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A178989
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a(n) = (k^k + k!) / (k*(k + 1)), where k = prime(n) - 1.
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1
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1, 1, 14, 1128, 90942080, 57157560576, 67818988957718528, 115047995548743401472, 674758653138775267142795264, 40819609745761407890621234130376982528, 221388314080552960064314183934017536000000, 79870389582370042643423622863118514819531536385179648
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OFFSET
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1,3
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COMMENTS
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According to the two theorems (Fermat and Wilson), k + 1 divides(k^k + k!) because k^k == 1 (mod k + 1) and k! == - 1 (mod k + 1) for any prime k + 1.
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LINKS
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EXAMPLE
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a(3) = 14 because prime(3) = 5 => p = 4 => (4^4 + 4!) / 4(4 + 1) = 280/20 = 14.
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MAPLE
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with(numtheory): for n from 1 to 20 do: p:=ithprime(n):q:=p-1:x:= (q^q + q!)/(q*p):
printf(`%d, `, x): od:
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MATHEMATICA
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f[n_] := Block[{k = Prime@ n - 1}, (k^k + k!)/(k (k + 1))]; Array[f, 10] (* Robert G. Wilson v, Jan 05 2011 *)
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PROG
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(PARI) a(n)={my(k=prime(n)-1); (k^k + k!) / (k*(k + 1))} \\ Andrew Howroyd, Apr 13 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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