|
%I #10 Dec 25 2013 02:40:34
%S 0,1,0,12,60,540,0,20160,181440,907200,19958400,359251200,1556755200,
%T 32691859200,0,10461394944000,177843714048000,1600593426432000,
%U 60822550204416000,608225502044160000,38318206628782080000,702500454861004800000,12926008369442488320000
%N Pt(n) mod n!, where Pt(n) is product of first n positive triangular numbers (A000217).
%C Pt(n) = n!*(n+1)! / 2^n.
%C Pt(n) mod n! = 0 if and only if 2^n divides (n+1)!, that is, n+1 is a power of 2. Thus indices of zeros are of the form 2^k-1.
%o (Python)
%o f=t=1
%o for n in range(1,33):
%o t*=n*(n+1)/2
%o f*=n
%o print str(t%f)+',',
%Y Cf. A000142, A000217.
%Y Cf. A006472 (triangular factorial, essentially equal to Pt(n)).
%Y Cf. A067667 (Pt(n)/n! for n's of the form 2^k-1).
%Y Cf. A069902, A007917.
%K nonn
%O 1,4
%A _Alex Ratushnyak_, Dec 03 2013
|
