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A061006
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a(n) = (n-1)! mod n.
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11
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0, 1, 2, 2, 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
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OFFSET
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1,3
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COMMENTS
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It appears that a(n) = (n!*h(n)) mod n, where h(n) = Sum_{k = 1..n} 1/k. - Gary Detlefs, Sep 04 2010
Indeed: It is easy to show n!*h(n) - (n-1)! = n*(n-1)!*h(n-1). Since (n-1)!*h(n-1) is integral, n!*h(n) == (n-1)! mod n. - Franz Vrabec, Apr 08 2017
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LINKS
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FORMULA
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a(4) = 2, a(p) = p - 1 for p prime (Wilson's_theorem), a(n) = 0 otherwise. Apart from n = 4, a(n) = (n-1)*A061007(n) = (n-1)*A010051(n).
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EXAMPLE
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a(4) = 2 since (4-1)! = 6 = 2 mod 4.
a(5) = 4 since (5-1)! = 24 = 4 mod 5.
a(6) = 0 since (6-1)! = 120 = 0 mod 6.
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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