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.

A187018 Coefficient of x^n in (1 + x + n*x^2)^n.

Original entry on oeis.org

1, 1, 5, 19, 145, 851, 7741, 58605, 600769, 5420035, 61026901, 628076153, 7648488145, 87388647373, 1138801242125, 14182492489651, 196218339243777, 2628971539313875, 38377805385510181, 547815690902283225, 8395817775835635601, 126725586542235932329
Offset: 0

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Author

Emanuele Munarini, Mar 02 2011

Keywords

Examples

			G.f. = 1 + x + 5*x^2 + 19*x^3 + 145*x^4 + 851*x^5 + 7741*x^6 + 58605*x^7 + ...
		

Crossrefs

Programs

  • Magma
    P:=PolynomialRing(Integers()); [ Coefficients((1+x+n*x^2)^n)[n+1]: n in [0..22] ]; // Klaus Brockhaus, Mar 03 2011
    
  • Maple
    A187018:= n -> simplify( n^(n/2)*GegenbauerC(n, -n, -1/(2*sqrt(n))) );
    1, seq(A187018(n), n = 1..30); # G. C. Greubel, May 31 2020
  • Mathematica
    Flatten[{1,Table[Sum[Binomial[n, k]*Binomial[n-k, n-2*k]*n^k, {k,0,Floor[n/2]}],{n,1,20}]}] (* Vaclav Kotesovec, Apr 17 2014 *)
    a[ n_]:= SeriesCoefficient[ (1 + x + n*x^2)^n, {x, 0, n}]; (* Michael Somos, Dec 12 2014 *)
    Table[If[n == 0, 1, Simplify[n^(n/2) GegenbauerC[n, -n, -1/(2 Sqrt[n])]]], {n, 0, 12}] (* Emanuele Munarini, Oct 20 2016 *)
  • Maxima
    a(n):=coeff(expand((1+x+n*x^2)^n),x,n);
    makelist(a(n),n,0,20);
    
  • PARI
    {a(n)=polcoeff(1/sqrt(1 - 2*x - (4*n-1)*x^2 +x*O(x^n)),n)}
    for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Dec 12 2014
    
  • PARI
    {a(n) = polcoef((1+x+n*x^2)^n, n)} \\ Seiichi Manyama, May 01 2019
    
  • Sage
    [1]+[ n^(n/2)*gegenbauer(n, -n, -1/(2*sqrt(n))) for n in (1..30)] # G. C. Greubel, May 31 2020

Formula

a(n) = [x^n] (1 + x + n*x^2)^n.
a(n) = n^(n/2)*GegenbauerPoly(n,-n,-1/(2*sqrt(n))). - Emanuele Munarini, Oct 20 2016
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(n-k, n-2*k)*n^k.
a(n) ~ 2^(n-1/2) * exp(sqrt(n)/2-1/8) * n^(n/2-1/2) / sqrt(Pi). - Vaclav Kotesovec, Apr 17 2014
a(n) = [x^n] 1/sqrt(1 - 2*x - (4*n-1)*x^2). - Paul D. Hanna, Dec 12 2014
a(n) = n! * [x^n] exp(x) * BesselI(0,2*sqrt(n)*x). - Ilya Gutkovskiy, May 31 2020