > 0, 2*a(n) + A001652(2*n-1) = A001653(2*n+2), e.g., 2*2811 + 119 = 5741. G.f.: -(11*x^2-22*x+3) / ((x-1)*(x^2-34*x+1)). - Colin Barker, Dec 14 2014 (Empirical g.f. confirmed for more terms and recurrence of source sequences. - Ray Chandler, Feb 05 2024)
MATHEMATICA
LinearRecurrence[{35, -35, 1}, {3, 83, 2811}, 20] (* Paolo Xausa, Feb 06 2024 *)
3, 71, 2379, 80783, 2744211, 93222359, 3166815963, 107578520351, 3654502875939, 124145519261543, 4217293152016491, 143263821649299119, 4866752642924153523, 165326326037771920631, 5616228332641321147899
FORMULA
a(n) = 2* A046176(n+1)+1, e.g., 2379 = 2*1189 + 1. G.f.: (x^2+34*x-3) / ((x-1)*(x^2-34*x+1)). - Colin Barker, Dec 14 2014 [adjusted for corrected term and empirical g.f. confirmed for more terms and recurrence of source sequences. - Ray Chandler, Feb 05 2024]
EXAMPLE
a(1) = 71 = 3*5 + 3*7 + 5*7.
MATHEMATICA
LinearRecurrence[{35, -35, 1}, {3, 71, 2379}, 20] (* Paolo Xausa, Feb 06 2024 *)
Numbers of the form 2*t^2-4 when t > 1 is a term in A001541.
+20
2
14, 574, 19598, 665854, 22619534, 768398398, 26102926094, 886731088894, 30122754096398, 1023286908188734, 34761632124320654, 1180872205318713598, 40114893348711941774, 1362725501650887306814, 46292552162781456489998, 1572584048032918633353214, 53421565080956452077519374
COMMENTS
Equivalently: integers k such that k$ / (k/2+1)! and k$ / (k/2+2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information).
FORMULA
a(n) = 2*(cosh(2*n*arcsinh(1)))^2 - 4.
EXAMPLE
A001541(1) = 3, then for t = 3, 2*t^2-4 = 14; also for k = 14, 14$ / 8! = 1309248519599593818685440000000^2 and 14$ / 9! = 436416173199864606228480000000^2. Hence, 14 is a term.
MAPLE
with(orthopoly):
sequence = (2*T(n, 3)^2-4, n=1..20);
MATHEMATICA
(2*#^2 - 4) & /@ LinearRecurrence[{6, -1}, {3, 17}, 17] (* Amiram Eldar, Dec 04 2021 *) LinearRecurrence[{35, -35, 1}, {14, 574, 19598}, 17] (* Ray Chandler, Mar 01 2024 *)
PROG
(PARI) a(n) = my(t=subst(polchebyshev(n), 'x, 3)); 2*t^2-4; \\ Michel Marcus, Dec 04 2021
Binomial transform of A001541 (with interpolated zeros).
+20
1
1, 1, 4, 10, 36, 116, 400, 1352, 4624, 15760, 53824, 183712, 627264, 2141504, 7311616, 24963200, 85229824, 290992384, 993510400, 3392055808, 11581203456, 39540700160, 135000395776, 460920178688, 1573679927296, 5372879343616
FORMULA
G.f.: (1-3x+2x^3)/(1-4x+8x^3-4x^4);
e.g.f.: exp(x)cosh(x)cosh(sqrt(2)x);
a(n) = ((sqrt(2))^n + (-sqrt(2))^n + (2+sqrt(2))^n + (2-sqrt(2))^n)/4.
MATHEMATICA
LinearRecurrence[{4, 0, -8, 4}, {1, 1, 4, 10}, 30] (* Harvey P. Dale, Jun 16 2017 *)
Numbers such that floor[a(n)^2/2] is a square ( A001541), written in binary.
+20
0
0, 1, 11, 10001, 1100011, 1001000001, 110100100011, 100110010010001, 11011111001000011, 10100010100100000001, 1110110011011111000011, 1010110010010010110010001, 111110110111010100110100011, 101101110011001101010001000001
Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).
(Formerly M1413 N0552)
+10
767
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149, 107578520350, 259717522849
COMMENTS
Sometimes also called lambda numbers.
Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/ A000129. Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002 a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1. Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2* A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003 Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - Amarnath Murthy, Mar 22 2003 This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003 Number of 132-avoiding two-stack sortable permutations.
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.
Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)
Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. WilsonThe Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004 (0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007 Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007 A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008 Related convergents (numerator/denominator):
Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008 a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008
Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008 In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)
and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)
and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then a(1,n) = a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
(End)
Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009 It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009 Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009 a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009 The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009 As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010 Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010 Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010 [u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010 An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010 Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011 Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + ( A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012 Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013 a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013 Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./_., ._\., where . is an empty square. - Jean M. Morales, Sep 18 2013 Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013 An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013 Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014 a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014 a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014 For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015 This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015 A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017 a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017 Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017 Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017 Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018 (1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019 Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020 Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023 Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023 a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023
REFERENCES
J. Austin and L. Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 76.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 941.
J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 53.
John Derbyshire, Prime Obsession, Joseph Henry Press, 2004, see p. 16.
S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.1.
Shaun Giberson and Thomas J. Osler, Extending Theon's Ladder to Any Square Root, Problem 3858, Elementa, No. 4 1996.
R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 43.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009, p. 90.
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, p. 34.
D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 62.
LINKS
Elena Barcucci, Antonio Bernini, and Renzo Pinzani, A Gray code for a regular language, Semantic Sensor Networks Workshop 2018, CEUR Workshop Proceedings (2018) Vol. 2113. M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28. A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252, 263. Shirley Law, Hopf Algebra of Sashes, in FPSAC 2014, Chicago, USA; Discrete Mathematics and Theoretical Computer Science (DMTCS) Proceedings, 2014, 621-632. Eric Weisstein's World of Mathematics, Matching.
FORMULA
G.f.: x/(1 - 2*x - x^2). - Simon Plouffe in his 1992 dissertation. G.f.: Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (2*k + x)/(1 + 2*k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 1 + k)/(1 + k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 3 - k)/(1 - k*x) ) may all be proved using telescoping series. - Peter Bala, Jan 04 2015 a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
For initial values a(0) and a(1), a(n) = ((a(0)*sqrt(2)+a(1)-a(0))*(1+sqrt(2))^n + (a(0)*sqrt(2)-a(1)+a(0))*(1-sqrt(2))^n)/(2*sqrt(2)). - Shahreer Al Hossain, Aug 18 2019 a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
a(n) = ((-i)^(n-1))*S(n-1, 2*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(-2, x)= -1. Binomial transform of expansion of sinh(sqrt(2)x)/sqrt(2). E.g.f.: exp(x)sinh(sqrt(2)x)/sqrt(2). - Paul Barry, May 09 2003 a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - Paul Barry, May 13 2003 Unreduced g.f.: x(1+x)/(1 - x - 3x^2 - x^3); a(n) = a(n-1) + 3*a(n-2) + a(n-2). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, inverse binomial transform of A052955. - Paul Barry, May 23 2004 a(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^k/2. - Paul Barry, Aug 28 2005 a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006 Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers ( A002203): For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k).
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 8*F(n)*F(m)*F(k). (End)
