OFFSET
0,2
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..57
FORMULA
Let f(n) = 2^((n+1)*(2*n-1)/2) * n^(log(n)/8) / Pi^((n-1)/2). For sufficiently large n 0.7675... < a(n)/f(n) < 0.7900... - Vaclav Kotesovec, Jul 11 2020
The above bounds of Vaclav Kotesovec can be recast as: |a(n)/f(n) - exp(-1/4)| <= (3*Pi)^(-2) for sufficiently large n. - Peter Luschny, Jul 12 2020
a(n) ~ exp(-1/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-2), exp(-4)) * 2^(n^2 + n/2) / Pi^(n/2) if n is even and a(n) ~ exp(-3/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-4), exp(-4)) * 2^(n^2 + n/2) * sqrt(n) / Pi^(n/2) if n is odd. - Vaclav Kotesovec, Jul 13 2020
MATHEMATICA
Unprotect[Power]; 0^0 = 1; a[n_] := Sum[n^k * Binomial[n, k]^n, {k, 0, n} ]; Array[a, 14, 0] (* Amiram Eldar, Jul 11 2020 *)
PROG
(PARI) {a(n) = sum(k=0, n, n^k*binomial(n, k)^n)}
(Magma) [(&+[n^j*Binomial(n, j)^n: j in [0..n]]): n in [0..20]]; // G. C. Greubel, Aug 26 2022
(SageMath) [sum(n^j*binomial(n, j)^n for j in (0..n)) for n in (0..20)] # G. C. Greubel, Aug 26 2022
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Seiichi Manyama, Jul 11 2020
STATUS
approved