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A055030
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(Sum(m^(p-1),m=1..p-1)+1)/p as p runs through the primes.
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11
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1, 2, 71, 9596, 1355849266, 1032458258547, 1653031004194447737, 3167496749732497119310, 22841077183004879532481321652, 1768861419039838982256898243427529138091, 10293527624511391856267274608237685758691696
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OFFSET
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1,2
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COMMENTS
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It is conjectured that (Sum(m^(n-1),m=1..n-1)+1)/n is an integer iff n is 1 or a prime.
Always an integer from little Fermat theorem. Converse is conjectured to be true: if p | (1+1^(p-1)+2^(p-1)+3^(p-1)+...+(p-1)^(p-1)) and p > 1, then p is prime. That was checked by Giuga up to p <= 10^1000. [Benoit Cloitre, Jun 09 2002]
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, A17.
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LINKS
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FORMULA
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MAPLE
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p := ithprime(n) ;
add(m^(p-1), m=1..p-1) ;
(1+%)/p ;
end proc:
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MATHEMATICA
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Array[(Sum[m^(# - 1), {m, # - 1}] + 1)/# &@ Prime@ # &, 11] (* Michael De Vlieger, Nov 04 2017 *)
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PROG
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(PARI) for(n=1, 20, print1((1+sum(i=1, prime(n)-1, i^(prime(n)-1)))/prime(n), ", ")) /* Benoit Cloitre, Jun 09 2002*/
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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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