A230405 a(n) = A000217(A230404(n+1)); the first differences of A219650.

1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 10, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3

Offset

0,3

Comments

Construction: Count the trailing zeros in the factorial base representation (A007623) of 2n+2 (2, 4, 6, 8, ...) and then take the corresponding triangular number from A000217.

Links

Antti Karttunen, Table of n, a(n) for n = 0..10079

Formula

a(n) = A000217(A230404(n+1)).

a(n) = A219650(n+1) - A219650(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=0..m} a(k) = 2. - Amiram Eldar, Jan 05 2024

Prog

(Scheme, two variants)
(define (A230405 n) (A000217 (A230404 (+ 1 n))))
(define (A230405 n) (- (A219650 (+ n 1)) (A219650 n)))

Crossrefs

First differences of A219650. Can be used to compute A219650 and A230412.

Cf. A000217, A007623, A230404.

Cf. also A230413.

Sequence in context: A353487 A356308 A228925 * A101685 A238737 A049653

Adjacent sequences: A230402 A230403 A230404 * A230406 A230407 A230408

Keyword

nonn

Author

Antti Karttunen, Oct 31 2013

Status

approved