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.

A202824 Expansion of e.g.f.: exp( (1+x)^4 - 1 ).

Original entry on oeis.org

1, 4, 28, 232, 2248, 24544, 295456, 3869632, 54555328, 821239552, 13115934976, 221076780544, 3915685846528, 72609585620992, 1405168845395968, 28302270409560064, 591874919018500096, 12824294700196052992, 287350628454224478208, 6647086535396002004992
Offset: 0

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Author

Paul D. Hanna, Dec 25 2011

Keywords

Examples

			E.g.f.: A(x) = 1 + 4*x + 28*x^2/2! + 232*x^3/3! + 2248*x^4/4! +...
where A(x) = exp(4*x + 6*x^2 + 4*x^3 + x^4).
		

Crossrefs

Programs

  • GAP
    List([0..20], n-> Sum([0..n], k-> (-1)^(n-k)*4^k*Bell(k)* Stirling1(n,k) )); # G. C. Greubel, Jul 25 2019
  • Magma
    [(&+[4^k*Bell(k)*StirlingFirst(n,k): k in [0..n]]): n in [0..20]]; // G. C. Greubel, Jul 25 2019
    
  • Mathematica
    CoefficientList[Series[Exp[(1+x)^4-1], {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, May 23 2013 *)
    Table[Sum[ (-1)^(n - k) Abs[StirlingS1[n, k]] 4^k BellB[k], {k, 0, n}], {n, 0, 100}] (* Emanuele Munarini, Aug 31 2017 *)
  • Maxima
    a(n) := sum((-1)^(n-k)*abs(stirling1(n,k))*4^k*belln(k),k,0,n);
    makelist(a(n),n,0,12); /* Emanuele Munarini, Aug 31 2017 */
    
  • PARI
    {a(n)=n!*polcoeff(exp((1+x +x*O(x^n))^4-1),n)}
    
  • PARI
    {Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
    {Bell(n)=n!*polcoeff(exp(exp(x+x*O(x^n))-1), n)}
    {a(n)=sum(k=0, n, Stirling1(n, k)*Bell(k) * 4^k)}
    
  • Sage
    [sum((-1)^(n-k)*4^k*bell_number(k)*stirling_number1(n,k) for k in (0..n)) for n in (0..20)] # G. C. Greubel, Jul 25 2019
    

Formula

a(n) = Sum_{k=0..n} Stirling1(n, k)*Bell(k) * 4^k.
a(n) ~ n^(3*n/4)*2^(n/2-1)*exp(-3/4+5/16*sqrt(2)*n^(1/4)+sqrt(2)*n^(3/4)-3/4*n+3/4*sqrt(n)). - Vaclav Kotesovec, May 23 2013
a(n+4) - 4*a(n+3) - 12*(n+3)*a(n+2) - 12*(n+2)*(n+3)*a(n+1) - 4*(n+1)*(n+2)*(n+3)*a(n) = 0. - Emanuele Munarini, Aug 31 2017