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A249802
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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.
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4
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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
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OFFSET
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1,1
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COMMENTS
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Variation on Schinzel's Hypothesis.
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LINKS
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EXAMPLE
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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.
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MAPLE
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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);
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PROG
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(PARI) a(n) = my(q=2); while(! isprime(n*(q-1)-1), q = nextprime(q+1)); q; \\ Michel Marcus, Nov 07 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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