A066989 a(n) = (n!)^3 * Sum_{i=1..n} 1/i^3.

1, 9, 251, 16280, 2048824, 444273984, 152759224512, 78340747014144, 57175952894078976, 57223737619918848000, 76212579497951858688000, 131758938842553681444864000, 289584291977410916858462208000, 794860754824699647616459210752000

Offset: 1

Author

Benoit Cloitre, Jan 27 2002

Keywords

• nonn

Comments

p^2 divides a(p-1) for prime p>5. - Alexander Adamchuk, Jul 11 2006

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..181 (terms 1..50 from T. D. Noe)

Crossrefs

Cf. A007408.

Cf. A001819, A143003, A143004, A143005, A143006.

Column k=3 of A291556.

Programs

Mathematica

f[k_] := k^3; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 22}] (* A066989 *)
(* Clark Kimberling, Dec 29 2011 *)
Table[(n!)^3 * Sum[1/i^3, {i, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Aug 27 2017 *)

Formula

Recurrence: a(1) = 1, a(2) = 9, a(n+2) = (2*n+3)*(n^2+3*n+3)*a(n+1) - (n+1)^6*a(n). b(n) = n!^3 satisfies the same recurrence with the initial conditions b(1) = 1, b(2) = 8. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1-1^6/(9-2^6/(35-3^6/(91-...-(n-1)^6/((2n-1)*(n^2-n+1)))))) for n >= 2, leading to the infinite continued fraction expansion zeta(3) = 1/(1-1^6/(9-2^6/(35-3^6/(91-...-(n-1)^6/((2n-1)*(n^2-n+1)-...))))). Compare with A001819. - Peter Bala, Jul 19 2008

a(n) ~ Zeta(3) * (2*Pi)^(3/2) * n^(3*n+3/2) / exp(3*n). - Vaclav Kotesovec, Aug 27 2017

Sum_{n>=1} a(n) * x^n / (n!)^3 = polylog(3,x) / (1 - x). - Ilya Gutkovskiy, Jul 14 2020