A249801 Take smallest prime q such that n*(q+1)+1 is prime (A249800), that is, the smallest prime q so that n = (p-1)/(q+1) with p prime; sequence gives values of p; or -1 if A249800(n) = -1.

5, 7, 13, 13, 31, 19, 29, 97, 37, 31, 67, 37, 53, 43, 61, 97, 103, 73, 229, 61, 127, 67, 139, 73, 101, 79, 109, 113, 233, 181, 373, 97, 199, 103, 211, 109, 149, 229, 157, 241, 739, 127, 173, 353, 181, 139, 283, 193, 197, 151, 307, 157, 743, 163, 331, 337, 229

Offset: 1

Author

Paolo P. Lava, Nov 06 2014

Keywords

• nonn
• easy

Comments

Variation on Schinzel's Hypothesis.

Examples

			For n=1 the minimum primes p and q are 5 and 3: (p-1)/(q+1) = (5-1)/(3+1) = 4/4 = 1. Therefore a(1)=5.
For n=2 the minimum primes p and q are 7 and 2: (p-1)/(q+1) = (7-1)/(2+1) = 6/3 = 2. Therefore a(2)=7. Etc.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Schinzel's Hypothesis.

Crossrefs

Cf. A060324, A062251, A064632, A249800, A249802-A249803.

Programs

Maple

with(numtheory): P:=proc(q) local k,n;
for n from 1 to q do for k from 1 to q do
if isprime(n*(ithprime(k)+1)+1) then print(n*(ithprime(k)+1)+1);
break; fi; od; od; end: P(10^5);

PARI

a(n) = my(q=2); while(! isprime(p=n*(q+1)+1), q = nextprime(q+1)); p; \\ Michel Marcus, Nov 07 2014