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A055030
(Sum(m^(p-1),m=1..p-1)+1)/p as p runs through the primes.
11
1, 2, 71, 9596, 1355849266, 1032458258547, 1653031004194447737, 3167496749732497119310, 22841077183004879532481321652, 1768861419039838982256898243427529138091, 10293527624511391856267274608237685758691696
OFFSET
1,2
COMMENTS
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]
For Sum(m^p, m=1..p-1)/p as p runs through the odd primes, see A219550. - Jonathan Sondow, Oct 31 2017
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, A17.
LINKS
K. MacMillan and J. Sondow, Proofs of power sum and binomial coefficient congruences via Pascal's identity, Amer. Math. Monthly, 118 (2011), 549-551.
FORMULA
a(n) = (1+A225578(n))/A000040(n). - R. J. Mathar, Jan 09 2017
MAPLE
A055030 := proc(n)
p := ithprime(n) ;
add(m^(p-1), m=1..p-1) ;
(1+%)/p ;
end proc:
seq(A055030(n), n=1..5) ; # R. J. Mathar, Jan 09 2017
MATHEMATICA
Array[(Sum[m^(# - 1), {m, # - 1}] + 1)/# &@ Prime@ # &, 11] (* Michael De Vlieger, Nov 04 2017 *)
PROG
(PARI) for(n=1, 20, print1((1+sum(i=1, prime(n)-1, i^(prime(n)-1)))/prime(n), ", ")) /* Benoit Cloitre, Jun 09 2002*/
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 11 2000
EXTENSIONS
Comments corrected by Jonathan Sondow, Jan 11 2012
STATUS
approved