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A119690 n! mod n*(n+1)/2. 5

%I #16 Sep 08 2022 08:45:25

%S 0,2,0,4,0,6,0,0,0,10,0,12,0,0,0,16,0,18,0,0,0,22,0,0,0,0,0,28,0,30,0,

%T 0,0,0,0,36,0,0,0,40,0,42,0,0,0,46,0,0,0,0,0,52,0,0,0,0,0,58,0,60,0,0,

%U 0,0,0,66,0,0,0,70,0,72,0,0,0,0,0,78,0,0,0,82,0,0,0,0,0,88,0,0,0,0,0,0

%N n! mod n*(n+1)/2.

%C All terms are even.

%C It appears that f(n)=(n!)^(2k+1) modulo n(n+1)/2 is n if n is one less than an odd prime, else f(n) is 0, for any integer k. See A175567 for related results involving an even power of n!. - _John W. Layman_, Jul 12 2010

%F a(n) = n if n+1 is an odd prime, a(n) = 0 otherwise.

%p P:=proc(n) local i, j, k; j:=1; k:=0; for i from 1 by 1 to n do j:=j*i; k:=k+i; print(j mod k); od; end: P(100);

%o (Magma) [ Factorial(n) mod (n*(n+1) div 2): n in [1..100] ];

%Y Cf. A006093, A010051, A175567, A175553.

%K easy,nonn

%O 1,2

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jun 09 2006

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Last modified April 27 18:09 EDT 2024. Contains 372020 sequences. (Running on oeis4.)