A267796 a(n) = (n+1)*4^(2n+1).

4, 128, 3072, 65536, 1310720, 25165824, 469762048, 8589934592, 154618822656, 2748779069440, 48378511622144, 844424930131968, 14636698788954112, 252201579132747776, 4323455642275676160, 73786976294838206464, 1254378597012249509888, 21250649172913403461632

Offset

0,1

Comments

The partial sums of A001246(n)/a(n) converge absolutely. This series is also the hypergeometric function 1/4 * 4F3(1/2,1/2,1,1;2,2,2;1). - Ralf Steiner, Feb 09 2016

Links

Colin Barker, Table of n, a(n) for n = 0..800

Index entries for linear recurrences with constant coefficients, signature (32,-256).

Formula

a(n) = A013709(n)*(n+1).

From Colin Barker, Mar 23 2017: (Start)

G.f.: 4 / (1 - 16*x)^2.

a(n) = 32*a(n-1) - 256*a(n-2) for n>1. (End)

From Amiram Eldar, Apr 17 2022: (Start)

a(n) = A193132(n+1)/3.

Sum_{n>=0} 1/a(n) = 4*log(16/15).

Sum_{n>=0} (-1)^n/a(n) = 4*log(17/16). (End)

Example

For n=3, a(3) = (3+1)*4^(2*3+1) = 4*4^7 = 65536.

Mathematica

Table[(n + 1) 4^(2 n + 1), {n, 0, 20}] (* Vincenzo Librandi, Feb 10 2016 *)

Prog

(PARI) a(n) = (n+1)*4^(2*n+1); \\ Michel Marcus, Jan 28 2016
(PARI) Vec(4 / (1 - 16*x)^2 + O(x^30)) \\ Colin Barker, Mar 23 2017
(Magma) [(n+1)*4^(2*n+1): n in [0..45]]; // Vincenzo Librandi, Feb 10 2016

Crossrefs

Cf. A001246, A013709, A193132, A267982.

Sequence in context: A270959 A291851 A128790 * A013823 A321233 A130318

Adjacent sequences: A267793 A267794 A267795 * A267797 A267798 A267799

Keyword

nonn,easy

Author

Ralf Steiner, Jan 24 2016

Extensions

More terms from Michel Marcus, Jan 28 2016

Status

approved