%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