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 A142984 a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n+1)*(n+2)*a(n). 5
 1, 4, 22, 136, 984, 8016, 73392, 742464, 8254080, 99838080, 1307301120, 18407831040, 277570298880, 4460506444800, 76131788544000, 1375048700928000, 26208041287680000, 525597067634688000, 11065538390925312000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. REFERENCES Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..250 FORMULA 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, 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-1)*n/4)))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-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)]). MAPLE a := n -> n!*(n+1)^2*sum ((-1)^(k+1)/(k^2*(k+1)^2), k = 1..n): seq(a(n), n = 1..20); PROG (Haskell) a142984 n = a142984_list !! (n-1) a142984_list = 1 : 4 : zipWith (+) (map (* 4) \$ tail a142984_list) (zipWith (*) (drop 2 a002378_list) a142984_list) -- Reinhard Zumkeller, Jul 17 2015 CROSSREFS Cf. A142983, A142985, A142986, A142987. Cf. A002378. Sequence in context: A069835 A007196 A091638 * A283055 A097593 A188686 Adjacent sequences: A142981 A142982 A142983 * A142985 A142986 A142987 KEYWORD easy,nonn AUTHOR Peter Bala, Jul 17 2008 STATUS approved

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Last modified January 30 07:42 EST 2023. Contains 359942 sequences. (Running on oeis4.)