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Coefficient of x^n in 1/(n+1) * (1 + x + n*x^2)^(n+1).
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%I #32 May 07 2019 08:17:07

%S 1,1,3,10,57,301,2251,15583,138209,1153603,11592451,111381348,

%T 1235739385,13276480803,159935056555,1884023828326,24356065951617,

%U 310189106485419,4266048524240323,58124516559463590,844705360693479801,12213285476055278959,186543178982826381387

%N Coefficient of x^n in 1/(n+1) * (1 + x + n*x^2)^(n+1).

%C Also coefficient of x^n in the expansion of 2/(1 - x + sqrt(1 - 2*x + (1 - 4*n)*x^2)).

%H Seiichi Manyama, <a href="/A307906/b307906.txt">Table of n, a(n) for n = 0..500</a>

%F a(n) = [x^n] (1 - x - sqrt(1 - 2*x + (1 - 4*n)*x^2))/(2*n*x^2).

%F a(n) = Sum_{k=0..floor(n/2)} n^k * binomial(n,k) * binomial(n-k,k)/(k+1) = Sum_{k=0..floor(n/2)} n^k * binomial(n,2*k) * A000108(k).

%F a(n) ~ exp(sqrt(n)/2 - 1/8) * 2^(n + 1/2) * n^((n-3)/2) / sqrt(Pi). - _Vaclav Kotesovec_, May 05 2019

%t Table[Hypergeometric2F1[1/2 - n/2, -n/2, 2, 4*n], {n, 0, 20}] (* _Vaclav Kotesovec_, May 05 2019 *)

%o (PARI) {a(n) = polcoef((1+x+n*x^2)^(n+1)/(n+1), n)}

%o (PARI) {a(n) = sum(k=0, n\2, n^k*binomial(n, k)*binomial(n-k, k)/(k+1))}

%o (PARI) {a(n) = sum(k=0, n\2, n^k*binomial(n, 2*k)*binomial(2*k, k)/(k+1))}

%Y Main diagonal of A306684.

%Y Cf. A000108, A001006, A187018, A247496, A292716.

%K nonn

%O 0,3

%A _Seiichi Manyama_, May 05 2019