a(n+1) = Sum_{k=0..n} binomial(n+1,2k+1) * 2^k = Sum_{k=0..n} A034867(n,k) * 2^k = (1/n!) * Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007 a(n) (n >= 3) is the determinant of the (n-1) X (n-1) tridiagonal matrix with diagonal entries 2, superdiagonal entries 1 and subdiagonal entries -1. - Emeric Deutsch, Aug 29 2008 fract((1+sqrt(2))^n) = (1/2)*(1 + (-1)^n) - (-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1 + (-1)^n) - (1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x). a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by Josh Inman, Mar 05 2024] a(n) = ((4+sqrt(18))*(1+sqrt(2))^n + (4-sqrt(18))*(1-sqrt(2))^n)/4 offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
If p[i] = Fibonacci(i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1] when i<=j, A[i,j]=-1 when i=j+1, and A[i,j]=0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, May 08 2010 a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - Gary Detlefs, Sep 09 2010 a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). (End)
G.f.: x/(1 - 2*x - x^2) = sqrt(2)*G(0)/4; G(k) = ((-1)^k) - 1/(((sqrt(2) + 1)^(2*k)) - x*((sqrt(2) + 1)^(2*k))/(x + ((sqrt(2) - 1)^(2*k + 1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 02 2011 In general, for n > k, a(n) = a(k+1)*a(n-k) + a(k)*a(n-k-1). See definition of Pell numbers and the formula for Sep 04 2008. - Charlie Marion, Jan 17 2012 (1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
(2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
G.f.: G(0)/(2+2*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 10 2013 G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013 a(n) = Sum_{r=0..n-1} Sum_{k=0..n-r-1} binomial(r+k,k)*binomial(k,n-k-r-1). - Peter Luschny, Nov 16 2013 a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - Charlie Marion, Mar 27 2014 a(k*n) = 2*a(k)*(a(k*n-k)+a(k*n-k-1)) + (-1)^k*a(k*n-2k). - Charlie Marion, Mar 30 2014 a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - Art DuPre, Apr 04 2014 a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - Art DuPre, Apr 04 2014 a(n) = F(n) + Sum_{k=1..n} F(k)*a(n-k), n >= 0 where F(n) the Fibonacci numbers A000045. - Ralf Stephan, May 23 2014 a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - Peter Luschny, Dec 17 2015 a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - Tony Foster III, May 07 2017 a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - Peter Luschny, Mar 07 2018 Some properties:
(1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319); (2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(3) Sum_{k=2..n+1} a(k)*a(k-1) = a(n+1)^2 if n is odd, else a(n+1)^2 - 1 if n is even;
(4) a(n) - a(n-2*k+1) = ( A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2); (5) Sum_{k=n..n+9} a(k) = 41* A001333(n+5). (End) a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
In general, for n > 0,
Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
Sum_{i=0..n} a(i)*J(n-i) = (a(n+1) + a(n) - J(n+2))/2 for J(n) = A001045(n). - Greg Dresden, Jan 05 2022 Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319. Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - R. J. Mathar, Feb 05 2024 a(n) = ((3^(n+1) + 1)^(n-1) mod (9^(n+1) - 2)) mod (3^(n+1) - 1). - Joseph M. Shunia, Jun 06 2024
EXAMPLE
G.f. = x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 70*x^6 + 169*x^7 + 408*x^8 + 985*x^9 + ...
From the comment on compositions with Fibonacci number of parts, F(n), there are F(1)=1 type of 1, F(2)=1 type of 2, F(3)=2 types of 3, F(4)=3 types of 4, F(5)=5 types of 5 and F(6)=8 types of 6.
The following table gives the number of compositions of n=6 with Fibonacci number of parts:
Composition, number of such compositions, number of compositions of this type:
6, 1, 8;
5+1, 2, 10;
4+2, 2, 6;
3+3, 1, 4;
4+1+1, 3, 9;
3+2+1, 6, 12;
2+2+2, 1, 1;
3+1+1+1, 4, 8;
2+2+1+1, 6, 6;
2+1+1+1+1, 5, 5;
1+1+1+1+1+1, 1, 1;
for a total of a(6)=70 compositions of n=6. (End).
MAPLE
A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008 A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
MATHEMATICA
Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *) LinearRecurrence[{2, 1}, {0, 1}, 60] (* Harvey P. Dale, Jan 04 2012 *) a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* Michael Somos, Jun 01 2013 *) a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* Michael Somos, Oct 30 2021 *)
PROG
(PARI) default(realprecision, 2000); for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[2, 1]; if (a > 10^(10^3 - 6), break); write("b000129.txt", n, " ", a)); \\ Harry J. Smith, Jun 12 2009 (PARI) {a(n) = imag( (1 + quadgen( 8))^n )}; /* Michael Somos, Jun 01 2013 */ (PARI) {a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* Michael Somos, Jun 01 2013 */ (PARI) {a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* Michael Somos, Oct 30 2021 */ (Sage) [lucas_number1(n, 2, -1) for n in range(30)] # Zerinvary Lajos, Apr 22 2009 (Haskell)
a000129 n = a000129_list !! n
a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
(Maxima)
a[0]:0$
a[1]:1$
a[n]:=2*a[n-1]+a[n-2]$
(Maxima) makelist((%i)^(n-1)*ultraspherical(n-1, 1, -%i), n, 0, 24), expand; /* Emanuele Munarini, Mar 07 2018 */ (Magma) [0] cat [n le 2 select n else 2*Self(n-1) + Self(n-2): n in [1..35]]; // Vincenzo Librandi, Aug 08 2015 (GAP) a := [0, 1];; for n in [3..10^3] do a[n] := 2 * a[n-1] + a[n-2]; od; A000129 := a; # Muniru A Asiru, Oct 16 2017 (Python)
from itertools import islice
def A000129_gen(): # generator of terms a, b = 0, 1
yield from [a, b]
while True:
a, b = b, a+2*b
yield b
CROSSREFS
a(n) = A054456(n-1, 0), n>=1 (first column of triangle). INVERT transform of Fibonacci numbers ( A000045). The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519. Cf. A034867, A131980, A133156, A143808, A135387, A153346, A001622, A006497, A014176 (growth power), A098316, A154325, A021083, A243399, A008555. Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), this sequence (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
KEYWORD
nonn,easy,core,cofr,nice,frac,changed
Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).
