A242927 -id:A242927 - cp's OEIS frontend

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

1, 2, 6, 42, 1806

Offset: 1

"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.
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
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
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
Squarefree numbers n such that b^n == 1 (mod n^2) for every b coprime to n. Squarefree terms of A341858. - Thomas Ordowski, Aug 05 2024
Conjecture: Numbers n such that gcd(d+1, n) > 1 for every proper divisor d of n. Verified up to 10^696. - David Radcliffe, May 29 2025

M. A. Alekseyev, J. M. Grau, and A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT], 2016-2018.
John H. Castillo and Jhony Fernando Caranguay Mainguez, The set of k-units modulo n, arXiv:1708.06812 [math.NT], 2017.
Yongyi Chen and Tae Kyu Kim, On Generalized Carmichael Numbers, arXiv:2103.04883 [math.NT], 2021.
J. Dyer-Bennet, A Theorem in Partitions of the Set of Positive Integers, Amer. Math. Monthly, 47(1940) pp. 152-4.
L. Halbeisen and N. Hungerbühler, On generalised Carmichael numbers, Hardy-Ramanujan Society, 1999, 22 (2), pp.8-22. (hal-01109575)
J. M. Grau, A. M. Oller-Marcen, and Jonathan Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv 1309.7941 [math.NT], 2013; Monatsh. Math., 177 (2015), 421-436.
B. C. Kellner, The equation denom(B_n) = n has only one solution, preprint 2005.
Don Reble, A014117 and related OEIS sequences (shows there are no other terms)
Jonathan Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + … + k^n = (k+1)^n modulo k and k^2, arXiv:1011.2154 [math.NT], 2010; see Prop. 2; Integers, 11 (2011), article A34.
Eric Weisstein's World of Mathematics, Bernoulli Number
D. Zagier, Problems posed at the St Andrews Colloquium, 1996

Squarefree terms of A124240. - Robert Israel and Thomas Ordowski, Jun 23 2017

Cf. A007018, A031971, A054377, A100016, A242927, A341858.

r[n_] := Reduce[ Mod[m^(n+1) - m, n] == 0, m, Integers]; ok[n_] := Range[n]-1 === Simplify[ Mod[ Flatten[ m /. {ToRules[ r[n][[2]] ]}], n], Element[C[1], Integers]]; ok[1] = True; A014117 = {}; Do[ If[ok[n], Print[n]; AppendTo[ A014117, n] ], {n, 1, 2000}] (* Jean-François Alcover, Dec 21 2011 *)
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 *)

[n for n in range(1, 2000) if all(pow(m, n+1, n) == m for m in range(n))] # David Radcliffe, May 29 2025

For n <= 5, a(n) = a(n-1)^2 + a(n-1) with a(0) = 1. - Raphie Frank, Nov 12 2012

a(n+1) = A007018(n) = A054377(n) = A100016(n) for n = 1, 2, 3, 4. - Jonathan Sondow, Oct 01 2013

A239723 Least number k such that n^k + (n+1)^k + ... + (n+k-1)^k is prime or 0 if no such number exists.

Original entry on oeis.org

2, 1, 1, 2, 1, 0, 1, 0, 2, 0, 1, 2, 1, 2, 0, 6, 1, 0, 1, 0, 6, 2, 1, 2, 2, 0, 0, 0, 1, 2, 1, 2, 0, 2, 2, 0, 1, 0, 2, 0, 1, 2, 1, 0, 0, 0, 1, 6, 0, 2, 0, 0, 1, 0, 0, 0, 0, 6, 1, 2, 1, 0, 0, 0, 2, 0, 1, 0, 2, 2, 1, 2, 1, 0, 0, 6, 0, 0, 1, 0, 0, 2, 1, 2, 2, 0, 2, 0, 1, 2, 6

Offset: 1

Derek Orr, May 30 2014

nonn

a(119) = 42. Is a(n) only equal to 0, 1, 2, 6, or 42?

a(n) = 0 is confirmed for k <= 500. See A242927.

			1^1 = 1 is not prime. 1^2+2^2 = 5 is prime. Thus a(1) = 2.

Derek Orr, Table of n, a(n) for n = 1..91

Cf. A242927.

a(n)=for(k=1,500,if(ispseudoprime(sum(i=0,k-1,(n+i)^k)),return(k)))
n=1;while(n<200,print1(a(n),", ");n+=1)

cp's OEIS Frontend

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

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