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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 (list; graph; refs; listen; history; text; internal format)
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*/
CROSSREFS
Sequence in context: A221959 A221553 A071871 * A361688 A185120 A217842
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 11 2000
EXTENSIONS
Comments corrected by Jonathan Sondow, Jan 11 2012
STATUS
approved

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Last modified April 23 02:41 EDT 2024. Contains 371906 sequences. (Running on oeis4.)