(Formerly M2665 N1064)
+10
357
1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199
COMMENTS
Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011 Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012 Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedmana(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017 Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002 a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1. a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference). Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310. For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004 For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004 Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004 Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006] The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators= A052542 and denominators here. - Clark Kimberling, Aug 26 2008 Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008 Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008 In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then b(1,n)=a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel ( A048211). - Sameen Ahmed Khan, Jun 28 2010 Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010 a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010 Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
(0 1 0 1)
(1 0 2 0)
(0 2 0 1).
(End)
For n >= 1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....2
.2..|..1.....2.....4
.3..|..1.....4.....4.....8
.4..|..1.....4....12.....8....16
.5..|..1.....6....12....32....16....32
.6..|..1.....6....24....32....80....32....64
.7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012 a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012 The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012 Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012 This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012 a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014 For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020 For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021 For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
._
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|_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024
REFERENCES
M. R. Bacon and C. K. Cook, Some properties of Oresme numbers and convolutions ..., Fib. Q., 62:3 (2024), 233-240.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
J. Devillet, J.‐L. Marichal, and B. Teheux, Classifications of quasitrivial semigroups, Semigroup Forum, 100 (2020), 743-764.
Maribel Díaz Noguera [Maribel Del Carmen Díaz Noguera], Rigoberto Flores, Jose L. Ramirez, and Martha Romero Rojas, Catalan identities for generalized Fibonacci polynomials, Fib. Q., 62:2 (2024), 100-111.
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
A. F. Horadam, R. P. Loh, and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.
Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
Kin Y. Li, Math Problem Book I, 2001, p. 24, Problem 159.
I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 102, Problem 10.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
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).
R. P. Stanley, Enumerative Combinatorics, Volume 1 (1986), p. 203, Example 4.1.2.
A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.
R. C. Tilley et al., The cell growth problem for filaments, Proc. Louisiana Conf. Combinatorics, ed. R. C. Mullin et al., Baton Rouge, 1970, 310-339.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.
LINKS
F. Harary and R. W. Robinson, Tapeworms, Unpublished manuscript, circa 1973. (Annotated scanned copy) Claude Soudieux, De l'infini arithmétique, Zurich, 1960. [Annotated scans of selected pages. Contains many sequences including A1333]
FORMULA
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation. a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1. a(n) = a(n-1) + A052542(n-1), n>1. a(n)/ A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003 E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003 a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003 For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)* A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003 For another recurrence see A000129. a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008 If p[1]=1, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010 For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012 Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016 a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021 Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
EXAMPLE
Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/ A000129. The 15 3 X 2 crossword grids, with white squares represented by an o:
ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...
MAPLE
A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n); with(numtheory): cf := cfrac (sqrt(2), 1000): [seq(nthnumer(cf, i), i=0..50)];
a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
A001333List := proc(m) local A, P, n; A := [1, 1]; P := [1, 1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022
MATHEMATICA
Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *) Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *) a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *) Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *) Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *) CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *) Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)
PROG
(PARI) {a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */ (PARI) {a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */ (PARI) a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021 (PARI) { default(realprecision, 2000); for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009 (Sage) from sage.combinat.sloane_functions import recur_gen2
it = recur_gen2(1, 1, 2, 1)
(Sage) [lucas_number2(n, 2, -1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009 (Haskell)
a001333 n = a001333_list !! n
a001333_list = 1 : 1 : zipWith (+)
a001333_list (map (* 2) $ tail a001333_list)
(Magma) [n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018 (Python)
from functools import cache
@cache
def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
CROSSREFS
See A040000 for the continued fraction expansion of sqrt(2). See also A078057 which is the same sequence without the initial 1. Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle). The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519. Second row of the array in A135597. Cf. Triangle A106513 (alternating row sums).
