A220295 - OEIS

%I #40 Sep 16 2017 00:35:30

%S 11,11,5,0,2,0,281,0,3,3,0,0,11,0,761,3,15233,0,0,2003,0,0,89,5,0,

%T 11369,0,431,0,3,0,2,15401,0,2393,0,0,0,14741,11,2,0,2,0,7901,0,0,0,

%U 11831,0,3,2,0,2,7211,235043,10781,0,0,3,0,29,0,31151,0,77471,0

%N Least prime q satisfying q^p == 1 (mod 2p+1) and p^q == 1 (mod 2q+1), or 0 if otherwise, where p = prime(n).

%C The numbers p and q that are the members of the solution (p,q) satisfying q^p == 1 (mod 2p+1) and p^q == 1 (mod 2q+1) where p and q are prime.

%C q^p == 1 (mod 2p+1) and p^q == 1 (mod 2q+1) has no solution if p is a prime of the form 6m+1 (A002476) => a(n) = 0.

%H T. D. Noe, <a href="/A220295/b220295.txt">Table of n, a(n) for n = 1..1000</a>

%e a(2) = 11 because for (p,q) = (3,11), 11^3 == 1 (mod 7) and 3^11 == 1 mod 23.

%p with(numtheory):T:=array(1..100):T[1]:=2:T[2]:=3:

%p          for n from 3 to 100 do :

%p             p:=ithprime(n):if irem(p,6)=5 then

%p            T[n]:=p:else T[n]:=0:

%p            fi:

%p          od:

%p           print(T):

%p                for a from 1 to 50 do:

%p                       p:=T[a]: if p= 0 then

%p                       printf(`%d, `,0):

%p                       else :

%p                        ii:=0:

%p                   for b from 1 to 10^8 while(ii=0) do:

%p                   q:=ithprime(b):  if  irem(p^q, 2*q+1) = 1 and irem(q^p, 2*p+1)=1 then

%p                   ii:=1: printf(`%d, `,q):

%p                  else :

%p                  fi:

%p                 od:

%p                 fi:

%p              od:

%t Table[p = Prime[n]; If[Mod[p, 6] == 1, 0, q = 2; While[! (PowerMod[p, q, 2 q + 1] == 1 && PowerMod[q, p, 2 p + 1] == 1), q = NextPrime[q]]; q], {n, 100}] (* _T. D. Noe_, Feb 22 2013 *)

%Y Cf. A002476, A003627, A198033.

%K nonn

%O 1,1

%A _Michel Lagneau_, Feb 19 2013