A361477 a(n) is the number of integers whose binary expansions have the same multiset of run-lengths as that of n.

1, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 3, 1, 3, 2, 1, 2, 3, 4, 3, 4, 1, 4, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 6, 6, 5, 6, 6, 4, 5, 1, 5, 6, 5, 4, 3, 2, 6, 6, 1, 6, 5, 6, 6, 1, 6, 4, 6, 2, 3, 2, 1, 2, 3, 4, 6, 12, 5, 12, 3, 12, 10, 6, 10, 4, 10, 12, 6, 4, 5, 6, 10

Offset

0,5

Comments

This sequence has similarities with A090706; here we consider multisets of run-lengths, there multisets of digits in binary expansions.

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Index entries for sequences related to binary expansion of n

Formula

a(n) = 1 iff n = 0 or n belongs to A140690.

Example

For n = 18:
- the binary expansion of 18 is "10010",
- the corresponding multiset of run-lengths is m = (1, 2, 1, 1),
- m has 4 terms: 3 times "1" and once "2",
- so a(18) = 4! / (3! * 1!) = 4.

Prog

(PARI) a(n) = { my (r=[]); while (n, my (v=valuation(n+n%2, 2)); n\=2^v; r=concat(v, r)); my (s=Set(r), f=vector(#s)); for (k=1, #r, f[setsearch(s, r[k])]++); (#r)! / prod(k=1, #f, f[k]!) }
(Python)
from math import factorial, prod
from itertools import groupby
from collections import Counter
def A361477(n): return factorial(len(c:=[len(list(g)) for k, g in groupby(bin(n)[2:])]))//prod(map(factorial, Counter(c).values())) # Chai Wah Wu, Mar 16 2023

Crossrefs

Cf. A090706, A101211, A140690.

Sequence in context: A308916 A353171 A182628 * A331366 A274225 A028334

Adjacent sequences: A361474 A361475 A361476 * A361478 A361479 A361480

Keyword

nonn,base

Author

Rémy Sigrist, Mar 13 2023

Status

approved