login
A Golomb-like recurrence that decreases infinitely often.
1

%I #27 Nov 27 2016 09:09:05

%S 2,8,6,4,2,24,22,20,18,16,14,12,10,8,6,4,2,72,70,68,66,64,62,60,58,56,

%T 54,52,50,48,46,44,42,40,38,36,34,32,30,28,26,24,22,20,18,16,14,12,10,

%U 8,6,4,2,216,214,212,210,208,206,204,202

%N A Golomb-like recurrence that decreases infinitely often.

%C The sequence is a solution to the recursion a(a(n) + n) = a(n) + 4n, which is similar to the Golomb recursion b(b(n) + kn) = 2b(n) + kn, k=1,2,...

%H G. C. Greubel, <a href="/A028352/b028352.txt">Table of n, a(n) for n = 1..1000</a>

%H E. J. Barbeau, J. Chew and S. Tanny, <a href="http://www.combinatorics.org/Volume_4/Abstracts/v4i1r16.html">A matrix dynamics approach to Golomb's recursion</a>, Electronic J. Combinatorics, #4.1 16 1997.

%F a(1)=1, a(2*3^m+r) = 8*3^m - 2r with m=0, 1, 2, ..., 0 <= r <= 4*3^m - 1. - _Ralf Stephan_, Jan 16 2003

%F a(n) = 12*3^floor(log(n/2)/log(3)) - 2*n. - _Benoit Cloitre_, Jan 23 2003

%t Table[12*3^Floor[Log[n/2]/Log[3]] - 2*n , {n,1,50}] (* _G. C. Greubel_, Nov 27 2016 *)

%o (PARI) a(n) = if(n==1,2,m=floor(log(n/2)/log(3));r=n-2*3^m;8*3^m-2*r)

%K nonn,easy

%O 1,1

%A _Simon Plouffe_

%E Edited by _Ralf Stephan_, Jan 16 2003