A284569 a(n) = LCM of the lengths of runs of 1-bits in binary representation of n.

1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 2, 6, 3, 3, 3, 6, 4, 4, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5, 2, 2, 2, 2, 2, 2, 2, 6, 2, 2, 2, 2, 2, 2, 6, 4, 3, 3, 3, 6, 3, 3, 6, 3, 4

Offset: 0

Author

Antti Karttunen, Apr 14 2017

Keywords

• nonn
• base

Examples

			For n = 27, in binary A007088(27) = "11011", the lengths of runs of 1-bits are [2,2], thus a(27) = lcm(2,2) = 2.
For n = 55, in binary A007088(55) = "110111", the lengths of runs of 1-bits are [2,3], thus a(55) = lcm(2,3) = 6.
		

Links

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

Crossrefs

Cf. A005940, A072411, A227349, A284562, A284559.

Cf. A003714 (positions of ones).

Differs from A227349 for the first time at n=27, where a(27)=2, while A227349(27)= 4.

Differs from A038374 for the first time at n=55, where a(55) = 6, while A038374(55) = 3.

Programs

Scheme

(define (A284569 n) (apply lcm (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2))))) ;; For bisect and binexp->runcount1list, see the Program section of A227349.
(define (A284569 n) (A072411 (A005940 (+ 1 n))))

Formula

a(n) = A072411(A005940(1+n)).

a(n) = A227349(n) / A284562(n).