OFFSET
1,4
COMMENTS
Pt(n) = n!*(n+1)! / 2^n.
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.
PROG
(Python)
f=t=1
for n in range(1, 33):
t*=n*(n+1)/2
f*=n
print str(t%f)+', ',
CROSSREFS
KEYWORD
nonn
AUTHOR
Alex Ratushnyak, Dec 03 2013
STATUS
approved