KEYWORD
nonn,cofr,easy,core,nice,frac,changed
Numbers k such that 2*k^2 - 1 is a square.
(Formerly M3955 N1630)
+10
205
1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409, 10181446324101389, 59341817924539925
COMMENTS
Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1 ( A069894). (x,y) = (a(n), a(n+1)) are the solutions with x < y of x/(yz) + y/(xz) + z/(xy)=3 with z=2. - Floor van Lamoen, Nov 29 2001 Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes ( A002593) is also a square. - Lekraj Beedassy, Jun 05 2002 Numbers n such that 2*n^2 = ceiling(sqrt(2)*n*floor(sqrt(2)*n)). - Benoit Cloitre, May 10 2003 Also, number of domino tilings in S_5 X P_2n. - Ralf Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}. In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005 Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n >0, define C(n) to be the largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - Charlie Marion, Sep 14 2005 Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova, Jan 10 2007 The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators = A002315 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008 Apparently Ljunggren shows that 169 is the last square term.
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p < r then s-r = p+q+1. - Mohamed Bouhamida, Aug 29 2009 If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = Y^2 with p < r then r = 3p+2q+1 and s = 4p+3q+2. - Mohamed Bouhamida, Sep 02 2009 Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500, ...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796, ...). - Gary W. Adamson, Jul 22 2010 a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010 The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5, ..., a(k-1), 0, a(k)-a(k-1), ..., a(k)-1, a(k)-1, ..., a(k)-a(k-1), 0, a(k-1), ..., 5, 1. See Bouhamida's Sep 01 2009 comment. - Charlie Marion, May 02 2011 (a(n+1), 2*b(n+1)) and (a(n+2), 2*b(n+1)), n >= 0, with b(n):= A001109(n), give the (u(2*n), v(2*n)) and (u(2*n+1), v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1, v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u < v and (u,v) -> (u,2*u+v) if u > v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, Mar 06 2012 Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - Colin Barker, Feb 04 2014 Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - Ralf Stephan, Feb 20 2014 Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - Colin Barker, Mar 04 2014 The value of the hypotenuse in each triple of the Tree of primitive Pythagorean triples (cf. Wikipedia link) starting with root (3,4,5) and recursively selecting the central branch at each triple node of the tree. - Stuart E Anderson, Feb 05 2015 Positive integers z such that z^2 is a centered square number ( A001844). - Colin Barker, Feb 12 2015 The aerated sequence (b(n)) n >= 1 = [1, 0, 5, 0, 29, 0, 169, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. - Peter Bala, Mar 25 2015 A002315(n-1)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. A002315(n-1)/a(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = y^2 + (y+1)^2. y-values are listed in A001652. Example: for x=29 and y=20, 28*29/2 + 29*30/2 = 20^2 + 21^2. - Bruno Berselli, Mar 19 2018 (a(n), a(n+1)), with a(0):= 1, give all proper positive solutions m1 = m1(n) and m2 = m2(n), with m1 < m2 and n >= 0, of the Markoff triple (m, m1, m2) (see A002559) for m = 2, i.e., m1^2 - 6*m1*m2 + m2^2 = -4. Hence the unique Markoff triple with largest value m = 2 is (1, 1, 2) (for general m from A002559 this is the famous uniqueness conjecture). For X = m2 - m1 and Y = m2 this becomes the reduced indefinite quadratic form representation X^2 + 4*X*Y - 4*Y^2 = -4, with discriminant 32, and the only proper fundamental solution (X(0), Y(0)) = (0, 1). For all nonnegative proper (X(n), Y(n)) solutions see ( A005319(n) = a(n+1) - a(n), a(n+1)), for n >= 0. (End) Each Pell(2*k+1) = a(k+1) number with k >= 3 appears as largest number of an ordered Markoff (Markov) triple [x, y, m] with smallest value x = 2 as [2, Pell(2*k-1), Pell(2*k+1)]. This known result follows also from all positive proper solutions of the Pell equation q^2 - 2*m^2 = -1 which are q = q(k) = A002315(k) and m = m(k) = Pell(2*k+1), for k >= 0. y = y(k) = m(k) - 2*q(k) = Pell(2*k-1), with Pell(-1) = 1. The k = 0 and 1 cases do not satisfy x=2 <= y(k) <= m(k). The numbers 1 and 5 appear also as largest Markoff triple members because they are also Fibonacci numbers, and for these triples x=1. - Wolfdieter Lang, Jul 11 2018 All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a= A001542(n), b= A005319(n), c= A001542(n+1), x= A001541(n), y=a(n+1), z= A002315(n) with 0 < n. - Michael Somos, Jun 26 2022
REFERENCES
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.
