A378803 a(n) = n^2 * binomial(4*n, n).
0, 4, 112, 1980, 29120, 387600, 4845456, 58017960, 673171200, 7625605680, 84766052800, 927990034972, 10032268963392, 107317291572400, 1137727464904800, 11968670068362000, 125062895892372480, 1299098807032012272, 13423997084049034560, 138068403550647828400, 1414126456884869728000
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
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Mathematica
a[n_] := n^2 * Binomial[4*n, n]; Array[a, 20, 0]
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PARI
a(n) = n^2 * binomial(4*n, n);
Formula
a(n) = n^2 * A005810(n).
a(n) = n * A378802(n).
a(n) == 0 (mod 4).
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 = 0.25947076781691783..., where c = sqrt(1 + (16/sqrt(3))*cos(arctan(sqrt(229/27))/3)) (Batir and Sofo, 2013, p. 336, Example 3).
Sum_{n>=1} (-1)^n/a(n) = -(3/2)*log((1-d)/(1+d))^2 + (3/4) * arctan(2*sqrt(d^2+2*d)/(d^2+2*d-1))^2 + (3/4) * arctan(2*sqrt(d^2-2*d)/(d^2-2*d-1))^2 = -0.24154452788843591937..., where d = sqrt(1 - (8/sqrt(3))*(((3*sqrt(3)+sqrt(283))/16)^(1/3) - (((3*sqrt(3)+sqrt(283))/16)^(-1/3)))) (Batir and Sofo, 2013, pp. 336-337, Example 4).