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%I #11 Feb 27 2020 23:16:02
%S 1,0,1,1,1,0,1,0,1,-1,1,1,1,0,1,1,1,0,1,2,1,0,1,0,1,-1,1,1,1,0,1,0,1,
%T -1,1,-1,1,0,1,-1,1,-2,1,0,1,0,1,1,1,0,1,2,1,0,1,0,1,-1,1,1,1,0,1,1,1,
%U -1,1,0,1,0,1,-3,1,-3,1,1,1,0,1,2,1,-2,1
%N Start with an empty stack S; for n = 1, 2, 3, ..., interpret the binary representation of n from left to right as follows: in case of bit 1, push the number 1 on top of S, in case of bit 0, replace the two numbers on top of S, say u on top of v, with v-u; a(n) gives the number on top of S after processing n.
%C This sequence is a variant of A330261.
%C After processing n, S has A268289(n) elements.
%C Every integer appears infinitely many times in the sequence:
%C - the proof is similar to that found in A330261,
%C - see A330265 for the values in order of appearance.
%H Rémy Sigrist, <a href="/A330262/b330262.txt">Table of n, a(n) for n = 1..8192</a>
%H Rémy Sigrist, <a href="/A330262/a330262.png">Scatterplot of the first 2^20 terms</a>
%H Rémy Sigrist, <a href="/A330262/a330262.gp.txt">PARI program for A330262</a>
%e The first terms, alongside the binary representation of n and the evolution of stack S, are:
%e n a(n) bin(n) S
%e - ---- ------ ------------------------------------------------------------
%e 1 1 1 () -> (1)
%e 2 0 10 (1) -> (1,1) -> (0)
%e 3 1 11 (0) -> (0,1) -> (0,1,1)
%e 4 1 100 (0,1,1) -> (0,1,1,1) -> (0,1,0) -> (0,1)
%e 5 1 101 (0,1) -> (0,1,1) -> (0,0) -> (0,0,1)
%e 6 0 110 (0,0,1) -> (0,0,1,1) -> (0,0,1,1,1) -> (0,0,1,0)
%e 7 1 111 (0,0,1,0) -> (0,0,1,0,1) -> (0,0,1,0,1,1) -> (0,0,1,0,1,1,1)
%o (PARI) See Links section.
%Y Cf. A268289, A308551, A330261, A330265.
%K sign,base
%O 1,20
%A _Rémy Sigrist_, Dec 07 2019
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