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A082647
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Number of ways n can be expressed as the sum of d consecutive positive integers where d>0 is a divisor of n.
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26
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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
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1,6
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COMMENTS
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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
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LINKS
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FORMULA
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G.f.: Sum_{k>0} x^(k*(2*k-1))/(1-x^(2*k-1)). - Vladeta Jovovic, Aug 25 2004
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EXAMPLE
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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.
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MAPLE
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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):
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PROG
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(PARI) a(n) = my(q = sqrt(2*n)); sumdiv(n, d, (d%2) && (d < q)); \\ Michel Marcus, Jul 04 2014
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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