A089357 a(n) = 2^(6*n).

1, 64, 4096, 262144, 16777216, 1073741824, 68719476736, 4398046511104, 281474976710656, 18014398509481984, 1152921504606846976, 73786976294838206464, 4722366482869645213696, 302231454903657293676544, 19342813113834066795298816

Offset

0,2

Comments

For n > 0, numbers M such that a(n) is the highest power of 2 in the Collatz (3x+1) iteration are given by 2^k*(a(n)-1)/3 for any k >= 0. Example: For n = 1, the numbers such that 64 is the highest power of 2 in the Collatz (3x+1) iteration are given by 2^k*(64-1)/3 = 21*2^k for any k >= 0. See A008908 for more information on the Collatz (3x+1) iteration. - Derek Orr, Sep 22 2014

Powers of 64. - Alexander Fraebel, Aug 29 2020

Links

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

Tanya Khovanova, Recursive Sequences

Index entries for linear recurrences with constant coefficients, signature (64).

Formula

G.f.: 1/(1-64*x). - Philippe Deléham, Nov 24 2008

a(n) = 63*a(n-1) + 64^(n-1), a(0)=1. - Vincenzo Librandi, Jun 07 2011

E.g.f.: exp(64*x). - Ilya Gutkovskiy, Jul 02 2016

a(n) = A000079(A008588(n)). - Wesley Ivan Hurt, Jul 02 2016

a(n) = 64*a(n-1). - Miquel Cerda, Oct 27 2016

Maple

seq(2^(6*n), n=0..14); # Nathaniel Johnston, Jun 26 2011

Mathematica

2^(6 Range[0, 20]) (* or *) NestList[64#&, 1, 20] (* Harvey P. Dale, Sep 28 2011 *)

Prog

(Magma) [2^(6*n): n in [0..20]]; // Vincenzo Librandi, Jun 07 2011
(Maxima) makelist(2^(6*n), n, 0, 20); /* Martin Ettl, Nov 12 2012 */
(PARI) a(n)=64^n \\ Charles R Greathouse IV, Jul 02 2016

Crossrefs

Cf. A000079, A000302, A001018, A001025, A008588, A008908, A009976.

Sequence in context: A229102 A222934 A267994 * A144320 A324493 A324490

Adjacent sequences: A089354 A089355 A089356 * A089358 A089359 A089360

Keyword

nonn,easy

Author

Douglas Winston (douglas.winston(AT)srupc.com), Dec 26 2003

Status

approved