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Formula for Wolfram's Rule 86 cellular automaton: a(n) = 4n XOR (2n OR n).
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%I #27 Jun 29 2022 10:27:49

%S 0,7,14,11,28,27,22,19,56,63,54,51,44,43,38,35,112,119,126,123,108,

%T 107,102,99,88,95,86,83,76,75,70,67,224,231,238,235,252,251,246,243,

%U 216,223,214,211,204,203,198,195,176,183,190,187,172,171,166,163,152,159,150,147,140,139,134,131,448,455,462,459

%N Formula for Wolfram's Rule 86 cellular automaton: a(n) = 4n XOR (2n OR n).

%C The sequence is injective: no value occurs more than once.

%C Fibbinary numbers (A003714) give all integers n>=0 for which a(n) = A048727(n) and for which a(n) = A269160(n).

%H Antti Karttunen, <a href="/A269161/b269161.txt">Table of n, a(n) for n = 0..16383</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Rule30.html">Rule 30</a>

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

%H <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>

%F a(n) = 4n XOR (2n OR n) = A003987(4*n, A003986(2*n, n)).

%F a(n) = 4*n XOR A163617(n).

%F Other identities. For all n >= 0:

%F a(2*n) = 2*a(n).

%F a(n) = A057889(A269160(A057889(n))). [Rule 86 is the mirror image of rule 30.]

%t a[n_] := BitXor[4n, BitOr[2n, n]]; Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Feb 23 2016 *)

%o (Scheme) (define (A269161 n) (A003987bi (* 4 n) (A003986bi (* 2 n) n))) ;; Where A003986bi and A003987bi are implementation of dyadic functions giving bitwise-OR (A003986) and bitwise-XOR (A003987) of their arguments.

%o (Python)

%o def A269161(n): return n<<2 ^ (n<<1 |n) # _Chai Wah Wu_, Jun 29 2022

%Y Cf. A003714, A003986, A003987, A057889, A163617.

%Y Cf. A265281 (iterates starting from 1).

%Y Cf. also A048727, A269160.

%K nonn

%O 0,2

%A _Antti Karttunen_, Feb 20 2016