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A069283
a(n) = -1 + number of odd divisors of n.
18
0, 0, 0, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 3, 0, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 3, 1, 0, 3, 1, 3, 2, 1, 1, 3, 1, 1, 3, 1, 1, 5, 1, 1, 1, 2, 2, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 1, 5, 0, 3, 3, 1, 1, 3, 3, 1, 2, 1, 1, 5, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 3, 1, 1, 5, 3, 1, 3, 1, 3, 1, 1, 2, 5, 2
OFFSET
0,10
COMMENTS
Number of nontrivial ways to write n as sum of at least 2 consecutive integers. That is, we are not counting the trivial solution n=n. E.g., a(9)=2 because 9 = 4 + 5 and 9 = 2 + 3 + 4. a(8)=0 because there are no integers m and k such that m + (m+1) + ... + (m+k-1) = 8 apart from k=1, m=8. - Alfred Heiligenbrunner, Jun 07 2004
Also number of sums of sequences of consecutive positive integers excluding sequences of length 1 (e.g., 9 = 2+3+4 or 4+5 so a(9)=2). (Useful for cribbage players.) - Michael Gilleland, Dec 29 2002
Let M be any positive integer. Then a(n) = number of proper divisors of M^n + 1 of the form M^k + 1.
This sequence gives the distinct differences of triangular numbers Ti giving n : n = Ti - Tj; none if n = 2^k. If factor a = n or a > (n/a - 1)/2 : i = n/a + (a - 1)/2; j = n/a - (a+1)/2. Else : i = n/2a + (2a - 1)/2; j = n/2a - (2a - 1)/2. Examples: 7 is prime; 7 = T4 - T2 = (1 + 2 + 3 + 4) - (1 + 2) (a = 7; n/a = 1). The odd factors of 35 are 35, 7 and 5; 35 = T18 - T16 (a = 35) = T8 - T1 (a = 7) = T5 - T7 (a = 5). 144 = T20 - T11 (a = 9) = T49 - T46 (a = 3). - M. Dauchez (mdzzdm(AT)yahoo.fr), Oct 31 2005
Also number of partitions of n into the form 1 + 2 + ...( k - 1) + k + k + ... + k for some k >= 2. Example: a(9) = 2 because we have [2, 2, 2, 2, 1] and [3, 3, 2, 1]. - Emeric Deutsch, Mar 04 2006
a(n) is the number of nontrivial runsum representations of n, and is also known as the politeness of n. - Ant King, Nov 20 2010
Also number of nonpowers of 2 dividing n, divided by the number of powers of 2 dividing n, n > 0. - Omar E. Pol, Aug 24 2019
a(n) only depends on the prime signature of A000265(n). - David A. Corneth, May 30 2020, corrected by Charles R Greathouse IV, Oct 31 2021
REFERENCES
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, see exercise 2.30 on p. 65.
LINKS
Tom M. Apostol, Sums of Consecutive Positive Integers, The Mathematical Gazette, Vol. 87, No. 508, (March 2003), pp. 98-101.
Alfred Heiligenbrunner, Sum of adjacent numbers (in German).
Henri Picciotto, Staircases.
Wikipedia, Polite Number.
FORMULA
a(n) = 0 if and only if n = 2^k.
a(n) = A001227(n)-1.
a(n) = 1 if and only if n = 2^k * p where k >= 0 and p is an odd prime. - Ant King, Nov 20 2010
G.f.: sum(k>=2, x^(k(k + 1)/2)/(1 - x^k) ). - Emeric Deutsch, Mar 04 2006
If n = 2^k p1^b1 p2^b2 ... pr^br, then a(n) = (1 + b1)(1 + b2) ... (1 + br) - 1. - Ant King, Nov 20 2010
Dirichlet g.f.: (zeta(s)*(1-1/2^s) - 1)*zeta(s). - Geoffrey Critzer, Feb 15 2015
a(n) = (A000005(n) - A001511(n))/A001511(n) = A326987(n)/A001511(n), with n > 0 in both formulas. - Omar E. Pol, Aug 24 2019
G.f.: Sum_{k>=1} x^(3*k) / (1 - x^(2*k)). - Ilya Gutkovskiy, May 30 2020
From David A. Corneth, May 30 2020: (Start)
a(2*n) = a(n).
a(n) = A001227(A000265(n)) - 1. (End)
Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma + log(2)/2 - 3/2)*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Dec 01 2023
EXAMPLE
a(14) = 1 because the divisors of 14 are 1, 2, 7, 14, and of these, two are odd, 1 and 7, and -1 + 2 = 1.
a(15) = 3 because the divisors of 15 are 1, 3, 5, 15, and of these, all four are odd, and -1 + 4 = 3.
a(16) = 0 because 16 has only one odd divisor, and -1 + 1 = 0.
Using Ant King's formula: a(90) = 5 as 90 = 2^1 * 3^2 * 5^1, so a(90) = (1 + 2) * (1 + 1) - 1 = 5. - Giovanni Ciriani, Jan 12 2013
x^3 + x^5 + x^6 + x^7 + 2*x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + ...
a(120) = 3 as the odd divisors of 120 are the odd divisors of 15 as 120 = 15*2^3. 15 has 4 odd divisors so that gives a(120) = 4 - 1 = 3. - David A. Corneth, May 30 2020
MAPLE
g:=sum(x^(k*(k+1)/2)/(1-x^k), k=2..20): gser:=series(g, x=0, 115): seq(coeff(gser, x, n), n=0..100); # Emeric Deutsch, Mar 04 2006
A069283 := proc(n)
A001227(n)-1 ;
end proc: # R. J. Mathar, Jun 18 2015
MATHEMATICA
g[n_] := Module[{dL = Divisors[2n], dP}, dP = Transpose[{dL, 2n/dL}]; Select[dP, ((1 < #[[1]] < #[[2]]) && (Mod[ #[[1]] - #[[2]], 2] == 1)) &] ]; Table[Length[g[n]], {n, 1, 100}]
Table[Length[Select[Divisors[k], OddQ[#] &]] - 1, {k, 100}] (* Ant King, Nov 20 2010 *)
Join[{0}, Times @@@ (#[[All, 2]] & /@ Replace[FactorInteger[Range[2, 50]], {2, a_} -> {2, 0}, Infinity] + 1) - 1] (* Horst H. Manninger, Oct 30 2021 *)
PROG
(Haskell)
a069283 0 = 0
a069283 n = length $ tail $ a182469_row n
-- Reinhard Zumkeller, May 01 2012
(PARI) {a(n) = if( n<1, 0, sumdiv( n, d, d%2) - 1)} /* Michael Somos, Aug 07 2013 */
(PARI) a(n) = numdiv(n >> valuation(n, 2)) - 1 \\ David A. Corneth, May 30 2020
(Magma) [0] cat [-1 + #[d:d in Divisors(n)| IsOdd(d)]:n in [1..100]]; // Marius A. Burtea, Aug 24 2019
(Python)
from sympy import divisor_count
def A069283(n): return divisor_count(n>>(~n&n-1).bit_length())-1 if n else 0 # Chai Wah Wu, Jul 16 2022
CROSSREFS
Cf. A095808 (sums of ascending and descending consecutive integers).
Sequence in context: A115413 A292435 A319094 * A319430 A285337 A328457
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Mar 13 2002
EXTENSIONS
Edited by Vladeta Jovovic, Mar 25 2002
STATUS
approved