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A142984
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a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).
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5
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1, 4, 22, 136, 984, 8016, 73392, 742464, 8254080, 99838080, 1307301120, 18407831040, 277570298880, 4460506444800, 76131788544000, 1375048700928000, 26208041287680000, 525597067634688000, 11065538390925312000
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OFFSET
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1,2
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COMMENTS
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This is the case m = 2 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
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Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
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LINKS
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FORMULA
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a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n^2.
Recurrence: a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 4 and b(2) = 18.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4))))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k^2*(k + 1)^2) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4 + ...))))) = 3 - 4*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
a(n) = n! * (((-1)^n * (2 * Psi(n/2 + 1) - 2 * Psi(n/2 + 3/2)) - 4 * log(2) + 3) * (n+1)^2 + (-1)^n * (2 * n + 1)). - Robert Israel, Mar 07 2024
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MAPLE
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a := n -> n!*(n+1)^2*sum ((-1)^(k+1)/(k^2*(k+1)^2), k = 1..n): seq(a(n), n = 1..20);
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PROG
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(Haskell)
a142984 n = a142984_list !! (n-1)
a142984_list = 1 : 4 : zipWith (+)
(map (* 4) $ tail a142984_list)
(zipWith (*) (drop 2 a002378_list) a142984_list)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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