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 A204187 a(n) = Sum_{m=1..n-1} m^(n-1) modulo n. 10
 0, 1, 2, 0, 4, 3, 6, 0, 6, 5, 10, 0, 12, 7, 10, 0, 16, 9, 18, 0, 14, 11, 22, 0, 20, 13, 18, 0, 28, 15, 30, 0, 22, 17, 0, 0, 36, 19, 26, 0, 40, 21, 42, 0, 21, 23, 46, 0, 42, 25, 34, 0, 52, 27, 0, 0, 38, 29, 58, 0, 60, 31, 42, 0, 52, 33, 66, 0, 46, 35, 70, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Equals 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. - Sondow 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 It seems that a(n)=n/2 iff n is of the form 4k+2. - Ivan Neretin, Sep 23 2016 REFERENCES 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. FORMULA a(p) = p - 1 if p is prime, and a(4n) = 0. a(n) + 1 == A201560(n) (mod n). 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 CROSSREFS Cf. A055023, A055030, A055031, A055032, A201560. Cf. A191677 (zeros). Sequence in context: A128263 A241384 A140254 * A095202 A291937 A243488 Adjacent sequences:  A204184 A204185 A204186 * A204188 A204189 A204190 KEYWORD nonn,easy AUTHOR Jonathan Sondow, Jan 12 2012 STATUS approved

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Last modified October 24 04:47 EDT 2020. Contains 337975 sequences. (Running on oeis4.)