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a(n) = A000217(A230404(n+1)); the first differences of A219650.
6

%I #16 Jan 05 2024 02:54:40

%S 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,

%T 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,

%U 3,1,1,6,1,1,3,1,1,3,1,1,3,1,1,6,1,1,3

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

%C 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.

%H Antti Karttunen, <a href="/A230405/b230405.txt">Table of n, a(n) for n = 0..10079</a>

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

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

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

%o (Scheme, two variants)

%o (define (A230405 n) (A000217 (A230404 (+ 1 n))))

%o (define (A230405 n) (- (A219650 (+ n 1)) (A219650 n)))

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

%Y Cf. A000217, A007623, A230404.

%Y Cf. also A230413.

%K nonn

%O 0,3

%A _Antti Karttunen_, Oct 31 2013