A348598 Least prime p of the form k^2+1 such that p == A002496(n) (mod A002496(n+1)) with p>A002496(n), or 0 if no such p exists.

17, 1297, 90001, 1008017, 147457, 2421137, 15952037, 1378277, 7203857, 107122501, 164968337, 34503877, 38688401, 4851958337, 1075577617, 197121601, 1044582401, 315559697, 70924211857, 730296577, 20705483237, 15103426817, 197740302401, 4587352901, 155964965777

Offset: 1

Author

Michel Lagneau, Mar 20 2022

Keywords

• nonn

Comments

a(n) == 1, 5 (mod 16).

Conjecture: Consider the smallest prime p of the form k^2+1 such that p is congruent to A002496(n) modulo q, q prime of the form m^2+1 > A002496(n). Then q = A002496(n+1).

Corollary: For any pair (A002496(n), A002496(n+1)), there exist two integers m, k such that A002496(m) = A002496(n) + k*A002496(n+1), m>n+1 and n=1,2,3,...

Examples (see A352582):

A002496(3) = A002496(1) + 3*A002496(2),

A002496(11) = A002496(2) + 76*A002496(3),

A002496(49) = A002496(3) + 2432*A002496(4),

A002496(113) = A002496(4) + 9980*A002496(5).

Examples

			a(2) = 1297 because 1297 == A002496(2) (mod A002496(3)) => 1297 == 5 (mod 17).
		

Links

• Michel Lagneau, a(n),n=1..90

Crossrefs

Cf. A002496, A352582.

Programs

Maple

with(numtheory):T:=array(1..30000):k:=0:
nn:=500000:
  for m from 1 to nn do:
   if isprime(m^2+1)
    then
     k:=k+1:T[k]:=m^2+1:
    else
   fi:
  od:
  for n from 1 to 32 do:
  ii:=0:r:=T[n]:q:=T[n+1]:
   for i from 1 to k while(ii=0) do:
     p:=T[i]:r1:=irem(p,q):
        if r1=r and p>q
         then
         ii:=1: printf(`%d, `,p)
         else
        fi:
       od:
      od: