A393984 a(n) = (n!)^4 * Sum_{1<=i<=j<=n} 1/(i*j)^2.

0, 1, 21, 1897, 515152, 334109776, 444139428096, 1086286396836096, 4512504650332962816, 29936169668265688498176, 302049012729161725378560000, 4454989520743493117582376960000, 92950790729047047971705255362560000, 2668743485551052734888225563381596160000

Offset

0,3

Links

Seiichi Manyama, Table of n, a(n) for n = 0..144

Eric Weisstein's World of Mathematics, Harmonic Number

Formula

a(n) = (n!)^4 * (H(n,2)^2 + H(n,4))/2 where H(n,r) = Sum_{k=1..n} 1/k^r.

a(n) = 2 * (n!)^6 * Sum_{k=1..n} (-1)^(k-1)/(k^4 * (n-k)! * (n+k)!).

a(n) = n^2 * (2*n^2-2*n+1) * a(n-1) - n^2 * (n-1)^6 * a(n-2) + ((n-1)!)^4 for n > 1.

a(n) ~ 7 * Pi^6 * n^(4*n+2) / (90 * exp(4*n)). - Vaclav Kotesovec, May 08 2026

Mathematica

Table[n!^4*(HarmonicNumber[n, 2]^2 + HarmonicNumber[n, 4])/2, {n, 0, 20}] (* Vaclav Kotesovec, May 08 2026 *)

Prog

(PARI) a(n) = 2*n!^6*sum(k=1, n, (-1)^(k-1)/(k^4*(n-k)!*(n+k)!));
(Python)
from math import factorial
from sympy import harmonic
def A393984(n): return int(factorial(n)**4*(harmonic(n, 2)**2+harmonic(n, 4))/2) # Chai Wah Wu, May 08 2026

Crossrefs

Cf. A001819, A060237, A203229, A394693.

Sequence in context: A305145 A296686 A163928 * A358803 A243685 A221122

Adjacent sequences: A393981 A393982 A393983 * A393985 A393986 A393987

Keyword

nonn

Author

Seiichi Manyama, May 08 2026

Status

approved