

A095808


Number of ways to write n in the form m + (m+1) + ... + (m+k1) + (m+k) + (m+k1) + ... + (m+1) + m with integers m>= 1, k>=1. Or, number of divisors a of 4n1 with 0 < (a1)^2 < 4n.


1



0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 2, 0, 0, 1, 1, 1, 1, 0, 0, 3, 0, 1, 2, 0, 1, 1, 0, 0, 2, 2, 0, 1, 1, 0, 3, 0, 1, 2, 0, 1, 1, 0, 0, 3, 1, 0, 2, 1, 0, 3, 1, 0, 1, 0, 2, 2, 0, 1, 1, 1, 1, 1, 0, 0, 5, 1, 1, 1, 0, 1, 1, 1, 0, 3, 1, 0, 2, 0, 1, 3, 0, 0, 2, 1, 1, 3
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OFFSET

1,16


COMMENTS

n = m + (m+1) + ... + (m+k1) + (m+k) + (m+k1) + ... + (m+1) + m means n = k^2 + m*(2k+1) or 4n1 = (2k+1)*(4m+2k1). So if 4n1 disparts into two odd factors a*b, then k = (a1)/2, m=(nk^2)/(2k+1) give the solution of the origin equation. We only count solutions with k^2 < n, such that m>0. This means we are taking into account only factors a < 2n+1.


LINKS



EXAMPLE

a(16)=2 because 16 = 5+6+5 and 16 = 1+2+3+4+3+2+1.
The trivial case 16=16 (k=0, m=n) is not counted. The cases m=0, e.g. 16 = 0+1+2+3+4+3+2+1+0 are not counted. The cases m<0 e.g. 16 = 4+3+2+1+0+1+2+3+4+5+6+5+4+3+2+1+0+1+2+3+4 are not counted.


MATHEMATICA

h1 = Table[count = 0; For[k = 1, k^2 < n, k++, If[Mod[n  k^2, 2k + 1] == 0, count++ ]]; count, {n, 100}]  or  h2 = Table[Length[Select[Divisors[4n  1], ((#  1)^2 < 4n) &]]  1, {n, 100}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



