A390118 Least positive integer m such that m*n + 1 = p^2 for some prime number p.

3, 4, 1, 2, 24, 4, 24, 1, 32, 12, 48, 2, 216, 12, 8, 3, 264, 16, 72, 6, 8, 24, 96, 1, 408, 108, 104, 6, 120, 4, 120, 9, 16, 132, 24, 8, 144, 36, 72, 3, 168, 4, 696, 12, 8, 48, 1680, 1, 192, 204, 88, 54, 216, 52, 144, 3, 24, 60, 2112, 2, 2208, 60, 80, 15, 96, 8, 1080, 66, 32, 12

Offset

1,1

Comments

a(n) always exists since there are (infinitely many) primes p congruent to 1 modulo n.

Conjecture: a(n) <= (n+5)*(n-1) for all n > 1 (verified for n <= 12000).

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..4000

Example

a(1) = 3 since 3*1 + 1 = 2^2 and 2 is prime.
a(2) = 4 since 4*2 + 1 = 3^2 and 3 is prime.
a(13) = 216 = (13+5)*(13-1) since 216*13 + 1 = 53^2 and 53 is prime.

Mathematica

PQ[n_]:=PQ[n]=PrimeQ[Sqrt[n]];
tab={}; Do[m=1; Label[bb]; If[PQ[m*n+1], tab=Append[tab, m]; Goto[aa]];
m=m+1; Goto[bb]; Label[aa], {n, 1, 70}]; Print[tab]

Prog

(PARI) a(n) = my(m=1); while (isprimepower(m*n+1) != 2, m++); m; \\ Michel Marcus, Nov 02 2025
(Python)
from math import isqrt
from sympy import isprime
from itertools import count
def a(n): return next(m for m in count(1) if (p:=isqrt(q:=m*n+1))**2==q and isprime(p))
print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Nov 02 2025

Crossrefs

Cf. A000040, A001248, A034693, A390117.

Sequence in context: A021297 A124909 A348354 * A356708 A281098 A090279

Adjacent sequences: A390115 A390116 A390117 * A390119 A390120 A390121

Keyword

nonn

Author

Zhi-Wei Sun, Oct 25 2025

Status

approved