A269161 Formula for Wolfram's Rule 86 cellular automaton: a(n) = 4n XOR (2n OR n).

0, 7, 14, 11, 28, 27, 22, 19, 56, 63, 54, 51, 44, 43, 38, 35, 112, 119, 126, 123, 108, 107, 102, 99, 88, 95, 86, 83, 76, 75, 70, 67, 224, 231, 238, 235, 252, 251, 246, 243, 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

Offset: 0

Author

Antti Karttunen, Feb 20 2016

Keywords

• nonn

Comments

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

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

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16383
• Eric Weisstein's World of Mathematics, Rule 30
• Index entries for sequences related to cellular automata
• Index to Elementary Cellular Automata

Crossrefs

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

Cf. A265281 (iterates starting from 1).

Cf. also A048727, A269160.

Programs

Mathematica

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

Python

def A269161(n): return n<<2 ^ (n<<1 |n) # Chai Wah Wu, Jun 29 2022

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.

Formula

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

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

Other identities. For all n >= 0:

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

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