A356215 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.

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

Offset: 0

Author

Rémy Sigrist, Jul 29 2022

Keywords

• nonn
• base

Comments

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

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).

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.

Examples

			For n = 11:
- we use rule 22 mod 16 = 6,
- the binary expansion of 6 is "0110", so we apply the following evolutions:
      11  10  01  00
       |   |   |   |
       v   v   v   v
       0   1   1   0
- the binary expansion of 11 (with a leading 0's) is    "...01011",
- the binary digit of a(11) at place value 2^0 is 0 (from     "11"),
- the binary digit of a(11) at place value 2^1 is 1 (from    "01"),
- the binary digit of a(11) at place value 2^2 is 1 (from   "10"),
- the binary digit of a(11) at place value 2^3 is 1 (from  "01"),
- the binary digit of a(11) at other places    is 0 (from "00"),
- so the binary expansion of a(11) is "1110",
- and a(11) = 14.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..8192
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to cellular automata

Crossrefs

Cf. A352528, A356195.

Programs

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) }

Formula

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

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