A378805 a(n) = n^2 * 2^n * binomial(4*n, n).
0, 8, 448, 15840, 465920, 12403200, 310109184, 7426298880, 172331827200, 3904310108160, 86800438067200, 1900523591622656, 41092173674053632, 879143252561100800, 18640526785000243200, 392189380800086016000, 8196121945202522849280, 170275478835299912515584, 3519020291600950115696640
Offset: 0
Links
- Amiram Eldar, Table of n, a(n) for n = 0..500
- Necdet Batir and Anthony Sofo, On some series involving reciprocals of binomial coefficients, Appl. Math. Comp., Vol. 220 (2013), pp. 331-338.
Programs
-
Mathematica
a[n_] := n^2 * 2^n * Binomial[4*n, n]; Array[a, 20, 0]
-
PARI
a(n) = n^2 * 2^n * binomial(4*n, n);
Formula
a(n) = 2^n * A378803(n).
a(n) = n * A378804(n).
a(n) == 0 (mod 8).
Sum_{n>=1} 1/a(n) = -(3/2)*log((c-1)/(c+1))^2 + (3/4) * arctan(2*sqrt(c^2+2*c)/(c^2+2*c-1))^2 + (3/4) * arctan(2*sqrt(c^2-2*c)/(c^2-2*c-1))^2 = 0.12729750445123620540..., where c = sqrt(1 + (16*sqrt(2/3))*cos(arctan(sqrt(485/27))/3)) (Batir and Sofo, 2013, p. 336, Example 1).