A051579 a(n) = (2*n+5)!!/5!!, related to A001147 (odd double factorials).

1, 7, 63, 693, 9009, 135135, 2297295, 43648605, 916620705, 21082276215, 527056905375, 14230536445125, 412685556908625, 12793252264167375, 422177324717523375, 14776206365113318125, 546719635509192770625

Offset: 0

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Row m=5 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..400
• A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Crossrefs

Cf. A000165, A001147(n+1), A002866(n+1), A051577, A051578 (rows m=0..4).

Cf. A051580, A051581, A051582, A051583, A178647.

Programs

GAP

List([0..20], n-> Product([0..n-1], j-> 2*j+7) ); # G. C. Greubel, Nov 12 2019

Magma

[1] cat [(&*[2*j+7: j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Nov 12 2019

Maple

df:=doublefactorial; seq(df(2*n+5)/df(5), n = 0..20); # G. C. Greubel, Nov 12 2019

Mathematica

Table[2^n*Pochhammer[7/2, n], {n,0,20}] (* G. C. Greubel, Nov 12 2019 *)

PARI

vector(20, n, prod(j=1,n-1, 2*j+5) ) \\ G. C. Greubel, Nov 12 2019

Sage

[product( (2*j+7) for j in (0..n-1)) for n in (0..20)] # G. C. Greubel, Nov 12 2019

Formula

a(n) = (2*n+5)!!/4!!.

E.g.f.: 1/(1-2*x)^(7/2).

a(n) ~ 8/15*sqrt(2)*n^3*2^n*e^-n*n^n*(1 + 107/24*n^-1 + ...). - Joe Keane (jgk(AT)jgk.org), Nov 23 2001

G.f.: G(0)/(10*x) -1/(5*x), where G(k)= 1 + 1/(1 - x*(2*k+5)/(x*(2*k+5) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 13 2013

From Peter Bala, May 26 2017: (Start)

a(n+1) = (2*n + 7)*a(n) with a(0) = 1.

O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 7*x)*A(x) - 1 with A(0) = 1.

G.f. as an S-fraction: A(x) = 1/(1 - 7*x/(1 - 2*x/(1 - 9*x/(1 - 4*x/(1 - 11*x/(1 - 6*x/(1 - ... - (2*n + 5)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).

Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 7*x/(1 - 9*x/(1 - 2*x/(1 - 11*x/(1 - 4*x/(1 - 13*x/(1 - 6*x/(1 - ... - (2*n + 7)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)

From Amiram Eldar, Dec 11 2022: (Start)

Sum_{n>=0} 1/a(n) = 15 * sqrt(e*Pi/2) * erf(1/sqrt(2)) - 20, where erf is the error function.

Sum_{n>=0} (-1)^n/a(n) = 15 * sqrt(Pi/(2*e)) * erfi(1/sqrt(2)) - 10, where erfi is the imaginary error function. (End)