A216090 Numbers n such that k^(n-1) == k (mod n) for every k = 1, 2, ..., n-1.
1, 2, 6, 10, 14, 22, 26, 30, 34, 38, 46, 58, 62, 74, 82, 86, 94, 106, 118, 122, 134, 142, 146, 158, 166, 178, 182, 194, 202, 206, 214, 218, 226, 254, 262, 274, 278, 298, 302, 314, 326, 334, 346, 358, 362, 382, 386, 394, 398, 422, 446, 454, 458, 466, 478, 482
Offset: 1
Keywords
Examples
a(4) = 10 because x^9 == 1, 2, ..., 9 (mod 10) with 9 distinct residues such that: 1^9 = 1 == 1 (mod 10); 2^9 = 512 == 2 (mod 10); 3^9 = 19683 == 3 (mod 10); 4^9 = 262144 == 4 (mod 10); 5^9 = 1953125 == 5 (mod 10); 6^9 = 10077696 == 6 (mod 10); 7^9 = 40353607 == 7 (mod 10); 8^9 = 134217728 == 8 (mod 10); 9^9 = 387420489 == 9 (mod 10).
Links
- Paolo P. Lava, Table of n, a(n) for n = 1..1000
Crossrefs
Programs
-
Maple
with(numtheory):for n from 1 to 500 do:j:=0:for i from 1 to n do: if irem(i^(n-1),n)=i then j:=j+1:else fi:od:if j=n-1 then printf(`%d, `, n):else fi:od:
-
Mathematica
f[n_] := And @@ Table[PowerMod[k, n - 1, n] == k, {k, n - 1}]; Select[Range[500], f] (* T. D. Noe, Sep 03 2012 *) -
PARI
isok(n) = {for (k=1, n-1, if (Mod(k, n)^(n-1) != Mod(k, n), return (0));); return (1);} \\ Michel Marcus, Apr 23 2017 -
Python
from sympy.ntheory.factor_ import core from sympy import primefactors def ok(n): if n<3: return True if core(n) == n: for p in primefactors(n): if (n - 2)%(p - 1): return False return True return False print([n for n in range(1, 501) if ok(n)]) # Indranil Ghosh, Apr 23 2017
Comments