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Take smallest prime q such that n(q-1)-1 is prime (A249802), that is, the smallest prime q so that n = (p+1)/(q-1) with p prime; sequence gives values of p; or -1 if A249802(n) = -1.
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%I #16 Feb 25 2023 15:18:04

%S 3,3,2,3,19,5,13,7,17,19,43,11,233,13,29,31,67,17,37,19,41,43,137,23,

%T 149,103,53,167,173,29,61,31,131,67,139,71,73,37,233,79,163,41,257,43,

%U 89,827,281,47,97,199,101,103,211,53,109,223,113,347,353,59,1097

%N Take smallest prime q such that n(q-1)-1 is prime (A249802), that is, the smallest prime q so that n = (p+1)/(q-1) with p prime; sequence gives values of p; or -1 if A249802(n) = -1.

%C Variation on Schinzel's Hypothesis.

%H Paolo P. Lava, <a href="/A249803/b249803.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SchinzelsHypothesis.html">Schinzel's Hypothesis</a>.

%e For n=1 the minimum primes p and q are 3 and 5: (p+1)/(q-1) = (3+1)/(5-1) = 4/4 = 1. Therefore a(1)=3.

%e For n=2 the minimum primes p and q are 3 and 3: (p+1)/(q-1) = (3+1)/(3-1) = 4/2 = 2. Therefore a(2)=3.

%p with(numtheory): P:=proc(q) local k,n;

%p for n from 1 to q do for k from 1 to q do

%p if isprime(n*(ithprime(k)-1)-1) then print(n*(ithprime(k)-1)-1);

%p break; fi; od; od; end: P(10^5);

%o (PARI) a(n) = my(q=2); while(! isprime(p=n*(q-1)-1), q = nextprime(q+1)); p; \\ _Michel Marcus_, Nov 07 2014

%Y Cf. A060324, A062251, A064632, A249800, A249801, A249802.

%K nonn,easy

%O 1,1

%A _Paolo P. Lava_, Nov 06 2014