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

0,4

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

Formula

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

Example

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

Prog

(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.
(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

Crossrefs

Cf. A000975 (positions of ones).

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

Sequence in context: A269783 A043276 A319416 * A284583 A064742 A227185

Adjacent sequences: A284556 A284557 A284558 * A284560 A284561 A284562

Keyword

nonn,base

Author

Antti Karttunen, Apr 14 2017

Status

approved