A142980 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).

1, 5, 29, 190, 1414, 11820, 110004, 1129200, 12686256, 154896480, 2043108000, 28958014080, 438997622400, 7088892491520, 121487996448000, 2202440792832000, 42113131054848000, 847071044402688000

Offset

1,2

Comments

This is the case m = 2 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Links

Seiichi Manyama, Table of n, a(n) for n = 1..448

Formula

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n) is the Ehrhart polynomial for the 2-dimensional cross polytope (a square).

Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).

The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.

Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + (n-1)^2/5)))), for n >= 2.

The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + n^2/(5 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4*k^4 + 1)) = log(2) - (1 - 1/2); the final equality is a result of Glaisher.

Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.

Maple

p := n -> 2*n^2+ 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Crossrefs

Cf. A024167, A142979, A142981, A142982.

Sequence in context: A367232 A078945 A113713 * A062191 A367240 A171267

Adjacent sequences: A142977 A142978 A142979 * A142981 A142982 A142983

Keyword

easy,nonn

Author

Peter Bala, Jul 17 2008

Status

approved