%I #34 Nov 05 2017 11:52:19
%S 1,2,71,9596,1355849266,1032458258547,1653031004194447737,
%T 3167496749732497119310,22841077183004879532481321652,
%U 1768861419039838982256898243427529138091,10293527624511391856267274608237685758691696
%N (Sum(m^(p-1),m=1..p-1)+1)/p as p runs through the primes.
%C It is conjectured that (Sum(m^(n-1),m=1..n-1)+1)/n is an integer iff n is 1 or a prime.
%C 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]
%C For Sum(m^p, m=1..p-1)/p as p runs through the odd primes, see A219550. - _Jonathan Sondow_, Oct 31 2017
%D R. K. Guy, Unsolved Problems in Number Theory, A17.
%H Seiichi Manyama, <a href="/A055030/b055030.txt">Table of n, a(n) for n = 1..76</a>
%H K. MacMillan and J. Sondow, <a href="http://arxiv.org/abs/1011.0076">Proofs of power sum and binomial coefficient congruences via Pascal's identity</a>, Amer. Math. Monthly, 118 (2011), 549-551.
%F a(n) = (1+A225578(n))/A000040(n). - _R. J. Mathar_, Jan 09 2017
%p A055030 := proc(n)
%p p := ithprime(n) ;
%p add(m^(p-1),m=1..p-1) ;
%p (1+%)/p ;
%p end proc:
%p seq(A055030(n),n=1..5) ; # _R. J. Mathar_, Jan 09 2017
%t Array[(Sum[m^(# - 1), {m, # - 1}] + 1)/# &@ Prime@ # &, 11] (* _Michael De Vlieger_, Nov 04 2017 *)
%o (PARI) for(n=1,20,print1((1+sum(i=1, prime(n)-1,i^(prime(n)-1)))/prime(n), ",")) /* _Benoit Cloitre_, Jun 09 2002*/
%Y Cf. A055031, A055032, A055023, A201560, A204187, A219550, A294507.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Jun 11 2000
%E Comments corrected by _Jonathan Sondow_, Jan 11 2012