A351494 a(n) is the least prime p such that n*(p+1)+1 is the square of a prime, or 0 if there is no such p.

2, 3, 7, 5, 23, 3, 23, 2, 31, 11, 47, 3, 479, 11, 7, 2, 263, 19, 71, 5, 7, 23, 839, 11, 887, 107, 103, 5, 1031, 3, 3119, 29, 239, 131, 23, 7, 599, 599, 71, 2, 167, 3, 17159, 11, 7, 47, 9239, 5, 191, 443, 199, 53, 839, 211, 311, 2, 23, 59, 2111, 5, 2207, 59, 79, 251, 263, 7, 2399, 149, 31, 11

Offset

1,1

Comments

a(n) is the least prime p of the form (q^2-(n+1))/n where q is a prime, or 0 if there is no such p.

If n+1 is prime, then a(n) <= n+1 as n*(n+1+1)+1 = (n+1)^2 is the square of a prime.

If n+1 = r^2 is a square, q^2-(n+1) = (q-r)*(q+r), so in order for p = (q-r)*(q+r)/n to be prime we need at least one of q-r and q+r to be a divisor of n. In particular, in this case we have q <= r+n. This can be used to show that a(n) = 0 for n = 143, 288, 323, 575, 728, 899, ....

Conjecture: all cases where a(n) = 0 arise in this way.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

a(5) = 23 because 23 is prime, 5*(23+1)+1 = 121 = 11^2, and 11 is prime, and no prime < 23 works.

Maple

f:= proc(n) local j, q, p, M, qmax;
  if issqr(n+1) then qmax:= n+sqrt(n+1) else qmax:= infinity fi;
  M:=sort(map(t -> rhs(op(t)), [msolve(q^2-1, n)]));
  for j from 0 do
    for m in M do
      q:= j*n+m;
      if q > qmax then return 0 fi;
      if isprime(q) then
         p:= (q^2-1)/n - 1;
         if isprime(p)  then return p fi
   fi od od
end proc:
f(1):= 2:
map(f, [$1..100]);

Prog

(PARI) a(n) = my(p=2, r); while (!(issquare(n*(p+1)+1, &r) && isprime(r)), p=nextprime(p+1)); p; \\ Michel Marcus, May 04 2022

Crossrefs

Sequence in context: A294639 A038026 A051860 * A155766 A153488 A275115

Adjacent sequences: A351491 A351492 A351493 * A351495 A351496 A351497

Keyword

nonn

Author

J. M. Bergot and Robert Israel, May 03 2022

Status

approved