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%I #32 Jul 14 2020 21:40:08
%S 1,9,251,16280,2048824,444273984,152759224512,78340747014144,
%T 57175952894078976,57223737619918848000,76212579497951858688000,
%U 131758938842553681444864000,289584291977410916858462208000,794860754824699647616459210752000
%N a(n) = (n!)^3 * Sum_{i=1..n} 1/i^3.
%C p^2 divides a(p-1) for prime p>5. - _Alexander Adamchuk_, Jul 11 2006
%H Seiichi Manyama, <a href="/A066989/b066989.txt">Table of n, a(n) for n = 1..181</a> (terms 1..50 from T. D. Noe)
%F 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
%F a(n) ~ Zeta(3) * (2*Pi)^(3/2) * n^(3*n+3/2) / exp(3*n). - _Vaclav Kotesovec_, Aug 27 2017
%F Sum_{n>=1} a(n) * x^n / (n!)^3 = polylog(3,x) / (1 - x). - _Ilya Gutkovskiy_, Jul 14 2020
%t f[k_] := k^3; t[n_] := Table[f[k], {k, 1, n}]
%t a[n_] := SymmetricPolynomial[n - 1, t[n]]
%t Table[a[n], {n, 1, 22}] (* A066989 *)
%t (* _Clark Kimberling_, Dec 29 2011 *)
%t Table[(n!)^3 * Sum[1/i^3, {i, 1, n}], {n, 1, 20}] (* _Vaclav Kotesovec_, Aug 27 2017 *)
%Y Cf. A007408.
%Y Cf. A001819, A143003, A143004, A143005, A143006.
%Y Column k=3 of A291556.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Jan 27 2002
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