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).
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022 David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.
LINKS
A. Blondin-Massé, S. Brlek, S. Labbé, and M. Mendès France, Fibonacci snowflakes, Special Issue dedicated to Paulo Ribenboim, Annales des Sciences Mathématiques du Québec 35, No 2 (2011). M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28. Giuseppe Lancia and Paolo Serafini, Polyhedra. Chapter 2 of Compact Extended Linear Programming Models (2018). EURO Advanced Tutorials on Operational Research. Springer, Cham., 11. Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
FORMULA
G.f.: x*(1-x)/(1-6*x+x^2).
a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.
Can be extended backwards by a(-n+1) = a(n).
a(n+1) = S(n, 6)-S(n-1, 6), n>=0, with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n+1) = T(2*n+1, sqrt(2))/sqrt(2), n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by Wolfdieter Lang, Mar 06 2012] a(n) ~ (1/4)*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = (((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1)) - ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n)) / (4*sqrt(2)). Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 12 2002 Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 4) = a(n). - Benoit Cloitre, Nov 10 2002 For n and j >= 1, Sum_{k=0..j} a(k)*a(n) - Sum_{k=0..j-1} a(k)*a(n-1) = A001109(j+1)*a(n) - A001109(j)*a(n-1) = a(n+j); e.g., (1+5+29)*5 - (1+5)*1=169. - Charlie Marion, Jul 07 2003 For n >= k >= 0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144. For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0..2*n-1} a(k) = 4* A001109(2n); e.g., 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416. Sum_{k=0..n} ((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144. (End)
Sum_{k=0...n} ((2k+1)*a(n-k)) = A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g., 1*169 + 3*29 + 5*5 + 7*1 = 288 = 17^2 - 1; 1*29 + 3*5 + 5*1 = 49 = 7^2. - Charlie Marion, Jul 18 2003 Sum_{k=0...n} a(k)*a(n) = Sum_{k=0..n} a(2k) and Sum_{k=0..n} a(k)*a(n+1) = Sum_{k=0..n} a(2k+1); e.g., (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741. - Charlie Marion, Sep 22 2003 For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003 Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003 Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n > 0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n)) = Sum_{k=0..n} c(2*k+1); e.g., 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1)) = Sum_{k=0..n} c(2*k); e.g., 119*120*169/(20+21+29) = 1+29+985+33461 = 34476. - Charlie Marion, Dec 01 2003 Also solutions x > 0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2). - Benoit Cloitre, Feb 15 2004 a(n)*a(n+3) = 24 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004 For n >= k, a(n)*a(n+2*k+1) - a(n+k)*a(n+k+1) = a(k)^2-1; e.g., 29*195025-985*5741 = 840 = 29^2-1; 1*169-5*29 = 24 = 5^2-1; a(n)*a(n+2*k)-a(n+k)^2 = A001542(k)^2; e.g., 169*195025-5741^2 = 144 = 12^2; 1*29-5^2 = 4 = 2^2. - Charlie Marion Jun 02 2004 For all k, a(n) is a factor of a((2n+1)*k+n). a((2*n+1)*k+n) = a(n)*(Sum_{j=0..k-1} (-1)^j*(a((2*n+1)*(k-j)) + a((2*n+1)*(k-j)-1))+(-1)^k); e.g., 195025 = 5*(33461+5741-169-29+1); 7645370045 = 169*(6625109+1136689-1).