A293574 a(n) = Sum_{k=0..n} n^(n-k)*binomial(n+k-1,k).
1, 2, 11, 82, 787, 9476, 139134, 2422218, 48824675, 1118286172, 28679699578, 814027423892, 25330145185646, 857375286365768, 31360145331198428, 1232586016712594010, 51805909208539809315, 2318588202311267591852, 110085368092924083334626, 5526615354023679440754396, 292501304641192746350100410
Offset: 0
Keywords
Examples
For n = 2 we have: ---------------------------- 0 1 [2] 3 4 5 ---------------------------- 1, 2, 4, 8, 16, 32, ... A000079 (powers of 2) 1, 3, 7, 15, 31, 63, ... A126646 (partial sums of A000079) 1, 4, [11], 26, 57, 120, ... A000295 (partial sums of A126646) ---------------------------- therefore a(2) = 11.
Programs
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Mathematica
Join[{1}, Table[Sum[n^(n - k) Binomial[n + k - 1, k], {k, 0, n}], {n, 1, 20}]] Table[SeriesCoefficient[1/((1 - x)^n (1 - n x)), {x, 0, n}], {n, 0, 20}] Join[{1, 2}, Table[n^(2 n)/(n - 1)^n - Binomial[2 n, n + 1] Hypergeometric2F1[1, 2 n + 1, n + 2, 1/n]/n, {n, 2, 20}]]
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PARI
a(n) = sum(k=0, n, n^(n-k)*binomial(n+k-1,k)); \\ Michel Marcus, Oct 12 2017
Formula
a(n) = [x^n] 1/((1 - x)^n*(1 - n*x)).
a(n) ~ exp(1) * n^n. - Vaclav Kotesovec, Oct 16 2017
Comments