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 A082647 Number of ways n can be expressed as the sum of d consecutive positive integers (where d>0 is a divisor of n). 14
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,6 COMMENTS Number of ways to present n as sum of 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, A237593. Sequence in context: A161111 A161046 A230404 * A214018 A304869 A161071 Adjacent sequences:  A082644 A082645 A082646 * A082648 A082649 A082650 KEYWORD easy,nonn AUTHOR Naohiro Nomoto, May 15 2003 STATUS approved

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Last modified January 17 17:23 EST 2019. Contains 319250 sequences. (Running on oeis4.)