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.

A076795 Partial sums of (2n-1)!!.

Original entry on oeis.org

0, 1, 2, 5, 20, 125, 1070, 11465, 146600, 2173625, 36633050, 691362125, 14440672700, 330674815925, 8236528396550, 221694575073425, 6411977928702800, 198310761891213425, 6530970632654064050
Offset: 0

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Author

Michael Somos, Nov 16 2002

Keywords

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 20*x^4 + 125*x^5 + 1070*x^6 + 11465*x^7 + 146600*x^8 + ...
		

Crossrefs

Programs

  • Mathematica
    Join[{0},Accumulate[Table[(2n-1)!!,{n,0,20}]]] (* Harvey P. Dale, Jan 27 2013 *)
    Round@Table[Sqrt[Pi/(2E)] Erfi[1/Sqrt[2]] + (2n-1)!! Im[Gamma[1/2-n, -1/2]]/((-2)^n Sqrt[2E]), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 27 2015 *)
    a[ n_] := Sum[ (2 k - 1)!!, {k, 0, n - 1}]; (* Michael Somos, Oct 29 2015 *)
  • PARI
    {a(n) = if( n<0, 0, sum(k=0, n-1, (2*k)! / (k! * 2^k)))};

Formula

E.g.f.: exp(x) * Integral_{t=0, x} exp(-t) / sqrt(1-2*t) dt.
a(n) = a(n-1) * (2*n - 2) - a(n-2) * (2*n - 3).
a(n) ~ 1/(sqrt(2) * n) * 2^n * (n/e)^n.
G.f.: A(x)=x/(1-x)*(1 + x/(U(0)-x)), where U(k)= (2*k+1)*x + 1 - (2*k+3)*x/U(k+1); (continued fraction Euler's 1st kind, 1-step ). - Sergei N. Gladkovskii, Jun 27 2012
G.f.: x/(1-x)/Q(0), where Q(k)= 1 - x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
G.f.: G(0)*x/(1-x), where G(k)= 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013
From Vladimir Reshetnikov, Oct 27 2015: (Start)
a(n) = Sum_{k=0..n-1} (2*k-1)!!.
a(n) = sqrt(Pi/2)*exp(-1/2)*erfi(1/sqrt(2))+(2*n-1)!!*Im(Gamma(1/2-n, -1/2))*exp(-1/2)/((-2)^n*sqrt(2)), where Gamma(a, x) is the upper incomplete Gamma function.
E.g.f.: sqrt(2)*(exp(x)*F(1/sqrt(2))-F(sqrt(1/2-x))), where F(x)=exp(-x^2)*erfi(x)*sqrt(Pi)/2 is the Dawson integral. (End)
0 = a(n)*(+2*a(n+1) - 3*a(n+2) + a(n+3)) + a(n+1)*(-a(n+1) + a(n+2) - a(n+3)) + a(n+2)*(+a(n+2)) for all n>=0. - Michael Somos, Oct 29 2015