|
%I #23 May 04 2017 08:42:28
%S 0,1,3,5,7,9,11,12,14,16,18,20,22,23,25,27,29,31,33,34,36,38,40,42,44,
%T 45,47,49,51,53,55,57,58,60,62,64,66,68,69,71,73,75,77,79,80,82,84,86,
%U 88,90,91,93,95,97,99,101,103,104,106,108,110,112,114,115,117,119,121,123,125,126,128,130,132
%N Defined by the properties that it starts with 0, and when you successively apply DIFF, RUNS, BISECT, RUNS you get (4,1,3,1) repeated infinitely often.
%C Here DIFF means take first differences, RUNS means list successive run lengths, and BISECT means take alternate terms.
%C This agrees with the Beatty sequence for the tribonacci constant (A158919) for n <= 17160 but thereafter is different. In fact A158919(17161) = 31564, whereas a(17161) = 31563.
%C This arose in an attempt to find recurrences for A158919 and several related sequences. The moral is that without a proof, apparent recurrences are worthless.
%H N. J. A. Sloane, <a href="/A276384/b276384.txt">Table of n, a(n) for n = 0..40000</a>
%F For n >= 1, a(n) = A276385(n)-n.
%e Seq. 0, 1, 3, 5, 7, 9, 11, 12, 14, 16, 18, 20, 22, 23, 25, 27, 29, ...
%e DIFF 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2 ...
%e RUNS 1, 5, 1, 5, 1, 5, 1, 5, 1, 6, 1, 5, 1, 5, 1, 5, 1, 6, 1, 5, 1, ...
%e BISECT 5, 5, 5, 5, 6, 5, 5, 5, 6, 5, 5, 5, 5, 6, 5, 5, 5, 6, 5, 5, ...
%e RUNS 4, 1, 3, 1, 4, 1, 3, 1, 4, 1, 3, 1, 4, 1, 3, 1, 4, 1, 3, 1, 4, ...
%p with(transforms): r1:=[]:
%p for n from 1 to 1000 do r1:=[op(r1), 1,5,1,5,1,5,1,5,1,6,1,5,1,5,1,5,1,6]; od:
%p r2:=[]: for n from 1 to nops(r1) do if r1[n]=1 then r2:=[op(r2),1]; else for i from 1 to r1[n] do r2:=[op(r2),2]; od: fi: od:
%p r3:=[0,op(PSUM(r2))]:
%Y Cf. A158919, A276385.
%K nonn
%O 0,3
%A _N. J. A. Sloane_, Sep 03 2016
|
