A035096 a(n) is the smallest k such that prime(n)*k+1 is prime.

1, 2, 2, 4, 2, 4, 6, 10, 2, 2, 10, 4, 2, 4, 6, 2, 12, 6, 4, 8, 4, 4, 2, 2, 4, 6, 6, 6, 10, 2, 4, 2, 6, 4, 8, 6, 10, 4, 14, 2, 2, 6, 2, 4, 18, 4, 10, 12, 24, 12, 2, 2, 6, 2, 6, 6, 8, 6, 4, 2, 6, 2, 4, 6, 6, 26, 6, 10, 6, 10, 14, 2, 6, 4, 12, 12, 24, 6, 8, 4, 2, 10, 2, 4, 10, 2, 8, 30

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Comments

These arithmetic progressions have prime differences. Note that both the terms of generated by this k values and the differences are primes as well.

This is one possible generalization of "the least prime problem in special arithmetic progressions" when n in the nk+1 form is replaced by n-th prime number.

Note that Dirichlet's theorem on primes in arithmetic progressions implies that a(n) always exists. - Max Alekseyev, Jul 11 2008

If a(n)=2, prime(n) is a Sophie Germain prime (A005384). Among the first 10^6 terms, the largest is a(330408) = 234. - Zak Seidov, Jan 28 2012

Examples

			a(15)=6 because the 15th prime is 47, and the smallest k such that 47k+1 is prime is k=6, for which 47k+1=283.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Math, Dirichlet's theorem.
• Index entries for sequences related to primes in arithmetic progressions.

Crossrefs

Smallest k such that k*n+1 is prime is A034693.

Sophie Germain primes are in A005384.

Cf. A000040, A035095. - Zak Seidov, Dec 27 2013

Cf. A117673.

Programs

Magma

S:=[];
k:=1;
for n in [1..90] do
  while not IsPrime(k*NthPrime(n)+1) do
       k:=k+1;
  end while;
  Append(~S, k);
  k:=1;
end for;
S; // Bruno Berselli, Apr 18 2013

Mathematica

Reap[Sow[1]; Do[p = Prime[n]; k = 2; While[! PrimeQ[k*p + 1], k = k + 2]; Sow[k], {n, 2, 10^4}]][[2, 1]] (* Zak Seidov, Jan 28 2012 *)
f[n_] := Block[{p = Prime@ n}, q = 1 + 2p; While[ !PrimeQ@ q, q += 2p]; (q - 1)/p]; f[1] = 1; Array[f, 88] (* Robert G. Wilson v, Dec 27 2014 *)

PARI

a(n) = if(n == 1, 1, my(t = 2*prime(n), m = t + 1); while(!isprime(m), m += t); 2*(m - 1)/t); \\ Amiram Eldar, Mar 19 2025

Formula

a(n) = (A035095(n)-1)/A000040(n). - Zak Seidov, Dec 27 2013