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Least number k such that k^k == k+1 (mod m), or 0 if no such k exists, where m = A007310(n).
2

%I #15 Sep 13 2017 03:12:02

%S 1,7,10,14,19,11,16,3,27,43,46,178,55,36,100,64,33,79,147,43,56,258,

%T 16,86,135,52,31,27,398,335,33,187,213,151,43,680,163,61,38,243,29,

%U 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

%N Least number k such that k^k == k+1 (mod m), or 0 if no such k exists, where m = A007310(n).

%C k^k == k+1 (mod m) does not have any solutions for m = 2 or 3, so only numbers in A007310 need be considered.

%C 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).

%C a(n) = 0 if and only if A007310(n) is in A191834. - _Robert Israel_, Sep 12 2017

%H Robert Israel, <a href="/A191833/b191833.txt">Table of n, a(n) for n = 1..10000</a>

%p f:= proc(n) local m,k;

%p m:= (6*n + (-1)^n - 3)/2;

%p for k from 1 to ilcm(m,numtheory:-phi(m)) do

%p if igcd(k,m) = 1 and k &^ k - k - 1 mod m = 0 then return k fi;

%p od:

%p 0

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Sep 12 2017

%t 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 *)

%o (PARI) a(n)=local(m);m=A007310(n);for(k=1,m^2,if(Mod(k,m)^k==k+1,return(k)));0

%Y Cf. A007310, A191834.

%K nonn

%O 1,2

%A _Franklin T. Adams-Watters_, Jun 17 2011