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%I #10 Nov 02 2014 15:57:37
%S 0,0,1,0,3,1,5,0,1,3,9,6,11,5,3,0,15,1,17,3,5,9,21,10,3,11,1,5,27,24,
%T 29,0,9,15,5,35,35,17,11,39,39,5,41,9,22,21,45,18,5,3,15,11,51,1,9,34,
%U 17,27,57,46,59,29,62,0,11,9,65,15,21,48,69,40,71,35,3,17,9,11,77,79,1
%N Number of trailing zeros in the factorial base representation of products of binomial coefficients: a(n) = A230403(A001142(n)).
%C a(n) = A249151(n)-1. Please see the comments and graph of that sequence.
%F a(n) = A230403(A001142(n)).
%o (Scheme) (define (A249150 n) (A230403 (A001142 n)))
%Y One less than A249151.
%Y Cf. A249423 (values k such that a(k) = k).
%Y Cf. A249425 (record positions).
%Y Cf. A249426 (record values).
%Y Cf. A001142, A187059, A230403.
%K nonn
%O 0,5
%A _Antti Karttunen_, Oct 25 2014
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