A348348 Smallest k such that the numbers j*k - 1 and j*k + 1 are prime for 1 <= j <= n.

4, 6, 6, 21968100, 100803789240, 683751016938990, 1651735848676253340

Offset: 1

Author

Pontus von Brömssen, Oct 13 2021

Keywords

• nonn
• more

Comments

The following heuristic argument suggests that a(n) exists for all n: For large (random) k and a specific j <= n, the probability that both j*k - 1 and j*k + 1 are prime should be of the order 1/(log k)^2 (a slight twist of the first Hardy-Littlewood conjecture). Assuming independence between different j, the probability that this holds for 1 <= j <= n is of the order 1/(log k)^(2*n). Since the sum over k of 1/(log k)^(2*n) diverges, this should hold for infinitely many k by the second Borel-Cantelli lemma (assuming independence between different k).

Examples

			a(1) = A014574(1) = 4.
a(2) = A066388(1) = 6.
a(3) = A118859(1) = 6.
a(4) = A118860(1) = 21968100.
a(5) = A349321(1) = 100803789240.
		

Links

• Wikipedia, Borel-Cantelli lemma
• Wikipedia, Twin prime

Crossrefs

Cf. A014574, A066388, A118859, A118860, A349321.

Programs

PARI

isok(k, n) = for (j=1, n, if (!isprime(j*k-1) || !isprime(j*k+1), return(0))); return(1);
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Jul 01 2022

Python

from sympy import isprime,nextprime
def A348348(n):
    p = 2
    while 1:
        p_next = nextprime(p)
        if p_next == p+2 and all(isprime(j*(p+1)-1) and isprime(j*(p+1)+1) for j in range(2,n+1)):
            return p+1
        p = p_next

Extensions

a(5), a(6) from Jon E. Schoenfield, Nov 14 2021

a(7) from Klaus Muuss, Jul 01 2022