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A338445 Numbers m with integer solution to x^x == (x+1)^(x+1) (mod m) with 1<=x<m. 2

%I #22 Nov 25 2020 13:12:54

%S 3,11,13,19,23,29,31,43,49,53,57,59,61,67,71,73,77,79,83,85,89,91,93,

%T 97,101,103,109,113,127,129,131,133,141,143,147,149,151,157,161,163,

%U 167,169,173,177,179,183,187,197,199,201,203,205,211,217,229,235,237,239

%N Numbers m with integer solution to x^x == (x+1)^(x+1) (mod m) with 1<=x<m.

%C Some values of m have multiple solutions.

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

%C All terms are odd. - _Robert Israel_, Nov 25 2020

%H Robert Israel, <a href="/A338445/b338445.txt">Table of n, a(n) for n = 1..10000</a>

%e 3 is a term because 1^1 == 2^2 (mod 3).

%e 11 is a term because 8^8 == 9^9 (mod 11).

%e 13 is a term because 8^8 == 9^9 (mod 13).

%p filter:= proc(n) local x,y,z;

%p y:= 1;

%p for x from 2 to n-1 do

%p z:= x &^ x mod n;

%p if z = y then return true fi;

%p y:= z

%p od;

%p false

%p end proc:

%p select(filter, [$2..1000]); # _Robert Israel_, Nov 25 2020

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

%o (PARI) isok(m)=sum(i=1, m-1, Mod(i,m)^i == Mod((i+1),m)^(i+1)) \\ _Andrew Howroyd_, Oct 28 2020

%Y Similar sequences: A174824, A239061, A239062, A239063.

%K nonn

%O 1,1

%A _Owen C. Keith_, Oct 28 2020

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Last modified September 15 04:39 EDT 2024. Contains 375931 sequences. (Running on oeis4.)