A136557 a(n) = Sum_{k=0..n} binomial(2^k + n-k-1, k).
1, 2, 6, 45, 1436, 171836, 68149425, 89431630740, 396956313475102, 6099399658235428041, 331007760926212498510464, 64484289650612910347505873728, 45677712418497545460138258802186905
Offset: 0
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 0..50
Programs
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Magma
[(&+[Binomial(2^k +n-k-1, k): k in [0..n]]): n in [0..20]]; // G. C. Greubel, Mar 15 2021
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Maple
A136557:= n-> add(binomial(2^k +n-k-1, k), k=0..n); seq(A136557(n), n=0..20); # G. C. Greubel, Mar 15 2021
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Mathematica
Table[Sum[Binomial[2^k+n-k-1, k], {k, 0, n}], {n, 0, 15}] (* Vaclav Kotesovec, Jul 02 2016 *)
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PARI
a(n)=sum(k=0,n,binomial(2^k+n-k-1,k))
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PARI
/* As coefficient of x^n in the g.f.: */ {a(n)=polcoeff(sum(i=0,n,((1+2^i*x+x*O(x^n))*(1-x-2^i*x^2))^-1*log(1+2^i*x+x*O(x^n))^i/i!),n)}
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Sage
[sum(binomial(2^k +n-k-1, k) for k in (0..n)) for n in (0..20)] # G. C. Greubel, Mar 15 2021
Formula
Equals antidiagonal sums of square array A136555.
G.f.: A(x) = Sum_{n>=0} (1+2^n*x)^-1 * (1-x-2^n*x^2)^-1 * log(1+2^n*x)^n / n!.
a(n) ~ 2^(n^2) / n!. - Vaclav Kotesovec, Jul 02 2016