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 A014117 Numbers n such that m^(n+1) == m (mod n) holds for all m. 40

%I

%S 1,2,6,42,1806

%N Numbers n such that m^(n+1) == m (mod n) holds for all m.

%C "Somebody incorrectly remembered Fermat's little theorem as saying that the congruence a^{n+1} = a (mod n) holds for all a if n is prime" (Zagier). The sequence gives the set of integers n for which this property is in fact true.

%C If i == j (mod n), then m^i == m^j (mod n) for all m. The latter congruence generally holds for any (m, n)=1 with i == j (mod k), k being the order of m modulo n, i.e., the least power k for which m^k == 1 (mod n). - _Lekraj Beedassy_, Jul 04 2002

%C Also, numbers n such that n divides denominator of the n-th Bernoulli number B(n) (cf. A106741). Also, numbers n such that 1^n + 2^n + 3^n + ... + n^n == 1 (mod n). Equivalently, numbers n such that B(n)*n == 1 (mod n). Equivalently, Sum_{prime p, (p-1) divides n} n/p == -1 (mod n). It is easy to see that for n > 1, n must be an even squarefree number. Moreover, the set P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This set is P = {2, 3, 7, 43}, implying that the sequence is finite and complete. - _Max Alekseyev_, Aug 25 2013

%C In 2005, B. C. Kellner proved E. W. Weisstein's conjecture that denom(B_n) = n only if n = 1806. - _Jonathan Sondow_, Oct 14 2013

%H M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:<a href="http://doi.org/10.1016/j.dam.2018.05.022">10.1016/j.dam.2018.05.022</a> arXiv:<a href="http://arxiv.org/abs/1602.02407">1602.02407</a> [math.NT], 2016-2018.

%H John H. Castillo and Jhony Fernando Caranguay Mainguez, <a href="https://arxiv.org/abs/1708.06812">The set of k-units modulo n</a>, arXiv:1708.06812 [math.NT], 2017.

%H J. Dyer-Bennet, <a href="http://www.jstor.org/stable/2304216">A Theorem in Partitions of the Set of Positive Integers</a>, Amer. Math. Monthly, 47(1940) pp. 152-4.

%H J. M. Grau, A. M. Oller-Marcen, and J. Sondow, <a href="http://arxiv.org/abs/1309.7941">On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m</a>, arXiv 1309.7941 [math.NT], 2013, Monatsh. Math., 177 (2015), 421-436.

%H B. C. Kellner, <a href="http://bernoulli.org/~bk/denombneqn.pdf">The equation denom(B_n) = n has only one solution</a>, preprint 2005.

%H Don Reble, <a href="/A014117/a014117.pdf">A014117 and related OEIS sequences</a> (shows there are no other terms)

%H J. Sondow and K. MacMillan, <a href="http://arxiv.org/abs/1011.2154">Reducing the Erdos-Moser equation 1^n + 2^n + … + k^n = (k+1)^n modulo k and k^2</a>, arXiv:1011.2154 [math.NT], 2010; see Prop. 2; Integers, 11 (2011), article A34.

%H E. W. Weisstein's Mathworld, <a href="http://mathworld.wolfram.com/BernoulliNumber.html">Bernoulli Number</a>

%H D. Zagier, <a href="http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Problems.html">Problems posed at the St Andrews Colloquium, 1996</a>

%F a(n) = a(n-1)^2 + a(n-1) with a(0) = 1. - _Raphie Frank_, Nov 12 2012 [This is wrong, since the sequence has only five terms. - _N. J. A. Sloane_, Apr 11 2017]

%F a(n+1) = A007018(n) = A054377(n) = A100016(n) for n = 1, 2, 3, 4. - _Jonathan Sondow_, Oct 01 2013

%t (* Checking up to n = 2000 *) r[n_] := Reduce[ Mod[m^(n+1) - m, n] == 0, m, Integers]; ok[n_] := Range[n]-1 === Simplify[ Mod[ Flatten[ m /. {ToRules[ r[n][] ]}], n], Element[C, Integers]]; ok = True; A014117 = {}; Do[ If[ok[n], Print[n]; AppendTo[ A014117, n] ], {n, 1, 2000}] (* _Jean-François Alcover_, Dec 21 2011 *)

%t Select[Range@ 2000, Function[n, Times @@ Boole@ Map[Function[m, PowerMod[m, n + 1, n] == Mod[m, n]], Range@ n] > 0]] (* _Michael De Vlieger_, Dec 30 2016 *)

%Y Squarefree terms of A124240. - _Robert Israel_ and _Thomas Ordowski_, Jun 23 2017

%Y Cf. A007018, A031971, A054377, A100016, A242927.

%K nonn,fini,full,nice

%O 1,2