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a(n) = n! modulo n*(n+1)*(n+2)/3.
1

%I #12 Apr 13 2024 02:00:43

%S 1,2,6,24,50,48,0,0,210,120,352,168,0,0,800,288,1122,360,0,0,2002,528,

%T 0,0,0,0,4032,840,4930,960,0,0,0,0,8400,1368,0,0,11440,1680,13202,

%U 1848,0,0,17250,2208,0,0,0,0,24752,2808,0,0,0,0,34162,3480,37760,3720,0,0,0,0

%N a(n) = n! modulo n*(n+1)*(n+2)/3.

%C It appears that a(1)=1, a(2)=2, a(3)=6, and, for n>3, a(n) = n*(n+2) if n+1 is prime, else a(n) = n*(n+1)*(n+5)/6 if n+2 is prime, else a(n)=0. This has been verified for n up to 1000.

%H Vincenzo Librandi, <a href="/A175624/b175624.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Mod[(n!), (n^3 + 3 n^2 + 2 n)/3], {n, 100}] (* _Vincenzo Librandi_, Jul 10 2014 *)

%o (PARI) a(n) = n! % (n*(n+1)*(n+2)/3); \\ _Michel Marcus_, Jul 09 2014

%o (Magma)

%o [Factorial(n) mod (2*Binomial(n+2,3)): n in [1..80]]; // _G. C. Greubel_, Apr 12 2024

%o (SageMath)

%o [factorial(n)%(2*binomial(n+2,3)) for n in range(1,81)] # _G. C. Greubel_, Apr 12 2024

%Y Cf. A061006, A072230, A119690, A120387, A175567.

%K nonn

%O 1,2

%A _John W. Layman_, Jul 27 2010