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
Seiichi Manyama, Table of n, a(n) for n = 1..76
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
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*/
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 11 2000
EXTENSIONS
Comments corrected by Jonathan Sondow, Jan 11 2012
STATUS
approved