- Charlie Marion, Jun 04 2004 a(n) = Sum_{k=0..n} binomial(n+k, 2*k)4^k. - Paul Barry, Aug 30 2004 [offset 0] a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k+1)*2^k. - Paul Barry, Sep 30 2004 [offset 0] a(n) = (-1)^n*U(2*n, i*sqrt(4)/2) = (-1)^n*U(2*n, i), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0] a(n) = Pell(2*n+1) = Pell(n)^2 + Pell(n+1)^2. - Paul Barry, Jul 18 2005 [offset 0] With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)+b^((2n+1)/2))/(2*sqrt(2)). a(n) = A001109(n+1)- A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003 If k is in the sequence, then the next term is floor(k*(3+2*sqrt(2))). - Lekraj Beedassy, Jul 19 2005 a(n) = Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0] a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*Pell(n-j+1), where Pell = A000129. - Paul Barry, May 19 2006 [offset 0] 6*a(n)*a(n+1) = a(n)^2+a(n+1)^2+4; e.g., 6*5*29 = 29^2+5^2+4; 6*169*985 = 169^2+985^2+4. - Charlie Marion, Oct 07 2007 2* A001541(k)*a(n)*a(n+k) = a(n)^2+a(n+k)^2+ A001542(k)^2; e.g., 2*3*5*29 = 5^2+29^2+2^2; 2*99*29*5741 = 2*99*29*5741=29^2+5741^2+70^2. - Charlie Marion, Oct 12 2007 In general, for n >= k, a(n+k) = 2* A001541(k)*a(n)-a(n-k); e.g., a(n+0) = 2*1*a(n)-a(n); a(n+1) = 6*a(n)-a(n-1); a(6+0) = 33461 = 2*33461-33461; a(5+1) = 33461 = 6*5741-985; a(4+2) = 33461 = 34*985-29; a(3+3) = 33461 = 198*169-1.
(End)
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = a(n-1)*k-((k-1)/(k^n)). - Charles L. Hohn, Mar 06 2011 Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(n) - a(n-1). - Charles L. Hohn, Apr 04 2011 G.f.: G(0)*(1-x)/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013 Sum_{n >= 2} 1/( a(n) - 1/a(n) ) = 1/4. - Peter Bala, Mar 25 2015 a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^floor(k/2). - David Pasino, Jul 09 2016 E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + 2*cosh(2*sqrt(2)*x))*exp(3*x)/2. - Ilya Gutkovskiy, Jul 09 2016 For n>1, a(n) is the numerator of the continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the denominators see A005319. - Greg Dresden, Sep 10 2019 a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/4). - Paul Weisenhorn, May 23 2020 a(n+1) = Sum_{k >= n} binomial(2*k,2*n)*(1/2)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
EXAMPLE
For k=1, 2*1^2 - 1 = 2 - 1 = 1 = 1^2.
For k=5, 2*5^2 - 1 = 50 - 1 = 49 = 7^2.
For k=29, 2*29^2 - 1 = 1682 - 1 = 1681 = 41^2.
... (End)
G.f. = x + 5*x^2 + 29*x^3 + 169*x^4 + 985*x^5 + 5741*x^6 + ... - Michael Somos, Jun 26 2022
MAPLE
a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006 A001653:=-(-1+5*z)/(z**2-6*z+1); # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's
MATHEMATICA
LinearRecurrence[{6, -1}, {1, 5}, 40] (* Harvey P. Dale, Jul 12 2011 *) a[ n_] := -(-1)^n ChebyshevU[2 n - 2, I]; (* Michael Somos, Jul 22 2018 *) Numerator[{1} ~Join~
Table[FromContinuedFraction[Flatten[Table[{1, 4}, n]]], {n, 1, 40}]]; (* Greg Dresden, Sep 10 2019 *)
PROG
(PARI) {a(n) = subst(poltchebi(n-1) + poltchebi(n), x, 3)/4}; /* Michael Somos, Nov 02 2002 */ (PARI) a(n)=([5, 2; 2, 1]^(n-1))[1, 1] \\ Lambert Klasen (lambert.klasen(AT)gmx.de), corrected by Eric Chen, Jun 14 2018 (PARI) {a(n) = -(-1)^n * polchebyshev(2*n-2, 2, I)}; /* Michael Somos, Jun 26 2022 */ (Haskell)
a001653 n = a001653_list !! n
a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
(Magma) I:=[1, 5]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 22 2014 (GAP) a:=[1, 5];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018
CROSSREFS
Cf. similar sequences listed in A238379.
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