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A014117
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Numbers n such that m^(n+1) = m mod n holds for all m.
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7
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OFFSET
| 1,2
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COMMENTS
| "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 (blekraj(AT)yahoo.com), Jul 04 2002
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REFERENCES
| J. Dyer-Bennet, "A Theorem in Partitions of the Set of Positive Integers", Amer. Math. Monthly, 47(1940) pp. 152-4.
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LINKS
| D. Zagier, Problems posed at the St Andrews Colloquium, 1996
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MATHEMATICA
| (* 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][[2]] ]}], n], Element[C[1], Integers]]; ok[1] = True; A014117 = {}; Do[ If[ok[n], Print[n]; AppendTo[ A014117, n] ], {n, 1, 2000}] (* From Jean-François Alcover, Dec 21 2011 *)
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CROSSREFS
| Sequence in context: A115961 A123137 * A188672 A054377 A007018 A100016
Adjacent sequences: A014114 A014115 A014116 * A014118 A014119 A014120
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KEYWORD
| nonn,fini,full,nice
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AUTHOR
| David Broadhurst (D.Broadhurst(AT)open.ac.uk)
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