a(n) = Sum_{i=2..2n} lambda(i, n), with lambda(p, 1) = 1 if p = 2, otherwise 0; lambda(p, n) = (p*(p-1)/2)*(lambda(p, n-1) + 2*lambda(p-1, n-1) + lambda(p-2, n-1)).
lambda(p, n) = Sum_k[( - 1)^(p + k) * C(p, k) * ((k - 1)*k/2)^n]. So if T(m, 0), T(m, 1), ..., T(m, m) is any row of
A035317 with m >= 2n - 1 then a(n) = Sum_j[(-1)^j * T(m, j) * ((m - j + 1)*(m - j)/2)^n]; e.g., a(2) = 13 = 1*6^2 - 3*3^2 + 4*1^2 - 2*0^2 = 1*10^2 - 4*6^2 + 7*3^2 - 6*1^2 + 3*0^2 = 1*15^2 - 5*10^2 + 11*6^2 - 13*3^2 + 9*1^2 - 3*0^2 etc. while a(3) = 409 = 1*15^3 - 5*10^3 + 11*6^3 - 13*3^3 + 9*1^3 - 3*0^3 etc. -
Henry Bottomley, Jan 03 2001
E.g.f.: Sum_{m>=0} exp(x*binomial(m,2))/2^(m+1). -
Vladeta Jovovic, Sep 24 2006
a(n) ~ n! * n^n * 2^(n-1) / (exp(n) * (log(2))^(2*n+1)). -
Vaclav Kotesovec, Mar 15 2014
a(n) = Sum_{k = 2..2*n} Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i*(i-1)/2)^n.
a(n) = (1/2^(n+1))*Sum_{k = 0..n} binomial(n,k)*
A000670(n+k) for n >= 1. (End)