A013709 a(n) = 4^(2n+1).

4, 64, 1024, 16384, 262144, 4194304, 67108864, 1073741824, 17179869184, 274877906944, 4398046511104, 70368744177664, 1125899906842624, 18014398509481984, 288230376151711744, 4611686018427387904

Offset

0,1

Comments

Also powers of 2 with singly even numbers (A016825) as exponents. - Alonso del Arte, Sep 03 2012

The partial sum of A000888(n) = Catalan(n)^2*(n + 1) resp. A267844(n) = Catalan(n)^2*(4n + 3) resp. A267987(n) = Catalan(n)^2*(4n + 4) divided by A013709(n) (this) a(n) = 2^(4n+2) absolutely converge to 1/Pi resp. 1 resp. 4/Pi. Thus this series is 1/Pi resp. 1 resp. 4/Pi. - Ralf Steiner, Jan 23 2016

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Tanya Khovanova, Recursive Sequences

Index to divisibility sequences

Index entries for linear recurrences with constant coefficients, signature (16).

Formula

a(n) = 16*a(n - 1), n > 0; a(0) = 4. G.f.: 4/(1 - 16*x). [Philippe Deléham, Nov 23 2008]

a(n) = 4^(2*n + 1) = 2^(4*n + 2). - Alonso del Arte, Sep 03 2012

a(n) = 4*A001025(n). - Michel Marcus, Jan 30 2016

Maple

A013709:=n->4^(2*n+1): seq(A013709(n), n=0..20); # Wesley Ivan Hurt, Jan 30 2016

Mathematica

2^(4 Range[0, 15] + 2) (* Alonso del Arte, Sep 03 2012 *)
NestList[16#&, 4, 20] (* Harvey P. Dale, Jun 03 2013 *)

Prog

(Magma) [4^(2*n+1): n in [0..20]]; // Vincenzo Librandi, May 26 2011
(PARI) a(n)=4<<(4*n) \\ Charles R Greathouse IV, Apr 07 2012

Crossrefs

Cf. A000888, A001025, A016825, A267844, A267987.

Sequence in context: A026301 A010042 A154021 * A139292 A152923 A085807

Adjacent sequences: A013706 A013707 A013708 * A013710 A013711 A013712

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved