|
|
A268466
|
|
Smallest m > 1 such that m^m == 1 (mod n).
|
|
2
|
|
|
2, 3, 2, 5, 4, 7, 6, 9, 8, 11, 5, 13, 3, 9, 4, 17, 4, 19, 9, 21, 8, 5, 22, 25, 24, 3, 26, 9, 7, 31, 6, 33, 10, 35, 6, 37, 9, 9, 8, 41, 10, 43, 6, 5, 8, 47, 46, 49, 18, 51, 4, 9, 13, 55, 12, 9, 20, 7, 29, 61, 15, 35, 8, 65, 8, 25, 22, 69, 22, 51, 5, 73, 18, 9, 26, 9, 12, 79, 24, 81
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
For n > 1, a(n) <= n + (-1)^n = A065190(n).
Conjecture: a(n) = n + (-1)^n for infinitely many n.
For m > 1, a(n) = m iff n is a divisor of m^m - 1 that is not a divisor of k^k - 1 for 1 < k < m.
In particular, a(m^m - 1) = m.
Is there any m such that this is the only n for which a(n) = m? (End)
|
|
LINKS
|
|
|
MAPLE
|
f:= proc(n) local k;
for k from 2 do if igcd(k, n) = 1 and k &^ k mod n = 1 then return k fi od
end proc:
|
|
MATHEMATICA
|
smg1[n_]:=Module[{m=2}, While[PowerMod[m, m, n]!=1, m++]; m]; Join[{2}, Array[ smg1, 80, 2]] (* Harvey P. Dale, Aug 13 2021 *)
|
|
PROG
|
(PARI) a(n) = {my(m = 2); while (Mod(m, n)^m != Mod(1, n), m++); m; } \\ Michel Marcus, Feb 05 2016
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|