A082647 Number of ways n can be expressed as the sum of d consecutive positive integers where d>0 is a divisor of n.

1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 3, 1, 1, 2, 1, 3, 2, 1, 1, 2, 2, 1, 3, 1, 1, 4, 1, 1, 2, 2, 2, 2, 1, 1, 3, 2, 2, 2, 1, 1, 3, 1, 1, 4, 1, 2, 3, 1, 1, 2, 3, 1, 3, 1, 1, 3, 1, 3, 2, 1, 2, 3, 1, 1, 3, 2, 1, 2, 2, 1, 4, 3, 1, 2, 1, 2, 2, 1, 2, 4, 2, 1, 2, 1, 2, 4

Offset: 1

Author

Naohiro Nomoto, May 15 2003

Keywords

• easy
• nonn

Comments

Number of ways to write n as the sum of an odd number of consecutive integers. - Vladeta Jovovic, Aug 28 2007

Number of odd divisors of n less than sqrt(2*n). - Vladeta Jovovic, Sep 16 2007

Conjecture: a(n) is also the number of subparts in an octant of the symmetric representation of sigma(n). - Omar E. Pol, Feb 22 2017

Examples

			For n=6: 6 has two ways -- (d=3; 3|6), 1+2+3=6; and (d=1; 1|6), 6=6 -- so a(6)=2.
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000
• M. D. Hirschhorn and P. M. Hirschhorn, Partitions into Consecutive Parts, Mathematics Magazine: 2003, Volume 76, Number 4, pp. 306-308.
• William Lowell Putnam Competition, Function A(k) in Problem B6, 2015.

Crossrefs

Cf. A001227, A054843, A082637, A082662, A237593, A285901.

Programs

Maple

N:= 1000: # to get a(1) to a(N)
g:= add(x^(k*(2*k-1))/(1-x^(2*k-1)), k=1..floor(sqrt(N/2))):
S:= series(g,x,N+1):
seq(coeff(S,x,n),n=1..N); # Robert Israel, Dec 08 2015

PARI

a(n) = my(q = sqrt(2*n)); sumdiv(n, d, (d%2) && (d < q)); \\ Michel Marcus, Jul 04 2014

Formula

G.f.: Sum_{k>0} x^(k*(2*k-1))/(1-x^(2*k-1)). - Vladeta Jovovic, Aug 25 2004

Conjecture: a(n) = A067742(n) + A131576(n). - Omar E. Pol, Feb 22 2017

Conjecture: a(n) = A001227(n) - A131576(n). - Omar E. Pol, Apr 18 2017