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A191833
Least number k such that k^k == k+1 (mod m), or 0 if no such k exists, where m = A007310(n).
2
1, 7, 10, 14, 19, 11, 16, 3, 27, 43, 46, 178, 55, 36, 100, 64, 33, 79, 147, 43, 56, 258, 16, 86, 135, 52, 31, 27, 398, 335, 33, 187, 213, 151, 43, 680, 163, 61, 38, 243, 29, 327, 39, 213, 2068, 72, 37, 799, 198, 223, 141, 887, 92, 304, 132, 250, 808, 217, 327, 192, 271, 538, 398, 187, 79, 38, 31, 1713, 0, 413, 24, 1287, 976, 501, 48
OFFSET
1,2
COMMENTS
k^k == k+1 (mod m) does not have any solutions for m = 2 or 3, so only numbers in A007310 need be considered.
In general, if there is a solution, the first is less than m * phi(m), where phi is the Euler totient function A000010, since the values loop from that point (at least for units).
a(n) = 0 if and only if A007310(n) is in A191834. - Robert Israel, Sep 12 2017
LINKS
MAPLE
f:= proc(n) local m, k;
m:= (6*n + (-1)^n - 3)/2;
for k from 1 to ilcm(m, numtheory:-phi(m)) do
if igcd(k, m) = 1 and k &^ k - k - 1 mod m = 0 then return k fi;
od:
0
end proc:
map(f, [$1..100]); # Robert Israel, Sep 12 2017
MATHEMATICA
A007310[n_] := 2*n + 2*Floor[n/2] - 1; a[n_] := (For[m = A007310[n]; k = 1, k <= m^2, k++, If[PowerMod[k, k, m] == Mod[k+1, m], Return[k]]]; 0); Table[a[n], {n, 1, 75}] (* Jean-François Alcover, Sep 13 2013 *)
PROG
(PARI) a(n)=local(m); m=A007310(n); for(k=1, m^2, if(Mod(k, m)^k==k+1, return(k))); 0
CROSSREFS
Sequence in context: A005526 A192292 A030123 * A020752 A134302 A229306
KEYWORD
nonn
AUTHOR
STATUS
approved