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A142990
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a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n).
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3
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1, 7, 57, 504, 4896, 51912, 598392, 7459200, 100085760, 1439061120, 22083719040, 360371773440, 6232667212800, 113901166310400, 2193425619840000, 44398776748032000, 942498015750144000, 20938290999865344000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| This is the case m = 2 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A1442988 (m=0). For other cases see A142989 (m=1) and A142991 (m=3).
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FORMULA
| a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (n^2-n+1)/3. Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 7, b(2) = 52. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/7)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/(7+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 24*log(2)-33/2.
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MAPLE
| p := n -> (n^2-n+1)/3: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);
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CROSSREFS
| Cf. A142979, A142983, A142988, A142989, A142991.
Sequence in context: A014990 A015565 A082413 * A202250 A147689 A176733
Adjacent sequences: A142987 A142988 A142989 * A142991 A142992 A142993
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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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