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A233004
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Pt(n) mod n!, where Pt(n) is product of first n positive triangular numbers (A000217).
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0
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0, 1, 0, 12, 60, 540, 0, 20160, 181440, 907200, 19958400, 359251200, 1556755200, 32691859200, 0, 10461394944000, 177843714048000, 1600593426432000, 60822550204416000, 608225502044160000, 38318206628782080000, 702500454861004800000, 12926008369442488320000
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OFFSET
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1,4
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COMMENTS
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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.
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LINKS
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PROG
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(Python)
f=t=1
for n in range(1, 33):
t*=n*(n+1)/2
f*=n
print str(t%f)+', ',
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CROSSREFS
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Cf. A006472 (triangular factorial, essentially equal to Pt(n)).
Cf. A067667 (Pt(n)/n! for n's of the form 2^k-1).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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