A220295 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).

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, 11369, 0, 431, 0, 3, 0, 2, 15401, 0, 2393, 0, 0, 0, 14741, 11, 2, 0, 2, 0, 7901, 0, 0, 0, 11831, 0, 3, 2, 0, 2, 7211, 235043, 10781, 0, 0, 3, 0, 29, 0, 31151, 0, 77471, 0

Offset

1,1

Comments

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.

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.

Links

T. D. Noe, Table of n, a(n) for n = 1..1000

Example

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

Maple

with(numtheory):T:=array(1..100):T[1]:=2:T[2]:=3:
         for n from 3 to 100 do :
            p:=ithprime(n):if irem(p, 6)=5 then
           T[n]:=p:else T[n]:=0:
           fi:
         od:
          print(T):
               for a from 1 to 50 do:
                      p:=T[a]: if p= 0 then
                      printf(`%d, `, 0):
                      else :
                       ii:=0:
                  for b from 1 to 10^8 while(ii=0) do:
                  q:=ithprime(b):  if  irem(p^q, 2*q+1) = 1 and irem(q^p, 2*p+1)=1 then
                  ii:=1: printf(`%d, `, q):
                 else :
                 fi:
                od:
                fi:
             od:

Mathematica

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 *)

Crossrefs

Cf. A002476, A003627, A198033.

Sequence in context: A291367 A061186 A135684 * A300289 A321108 A126610

Adjacent sequences: A220292 A220293 A220294 * A220296 A220297 A220298

Keyword

nonn

Author

Michel Lagneau, Feb 19 2013

Status

approved