A082647 - OEIS

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

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OFFSET

1,6

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

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.

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

EXAMPLE

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.

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

PROG

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

CROSSREFS

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

Sequence in context: A341674 A363922 A230404 * A367627 A214018 A304869

Adjacent sequences: A082644 A082645 A082646 * A082648 A082649 A082650

KEYWORD

easy,nonn

AUTHOR

Naohiro Nomoto, May 15 2003

STATUS

approved