A284559 a(n) = LCM of run lengths in binary representation of n.

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

Offset: 0

Author

Antti Karttunen, Apr 14 2017

Keywords

• nonn
• base

Examples

			For n=12, A007088(12) = "1100" in binary, the run lengths are [2,2], thus a(12) = lcm(2,2) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10922
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to lcm's

Crossrefs

Cf. A000975 (positions of ones).

Cf. A007088, A167489, A227349, A284558, A284569, A284579.

Programs

Python

from math import lcm
from itertools import groupby
def a(n): return lcm(*(len(list(g)) for k, g in groupby(bin(n)[2:])))
print([a(n) for n in range(87)]) # Michael S. Branicky, Oct 15 2022

Scheme

(define (A284559 n) (apply lcm (binexp->runcount1list n)))
;; Or:
(define (A284559 n) (reduce lcm 1 (binexp->runcount1list n))) ;; For binexp->runcount1list, see the Program section of A227349.

Formula

a(n) = A167489(n) / A284558(n).