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A119690
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n! mod n*(n+1)/2.
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5
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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, 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, 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
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OFFSET
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1,2
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COMMENTS
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All terms are even.
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
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LINKS
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FORMULA
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a(n) = n if n+1 is an odd prime, a(n) = 0 otherwise.
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MAPLE
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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);
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PROG
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(Magma) [ Factorial(n) mod (n*(n+1) div 2): n in [1..100] ];
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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