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A066989
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(n!)^3*sum(i=1,n,1/i^3).
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7
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1, 9, 251, 16280, 2048824, 444273984, 152759224512, 78340747014144, 57175952894078976, 57223737619918848000, 76212579497951858688000, 131758938842553681444864000, 289584291977410916858462208000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| p^2 divides a(p-1) for prime p>5. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..50
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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 (pbala(AT)toucansurf.com), Jul 19 2008
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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 *)
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CROSSREFS
| Cf. A007408.
Cf. A001819, A143003, A143004, A143005, A143006.
Sequence in context: A012098 A012072 A007408 * A160501 A075987 A135099
Adjacent sequences: A066986 A066987 A066988 * A066990 A066991 A066992
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 27 2002
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