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A249802
a(n) is the smallest prime q such that n(q-1)-1 is prime, that is, the smallest prime q so that n = (p+1)/(q-1) with p prime; or a(n) = -1 if no such q exists.
4
5, 3, 2, 2, 5, 2, 3, 2, 3, 3, 5, 2, 19, 2, 3, 3, 5, 2, 3, 2, 3, 3, 7, 2, 7, 5, 3, 7, 7, 2, 3, 2, 5, 3, 5, 3, 3, 2, 7, 3, 5, 2, 7, 2, 3, 19, 7, 2, 3, 5, 3, 3, 5, 2, 3, 5, 3, 7, 7, 2, 19, 2, 5, 3, 7, 3, 7, 2, 3, 3, 5, 2, 67, 2, 3, 3, 5, 5, 3, 2, 11, 3, 5, 2, 7, 11
OFFSET
1,1
COMMENTS
Variation on Schinzel's Hypothesis.
LINKS
Eric Weisstein's World of Mathematics, Schinzel's Hypothesis
EXAMPLE
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)=5.
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. Etc.
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(ithprime(k)); break; fi;
od; od; end: P(10^5);
PROG
(PARI) a(n) = my(q=2); while(! isprime(n*(q-1)-1), q = nextprime(q+1)); q; \\ Michel Marcus, Nov 07 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paolo P. Lava, Nov 06 2014
STATUS
approved