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A142986
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a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n+1)*(n+2)*a(n).
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4
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1, 8, 70, 656, 6648, 72864, 862128, 10977408, 149892480, 2187106560, 33985025280, 560578268160, 9786290088960, 180315565516800, 3497645442816000, 71256899266560000, 1521414754578432000, 33975929212194816000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| This is the case m = 4 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n+1)*(n+2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.
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REFERENCES
| Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
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FORMULA
| a(n) = n!*p(n+1)*sum {k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (n^4+2*n^2)/3 = A014820(n). Recurrence: a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1)+(n+1)*(n+2)*a(n). The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 8, b(2) = 66. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(8 +1*2/(8 +2*3/(8 +3*4/(8 +...+(n-1)*n/8)))), for n >=2. The behaviour of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(8 +1*2/(8 +2*3/(8 +3*4/(8 +...+n*(n+1)/(8 +...))))) = 17/3 - 8*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
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MAPLE
| p := n -> (n^4+2*n^2)/3: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);
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CROSSREFS
| Cf. A014820, A142983, A142984, A142985, A142987.
Sequence in context: A056631 A190560 A152263 * A123511 A053729 A020532
Adjacent sequences: A142983 A142984 A142985 * A142987 A142988 A142989
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KEYWORD
| easy,nonn
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AUTHOR
| Peter Bala (pbala(AT)toucansurf.com), Jul 17 2008
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