A245066 Central terms of triangles A001497 and A001498.

1, 3, 45, 1260, 51975, 2837835, 192972780, 15713497800, 1490818103775, 161505294575625, 19671344879311125, 2660996470946814000, 395823225053338582500, 64214706279807005422500, 11283441246308945238525000, 2134827083801652439128930000, 432702764548047428755944999375

Offset

0,2

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..250

Formula

a(n) = A001497(2*n,n) = A001498(2*n,n).

From Michael Somos, Jul 11 2014: (Start)

O.g.f. A(x) satisfies 0 = 6*A(x) + (-2 + 54*x) * A'(x) + 27*x^2 * A''(x).

E.g.f. A(x) satisfies 0 = 6*A(x) + (-2 + 54*x) * A'(x) + (-2*x + 27*x^2) * A''(x).

a(n) = (3*n)! / (2^n * n!^2). (End)

a(n) = (2*n-1)!! * [x^(2*n)] x^n/(1 - x)^(2*n+1). - Ilya Gutkovskiy, Nov 24 2017

a(n) ~ 3^(3*n+1/2) * n^(n-1/2) / (2^(n+1/2) * exp(n) * sqrt(Pi)). - Amiram Eldar, Sep 21 2025

Example

G.f. = 1 + 3*x + 45*x^2 + 1260*x^3 + 51975*x^4 + 2837835*x^5 + ...

Mathematica

a[n_] := (3*n)! / (2^n * n!^2); Array[a, 20, 0] (* Amiram Eldar, Sep 21 2025 *)

Prog

(Haskell)
a245066 n = a001497 (2 * n) n
(PARI) {a(n) = if( n<0, 0, (3*n)! / (2^n * n!^2))}; /* Michael Somos, Jul 11 2014 */

Crossrefs

Cf. A001497, A001498.

Sequence in context: A361046 A241242 A009088 * A352409 A187662 A113065

Adjacent sequences: A245063 A245064 A245065 * A245067 A245068 A245069

Keyword

nonn,easy

Author

Reinhard Zumkeller, Jul 11 2014

Status

approved