OFFSET
1,3
COMMENTS
a(n) = n - 1 if n is 1 or a prime, by Fermat's little theorem. It is conjectured that the converse is also true; see A055032 and A201560 and note that a(n) = n-1 <==> A055032(n) = 1 <==> A201560(n) = 0.
As of 1991, Giuga and Bedocchi had verified no composite n < 10^1700 satisfies a(n) = n - 1 (Ribemboim, 1991). - Alonso del Arte, May 10 2013
REFERENCES
Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 6 of Hong Kong Mathematical Olympiad 2007 (find a(7)), page 134.
Richard K. Guy, Unsolved Problems in Number Theory, A17.
Paulo Ribemboim, The Little Book of Big Primes. New York: Springer-Verlag (1991): 17.
LINKS
Ivan Neretin, Table of n, a(n) for n = 1..10000
John Clark, On a conjecture involving Fermat's Little Theorem, Thesis, 2008, University of South Florida.
Hong Kong Mathematics Olympiad (2007-2008), Final Event 2 (Group), problem 2, p. 437.
Bernard Schott, Proof that a(n)=n/2 if n is of the form 4k+2.
FORMULA
a(p) = p - 1 if p is prime, and a(4n) = 0.
a(n) + 1 == A201560(n) (mod n).
a(n) = n/2 iff n is of the form 4k+2 (conjectured). - Ivan Neretin, Sep 23 2016
a(4*k+2) = 2*k+1; for a proof see corresponding link. - Bernard Schott, Dec 29 2021
EXAMPLE
Sum(m^3, m = 1 .. 3) = 1^3 + 2^3 + 3^3 = 36 == 0 (mod 4), so a(4) = 0.
MATHEMATICA
Table[Mod[Sum[i^(n - 1), {i, n - 1}], n], {n, 75}] (* Alonso del Arte, May 10 2013 *)
PROG
(PARI) a(n) = lift(sum(i=1, n, Mod(i, n)^(n-1))); \\ Michel Marcus, Feb 23 2020
(Python)
def a(n): return sum(pow(m, n-1, n) for m in range(1, n))%n
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Jan 02 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jonathan Sondow, Jan 12 2012
STATUS
approved