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The binary expansion of a(n) is obtained by applying the elementary cellular automaton with rule (2*n) mod 256 to the binary expansion of n.
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%I #12 Mar 20 2022 18:29:29

%S 0,1,1,2,0,4,2,6,2,11,5,12,5,12,4,12,0,17,9,26,4,21,14,30,2,19,3,18,9,

%T 25,8,24,0,33,17,50,0,36,19,54,2,35,21,52,7,38,21,52,8,41,9,42,28,61,

%U 31,62,6,39,7,38,19,51,17,48,0,65,33,98,0,68,34,103

%N The binary expansion of a(n) is obtained by applying the elementary cellular automaton with rule (2*n) mod 256 to the binary expansion of n.

%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+2), 2^(k+1) and 2^k (and of (2*n) mod 256).

%C We use even elementary cellular automaton rules, so "000" 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="/A352528/b352528.txt">Table of n, a(n) for n = 0..8192</a>

%H Rémy Sigrist, <a href="/A352528/a352528.png">Scatterplot of the first 2^20 terms</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</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>

%e For n = 13:

%e - we apply rule 26,

%e - the binary expansion of 26 being "00011010", we apply the following evolutions:

%e 111 110 101 100 011 010 001 000

%e 0 0 0 1 1 0 1 0

%e - the binary expansion of 13 (with leading zeros) is "...0001101",

%e - the binary digit of a(13) at place value 2^0 is 0 (from "101"),

%e - the binary digit of a(13) at place value 2^1 is 0 (from "110"),

%e - the binary digit of a(13) at place value 2^2 is 1 (from "011"),

%e - the binary digit of a(13) at place value 2^3 is 1 (from "001"),

%e - the other binary digits of a(13) are 0 (from "000"),

%e - so the binary expansion of a(13) is "1100",

%e - so a(13) = 12.

%o (PARI) a(n) = { my (v=0, m=n); for (k=0, oo, if (m==0, return (v), bittest(2*n, m%8), v+=2^k); m\=2) }

%Y Cf. A269160, A269161, A269173, A269174.

%K nonn,base

%O 0,4

%A _Rémy Sigrist_, Mar 19 2022