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.

Showing 1-2 of 2 results.

A184892 a(n) = C(2n,n) * (5^n/n!^2) * Product_{k=0..n-1} (5k+1)*(5k+4).

Original entry on oeis.org

1, 40, 8100, 2310000, 768075000, 278719056000, 107022956040000, 42753018765600000, 17585519046944062500, 7397979398239787500000, 3168258657090171394750000, 1376657183877933677265000000
Offset: 0

Views

Author

Paul D. Hanna, Jan 25 2011

Keywords

Examples

			G.f.: A(x) = 1 + 40*x + 8100*x^2 + 2310000*x^3 +...
A(x)^(1/2) = 1 + 20*x + 3850*x^2 + 1078000*x^3 +...+ A184891(n)*x^n +...
		

Crossrefs

Programs

  • Mathematica
    Table[Binomial[2*n, n] * 5^n / n!^2 * Product[(5*k + 1)*(5*k + 4), {k, 0, n - 1}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 07 2020 *)
  • PARI
    {a(n)=(2*n)!/n!^2*(5^n/n!^2)*prod(k=0,n-1,(5*k+1)*(5*k+4))}

Formula

Self-convolution of A184891, where
. A184891(n) = (5^n/n!^2) * Product_{k=0..n-1} (10k+1)*(10k+4).
a(n) ~ sqrt(5 - sqrt(5)) * 2^(2*n - 3/2) * 5^(3*n) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Oct 07 2020

A184889 a(n) = (5^n/n!^2) * Product_{k=0..n-1} (10k+2)*(10k+3).

Original entry on oeis.org

1, 30, 5850, 1644500, 542685000, 196017822000, 75031266310000, 29905319000700000, 12279871614662437500, 5159062111690898125000, 2207046771381366217875000, 958150139674902210123750000
Offset: 0

Views

Author

Paul D. Hanna, Jan 25 2011

Keywords

Examples

			G.f.: A(x) = 1 + 30*x + 5850*x^2 + 1644500*x^3 +...
A(x)^2 = 1 + 60*x + 12600*x^2 + 3640000*x^3 +...+ A184890(n)*x^n +...
		

Crossrefs

Programs

  • Mathematica
    FullSimplify[Table[500^n * Gamma[n+1/5] * Gamma[n+3/10] / (Gamma[n+1]^2 * Gamma[1/5] * Gamma[3/10]), {n, 0, 15}]] (* Vaclav Kotesovec, Jul 03 2014 *)
    Join[{1},With[{nn=15},Table[5^n/(n!)^2,{n,nn}] Rest[FoldList[Times,1, Table[ (10k+2)(10k+3),{k,0,nn-1}]]]]] (* Harvey P. Dale, Sep 20 2014 *)
  • PARI
    {a(n)=(5^n/n!^2)*prod(k=0,n-1,(10*k+2)*(10*k+3))}

Formula

Self-convolution yields Sum_{k=0..n} a(n-k)*a(k) = A184890(n) where A184890(n) = C(2n,n) * (5^n/n!^2)*Product_{k=0..n-1} (5k+2)*(5k+3).
Showing 1-2 of 2 results.