A320568 a(n) = Sum_{k=1..n} (-1)^(n-k)*binomial(n,k)*sigma(k).
1, 1, -2, 5, -14, 40, -111, 293, -731, 1726, -3882, 8408, -17796, 37423, -79337, 170917, -373812, 823585, -1809844, 3934750, -8424747, 17749392, -36874749, 75862016, -155359339, 318331170, -655146929, 1356952623, -2828151136, 5920984735, -12420140296, 26036563525
Offset: 1
Keywords
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..3000
- Vaclav Kotesovec, Graph abs(a(n)) / (n*2^n)
- N. J. A. Sloane, Transforms
Programs
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GAP
Flat(List([1..10],n->Sum([1..n],k->(-1)^(n-k)*Binomial(n,k)*Sigma(k)))); # Muniru A Asiru, Oct 15 2018
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Maple
seq(add((-1)^(n-k)*binomial(n,k)*sigma(k),k=1..n),n=1..32); # Paolo P. Lava, Oct 25 2018
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Mathematica
Table[Sum[(-1)^(n - k) Binomial[n, k] DivisorSigma[1, k], {k, n}], {n, 32}] nmax = 32; Rest[CoefficientList[Series[1/(1 + x) Sum[k x^k/((1 + x)^k - x^k), {k, 1, nmax}], {x, 0, nmax}], x]]
Formula
G.f.: (1/(1 + x))*Sum_{k>=1} k*x^k/((1 + x)^k - x^k).
L.g.f.: Sum_{k>=1} sigma(k)*x^k/(k*(1 + x)^k) = Sum_{n>=1} a(n)*x^n/n.
Conjecture: a(n) ~ (-1)^n * Pi^2/48 * n * 2^n. - Vaclav Kotesovec, Oct 16 2018
Comments