A160967 Numbers of the form (4^k - 1)/3 or 2^k.

0, 1, 2, 4, 5, 8, 16, 21, 32, 64, 85, 128, 256, 341, 512, 1024, 1365, 2048, 4096, 5461, 8192, 16384, 21845, 32768, 65536, 87381, 131072, 262144, 349525, 524288, 1048576, 1398101, 2097152, 4194304, 5592405, 8388608, 16777216, 22369621, 33554432

Offset: 0

Author

Reinhard Zumkeller, May 31 2009

Keywords

• nonn
• easy

Comments

A139391(a(n)) = 1 for n > 0.

Except for the first seven terms, the 3x + 1 trajectory for every number in this sequence includes 32. - Alonso del Arte, Jan 01 2015

For n>0, the Collatz-function starting with a(n) will terminate at 1. This is because all numbers of the form 2^k will terminate at 1, and ((4^k - 1)/3)*3 + 1 = 4^k = 2^2k. - Bob Selcoe, Apr 03 2015

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-4).

Crossrefs

Union of A002450 and A000079.

Programs

Haskell

a160967 n = a160967_list !! n
a160967_list = m a000079_list a002450_list where
   m xs'@(x:xs) ys'@(y:ys) | x < y     = x : m xs ys'
                           | x == y    = x : m xs ys
                           | otherwise = y : m xs' ys
-- Reinhard Zumkeller, Oct 03 2012

Magma

I:=[0,1,2,4,5,8,16]; [n le 7 select I[n] else 5*Self(n-3)-4*Self(n-6): n in [1..40]]; // Vincenzo Librandi, Apr 03 2015

Maple

0, seq(op([(4^j-1)/3, 4^j/2,4^j]),j=1..20); # Robert Israel, Jan 01 2015

Mathematica

Union[2^Range[0, 25], (4^Range[0, 13] - 1)/3] (* Alonso del Arte, Jan 01 2015 *)
Join[{0}, LinearRecurrence[{0, 0, 5, 0, 0, -4}, {1, 2, 4, 5, 8, 16}, 50]] (* Vincenzo Librandi, Apr 03 2015 *)

Formula

G.f.: x*(1 + 2*x + 4*x^2 - 2*x^4 - 4*x^5)/(1-5*x^3+4*x^6).

a(n+6) = 5*a(n+3) - 4*a(n) for n >= 1. - Robert Israel, Jan 01 2015