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A338445 Numbers m with integer solution to x^x == (x+1)^(x+1) (mod m) with 1<=x

%I A338445 #22 Nov 25 2020 13:12:54
%S A338445 3,11,13,19,23,29,31,43,49,53,57,59,61,67,71,73,77,79,83,85,89,91,93,
%T A338445 97,101,103,109,113,127,129,131,133,141,143,147,149,151,157,161,163,
%U A338445 167,169,173,177,179,183,187,197,199,201,203,205,211,217,229,235,237,239
%N A338445 Numbers m with integer solution to x^x == (x+1)^(x+1) (mod m) with 1<=x<m.
%C A338445 Some values of m have multiple solutions.
%C A338445 For example, for m = 49, 25^25 == 26^26 (mod 49) and 37^37 == 38^38 (mod 49).
%C A338445 All terms are odd. - _Robert Israel_, Nov 25 2020
%H A338445 Robert Israel, <a href="/A338445/b338445.txt">Table of n, a(n) for n = 1..10000</a>
%e A338445 3 is a term because 1^1 == 2^2 (mod 3).
%e A338445 11 is a term because 8^8 == 9^9 (mod 11).
%e A338445 13 is a term because 8^8 == 9^9 (mod 13).
%p A338445 filter:= proc(n) local x,y,z;
%p A338445   y:= 1;
%p A338445   for x from 2 to n-1 do
%p A338445     z:= x &^ x mod n;
%p A338445     if z = y then return true fi;
%p A338445     y:= z
%p A338445   od;
%p A338445   false
%p A338445 end proc:
%p A338445 select(filter, [$2..1000]); # _Robert Israel_, Nov 25 2020
%t A338445 seqQ[n_] := AnyTrue[Range[n - 1], PowerMod[#, #, n] == PowerMod[# + 1, # + 1, n] &]; Select[Range[240], seqQ] (* _Amiram Eldar_, Oct 28 2020 *)
%o A338445 (PARI) isok(m)=sum(i=1, m-1, Mod(i,m)^i == Mod((i+1),m)^(i+1)) \\ _Andrew Howroyd_, Oct 28 2020
%Y A338445 Similar sequences: A174824, A239061, A239062, A239063.
%K A338445 nonn
%O A338445 1,1
%A A338445 _Owen C. Keith_, Oct 28 2020