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A175567 (n!)^2 modulo n(n+1)/2.

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%I A175567 #10 Aug 27 2016 14:08:48
%S A175567 0,1,0,6,0,15,0,0,0,45,0,66,0,0,0,120,0,153,0,0,0,231,0,0,0,0,0,378,0,
%T A175567 435,0,0,0,0,0,630,0,0,0,780,0,861,0,0,0,1035,0,0,0,0,0,1326,0,0,0,0,
%U A175567 0,1653,0,1770,0,0,0,0,0,2145,0,0,0,2415,0,2556,0,0,0,0,0,3003,0,0,0,3321
%N A175567 (n!)^2 modulo n(n+1)/2.
%C A175567 It appears that if n is one less than an odd prime then (n!)^2 modulo n(n+1)/2 is n(n-1)/2 else 0. This result appears to hold for any even power of n!. See A119690 for similar results related to odd powers of n!.
%H A175567 Vincenzo Librandi, <a href="/A175567/b175567.txt">Table of n, a(n) for n = 1..1000</a>
%t A175567 Table[Mod[(n!)^2, (n^2 + n)/2], {n, 100}] (* _Vincenzo Librandi_, Jul 10 2014 *)
%t A175567 Table[PowerMod[n!,2,(n(n+1))/2],{n,100}] (* _Harvey P. Dale_, Aug 27 2016 *)
%o A175567 (PARI) a(n) = (n!)^2 % (n*(n+1)/2); \\ _Michel Marcus_, Jul 09 2014
%Y A175567 Cf. A119690, A169690, A175553.
%K A175567 nonn
%O A175567 1,4
%A A175567 _John W. Layman_, Jul 12 2010