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a(n) = hypergeom([1, -n, -n-1], [2], 1).
3

%I #20 Jan 09 2022 09:25:21

%S 1,2,6,25,135,896,7042,63841,654901,7491574,94470926,1301130777,

%T 19423173211,312256205652,5376809244458,98700795776641,

%U 1923638785344457,39661911384761866,862362968121278038,19717031047061570777,472849461034147171791,11866892471399392308232

%N a(n) = hypergeom([1, -n, -n-1], [2], 1).

%H Reinhard Zumkeller, <a href="/A247499/b247499.txt">Table of n, a(n) for n = 0..250</a>

%F a(n) = n!*hypergeom([-n-1], [2], -1) - 1/((n+1)*(n+2)). (original name)

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

%F From _Vaclav Kotesovec_, Jul 05 2018: (Start)

%F Recurrence: (n-3)*(n+2)*a(n) = 2*(n^3 - n^2 - 5*n - 1)*a(n-1) - (n-1)*(n^3 - n^2 - 3*n - 2)*a(n-2) + (n-2)^2*(n-1)^2*a(n-3).

%F a(n) ~ exp(2*sqrt(n) - n - 1/2) * n^(n - 1/4) / sqrt(2). (End)

%p A247499 := n -> hypergeom([1, -n, -n-1], [2], 1):

%p seq(simplify(A247499(n)), n = 0..21);

%t Table[Sum[n!/k!*Binomial[n+2, k+1]/(n+2), {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Jul 05 2018 *)

%o (Haskell)

%o a247499 = sum . a247500_row -- _Reinhard Zumkeller_, Oct 19 2014

%Y Row sums of A247500.

%K nonn

%O 0,2

%A _Peter Luschny_, Oct 17 2014

%E Name updated by _Peter Luschny_, Jan 09 2022