A350550 a(n) is the least prime q such that there exists a prime p with p^2 + n = (n+1)*q^2, or 0 if there is no such q.

5, 3, 0, 5, 3, 5, 2, 0, 13, 7, 5, 2911343369048029930623841, 11, 3, 2

Offset: 1

Author

J. M. Bergot and Robert Israel, Jan 04 2022

Keywords

• nonn
• hard
• more

Comments

a(16) > 10^1000 if it is not 0.

If it is not 0, then a(16) = A199772(k) where k is the smallest index such that both q = A199772(k) and p = A199773(k) are prime. If such an index exists, a(16) > 10^10000. - Jon E. Schoenfield, Jan 11 2022

Examples

			a(3) = 0 as the only positive integer solution of p^2 + 3 = 4*q^2 is p=1, q=1, and 1 is not prime.
a(4) = 5 as 11^2 + 4 = 125 = (4+1)*5^2 with 11 and 5 prime.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..179 with conjectured 0 values as -1.

Crossrefs

Cf. A199772, A199773, A350544.

Programs

Maple

# Returned values of -1 indicate that either a(n) = 0 or a(n) > 10^1000.
f:= proc(n) local m, x, y, S, cf, i, c, a, b, A, M, Sp;
  m:= n+1;
  if issqr(m) then
    S:= [isolve(x^2+n=m*y^2)];
    S:= map(t -> subs(t, [x, y]), S);
    S:= select(t -> andmap(isprime, t), S);
    if S = [] then return 0
    else return min(map(t -> t[2], S))
    fi;
  fi;
  cf:= NumberTheory:-ContinuedFraction(sqrt(m));
  for i from 1 do
    c:= Convergent(cf, i);
    if numer(c)^2 - m*denom(c)^2 = 1 then break fi
  od;
  a:= numer(c); b:= denom(c);
  A:= <|>;
  M:= floor(sqrt(n)*(1+sqrt(a+b*sqrt(m)))/(2*sqrt(m)));
  S:= select(t -> issqr(m*t^2-m+1), [$0..M]);
  S:= select(t -> igcd(t[1], t[2])=1, map(t -> , S));
  S:= map(t -> (t, <-t[1], t[2]>), S);
  if nops(S) = 0 then return 0 fi;
  for i from 0 do
    Sp:= select(t -> isprime(t[1]) and isprime(t[2]), S);
    if nops(Sp)>0 then return min(map(t -> t[2], Sp)) fi;
    S:= map(t -> A.t, S);
    if min(map(t -> t[2], S))>10^1000 then break fi;
  od;
  -1
end proc:
map(f, [$1..20]);

Formula

A350544(n)^2 + n = (n+1)*a(n)^2 if a(n) > 0.