A230405 - OEIS

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

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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.