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

Author

Antti Karttunen, Oct 31 2013

Keywords

• nonn

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

Crossrefs

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

Cf. A000217, A007623, A230404.

Cf. also A230413.

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