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%I #54 Jan 24 2021 21:36:45
%S 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,
%T 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,
%U 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
%N Number of ways n can be expressed as the sum of d consecutive positive integers where d>0 is a divisor of n.
%C Number of ways to write n as the sum of an odd number of consecutive integers. - _Vladeta Jovovic_, Aug 28 2007
%C Number of odd divisors of n less than sqrt(2*n). - _Vladeta Jovovic_, Sep 16 2007
%C 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
%H Peter Kagey, <a href="/A082647/b082647.txt">Table of n, a(n) for n = 1..10000</a>
%H M. D. Hirschhorn and P. M. Hirschhorn, <a href="http://www.maa.org/sites/default/files/3004420056860.pdf.bannered.pdf">Partitions into Consecutive Parts</a>, Mathematics Magazine: 2003, Volume 76, Number 4, pp. 306-308.
%H William Lowell Putnam Competition, <a href="http://kskedlaya.org/putnam-archive/2015.pdf">Function A(k) in Problem B6</a>, 2015.
%F G.f.: Sum_{k>0} x^(k*(2*k-1))/(1-x^(2*k-1)). - _Vladeta Jovovic_, Aug 25 2004
%F Conjecture: a(n) = A067742(n) + A131576(n). - _Omar E. Pol_, Feb 22 2017
%F Conjecture: a(n) = A001227(n) - A131576(n). - _Omar E. Pol_, Apr 18 2017
%e 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.
%p N:= 1000: # to get a(1) to a(N)
%p g:= add(x^(k*(2*k-1))/(1-x^(2*k-1)), k=1..floor(sqrt(N/2))):
%p S:= series(g,x,N+1):
%p seq(coeff(S,x,n),n=1..N); # _Robert Israel_, Dec 08 2015
%o (PARI) a(n) = my(q = sqrt(2*n)); sumdiv(n, d, (d%2) && (d < q)); \\ _Michel Marcus_, Jul 04 2014
%Y Cf. A001227, A054843, A082637, A082662, A237593, A285901.
%K easy,nonn
%O 1,6
%A _Naohiro Nomoto_, May 15 2003
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