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

3, 11, 13, 19, 23, 29, 31, 43, 49, 53, 57, 59, 61, 67, 71, 73, 77, 79, 83, 85, 89, 91, 93, 97, 101, 103, 109, 113, 127, 129, 131, 133, 141, 143, 147, 149, 151, 157, 161, 163, 167, 169, 173, 177, 179, 183, 187, 197, 199, 201, 203, 205, 211, 217, 229, 235, 237, 239

Offset: 1

Author

Owen C. Keith, Oct 28 2020

Keywords

• nonn

Comments

Some values of m have multiple solutions.

For example, for m = 49, 25^25 == 26^26 (mod 49) and 37^37 == 38^38 (mod 49).

All terms are odd. - Robert Israel, Nov 25 2020

Examples

			3 is a term because 1^1 == 2^2 (mod 3).
11 is a term because 8^8 == 9^9 (mod 11).
13 is a term because 8^8 == 9^9 (mod 13).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Similar sequences: A174824, A239061, A239062, A239063.

Programs

Maple

filter:= proc(n) local x,y,z;
  y:= 1;
  for x from 2 to n-1 do
    z:= x &^ x mod n;
    if z = y then return true fi;
    y:= z
  od;
  false
end proc:
select(filter, [$2..1000]); # Robert Israel, Nov 25 2020

Mathematica

seqQ[n_] := AnyTrue[Range[n - 1], PowerMod[#, #, n] == PowerMod[# + 1, # + 1, n] &]; Select[Range[240], seqQ] (* Amiram Eldar, Oct 28 2020 *)

PARI

isok(m)=sum(i=1, m-1, Mod(i,m)^i == Mod((i+1),m)^(i+1)) \\ Andrew Howroyd, Oct 28 2020