A087787 a(n) = Sum_{k=0..n} (-1)^(n-k)*A000041(k).
1, 0, 2, 1, 4, 3, 8, 7, 15, 15, 27, 29, 48, 53, 82, 94, 137, 160, 225, 265, 362, 430, 572, 683, 892, 1066, 1370, 1640, 2078, 2487, 3117, 3725, 4624, 5519, 6791, 8092, 9885, 11752, 14263, 16922, 20416, 24167, 29007, 34254, 40921, 48213, 57345, 67409
Offset: 0
Keywords
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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Mathematica
Table[Sum[(-1)^(n-k)*PartitionsP[k], {k,0,n}], {n,0,50}] (* Vaclav Kotesovec, Aug 16 2015 *) (* more efficient program *) sig = 1; su = 1; Flatten[{1, Table[sig = -sig; su = su + sig*PartitionsP[n]; Abs[su], {n, 1, 50}]}] (* Vaclav Kotesovec, Nov 06 2016 *) Table[Length[Select[IntegerPartitions[n], EvenQ[Count[#,1]]&]],{n,0,30}] (* Gus Wiseman, May 20 2024 *)
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Python
from sympy import npartitions def A087787(n): return sum(-npartitions(k) if n-k&1 else npartitions(k) for k in range(n+1)) # Chai Wah Wu, Oct 25 2023
Formula
G.f.: 1/(1+x)*1/Product_{k>0} (1-x^k).
a(n) = 1/n*Sum_{k=1..n} (sigma(k)+(-1)^k)*a(n-k).
a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*sqrt(3)*n) * (1 + (11*Pi/(24*sqrt(6)) - sqrt(3/2)/Pi)/sqrt(n) - (11/16 + (23*Pi^2)/6912)/n). - Vaclav Kotesovec, Nov 05 2016
Alternating partial sums of A000041. - Gus Wiseman, May 20 2024
Comments