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Central terms of triangles A001497 and A001498.
4

%I #16 Nov 24 2017 23:18:18

%S 1,3,45,1260,51975,2837835,192972780,15713497800,1490818103775,

%T 161505294575625,19671344879311125,2660996470946814000,

%U 395823225053338582500,64214706279807005422500,11283441246308945238525000,2134827083801652439128930000

%N Central terms of triangles A001497 and A001498.

%H Reinhard Zumkeller, <a href="/A245066/b245066.txt">Table of n, a(n) for n = 0..250</a>

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

%F O.g.f. A(x) satisfies 0 = 6*A(x) + (-2 + 54*x) * A'(x) + 27*x^2 * A''(x). - _Michael Somos_, Jul 11 2014

%F E.g.f. A(x) satisfies 0 = 6*A(x) + (-2 + 54*x) * A'(x) + (-2*x + 27*x^2) * A''(x). - _Michael Somos_, Jul 11 2014

%F a(n) = (3*n)! / (2^n * n!^2). - _Michael Somos_, Jul 11 2014

%F a(n) = (2*n-1)!! * [x^(2*n)] x^n/(1 - x)^(2*n+1). - _Ilya Gutkovskiy_, Nov 24 2017

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

%o (Haskell)

%o a245066 n = a001497 (2 * n) n

%o (PARI) {a(n) = if( n<0, 0, (3*n)! / (2^n * n!^2))}; /* _Michael Somos_, Jul 11 2014 */

%K nonn

%O 0,2

%A _Reinhard Zumkeller_, Jul 11 2014