%I #10 Jul 31 2022 19:54:45

%S 0,1,1,2,0,5,3,7,0,9,5,14,4,13,7,15,0,17,9,26,0,21,11,31,0,17,5,22,12,

%T 29,15,31,0,33,17,50,0,37,19,55,0,41,21,62,4,45,23,63,0,33,9,42,16,53,

%U 27,63,0,33,5,38,28,61,31,63,0,65,33,98,0,69,35,103

%N The binary expansion of a(n) is obtained by applying the elementary cellular automaton with rule (2*n) mod 16 to the binary expansion of n.

%C This sequence is a variant of A352528; here the cellular automaton maps 2 cells into 1, there 3 cells into 1.

%C The binary digit of a(n) at place value 2^k is a function of the binary digits of n at place values 2^(k+1) and 2^k (and of (2*n) mod 256).

%C We use even elementary cellular automaton rules, so "00" will always evolve to "0", and the binary expansion of a(n) will have finitely many 1's and will be correctly defined.

%H Rémy Sigrist, <a href="/A356215/b356215.txt">Table of n, a(n) for n = 0..8192</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>

%F a(2^k-1) = 2^k-1 for any k <> 2.

%F a(2^k) = 0 for any k > 1.

%e For n = 11:

%e - we use rule 22 mod 16 = 6,

%e - the binary expansion of 6 is "0110", so we apply the following evolutions:

%e 11 10 01 00

%e | | | |

%e v v v v

%e 0 1 1 0

%e - the binary expansion of 11 (with a leading 0's) is "...01011",

%e - the binary digit of a(11) at place value 2^0 is 0 (from "11"),

%e - the binary digit of a(11) at place value 2^1 is 1 (from "01"),

%e - the binary digit of a(11) at place value 2^2 is 1 (from "10"),

%e - the binary digit of a(11) at place value 2^3 is 1 (from "01"),

%e - the binary digit of a(11) at other places is 0 (from "00"),

%e - so the binary expansion of a(11) is "1110",

%e - and a(11) = 14.

%o (PARI) a(n) = { my (v=0, m=n); for (k=0, oo, if (m==0, return (v), bittest(2*n, m%4), v+=2^k); m\=2) }

%Y Cf. A352528, A356195.

%K nonn,base

%O 0,4

%A _Rémy Sigrist_, Jul 29 2022