cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A133991 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(2^k+n-1,n).

Original entry on oeis.org

1, 3, 17, 193, 5427, 463023, 134675759, 139917028089, 527871326293913, 7281357469833220843, 368715613115281663650597, 68787958348542935934247206953, 47453320297069210448891035137347047
Offset: 0

Views

Author

Paul D. Hanna and Vladeta Jovovic, Jan 21 2008

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n-k) * StirlingS1[n, k]*(2^k + 1)^n, {k, 0, n}]/n!, {n, 0, 12}] (* Vaclav Kotesovec, Jun 08 2019 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(2^k+n-1, n)); \\ Seiichi Manyama, Feb 24 2023

Formula

a(n) = (1/n!)*Sum_{k=0..n} (-1)^(n-k)*Stirling1(n,k)*(2^k+1)^n.
G.f.: Sum_{n>=0} (-log(1 - (2^n+1)*x))^n/n!.
a(n) ~ 2^(n^2) / n!. - Vaclav Kotesovec, Jun 08 2019

Extensions

More terms from Alexis Olson (AlexisOlson(AT)gmail.com), Nov 14 2008

A136649 Binomial transform of A014070: a(n) = Sum_{k=0..n} C(n,k)*C(2^k,k).

Original entry on oeis.org

1, 3, 11, 81, 2089, 211107, 76211147, 95054910473, 410422012327681, 6211807332775516467, 334327967114349983723899, 64835852334793138873642165105, 45812640033676518721399820389451657
Offset: 0

Views

Author

Paul D. Hanna and Vladeta Jovovic, Jan 21 2008

Keywords

Crossrefs

Cf. A014070 (C(2^n, n)), A134173.
Partial sums of A180687.

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]*Binomial[2^k,k],{k,0,n}],{n,0,15}] (* Vaclav Kotesovec, Jul 02 2016 *)
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)*binomial(2^k,k))}
    
  • PARI
    /* Using the g.f.: */ {a(n)=local(X=x+x*O(x^n));polcoeff(sum(k=0,n,(log(1+(2^k-1)*X)-log(1-X))^k/k!)/(1-X),n)}

Formula

G.f.: A(x) = (1/(1-x))*Sum_{n>=0} [log(1 + (2^n-1)*x) - log(1-x)]^n / n!.
From Vaclav Kotesovec, Jul 02 2016: (Start)
a(n) ~ binomial(2^n,n).
a(n) ~ 2^(n^2) / n!.
a(n) ~ 2^(n^2 - 1/2) * exp(n) / (sqrt(Pi) * n^(n+1/2)). (End)

Extensions

Edited by Charles R Greathouse IV, Oct 28 2009
Showing 1-2 of 2 results.