login
A356195
The binary expansion of a(n) is obtained by applying the totalistic cellular automaton with rule 2*n to the binary expansion of n.
2
0, 1, 0, 3, 0, 6, 3, 7, 0, 14, 3, 14, 0, 9, 7, 15, 0, 30, 3, 30, 0, 25, 7, 30, 0, 16, 12, 29, 7, 23, 15, 31, 0, 62, 3, 62, 0, 57, 7, 62, 0, 48, 12, 61, 7, 55, 15, 62, 0, 32, 28, 60, 7, 38, 28, 61, 0, 33, 19, 51, 15, 47, 31, 63, 0, 126, 3, 126, 0, 121, 7, 126
OFFSET
0,4
COMMENTS
To compute the binary expansion of a(n):
- we scan the binary digits of n from right to left,
- at some position k >= 0 (0 corresponding to the least significant bit):
- we count the number of 1's at positions >= k, say we have w 1's,
- if 2^w appears in the binary expansion of 2*n,
then we insert a 1,
otherwise we insert a 0,
- as we are considering an even automaton (with rule 2*n),
once scanning the leading 0's of n, we will only insert 0's,
- and the result will have finitely many 1's.
More formally: 2^k appears in the binary expansion of a(n) iff 2^A000120(floor(n/2^k)) appears in the binary expansion of 2*n.
FORMULA
a(n) = n iff n belongs to A000225.
a(n) = 0 iff n AND A038573(n) = 0 (where AND denotes the bitwise AND operator).
EXAMPLE
For n = 43:
- the binary expansion of 2*43 is "1010110",
- so we apply the following totalistic cellular automaton:
w | >=7 6 5 4 3 2 1 0
out | 0 1 0 1 0 1 1 0
- scanning the binary expansion of n, we obtains:
bin(n) | 1 0 1 0 1 1
w | 1 1 2 2 3 4
bin(a(n)) | 1 1 1 1 0 1
- so a(n) = 61.
PROG
(PARI) a(n) = { my (v=0, m=n); for (k=0, oo, if (m==0, return (v), bittest(2*n, hammingweight(m)), v+=2^k); m\=2) }
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Jul 29 2022
STATUS
approved