

A225302


Smallest prime p such that n*(p1)1 and n*(p+1)+1 are both prime, or 0 if no such p exists.


1



5, 5, 3, 2, 5, 2, 3, 11, 3, 3, 5, 2, 0, 2, 3, 0, 5, 5, 0, 2, 5, 3, 11, 2, 0, 5, 3, 0, 7, 7, 0, 2, 5, 3, 5, 11, 3, 5, 7, 0, 61, 2, 0, 7, 3, 0, 13, 11, 3, 11, 11, 0, 19, 2, 0, 5, 3, 0, 19, 2, 0, 17, 5, 3, 7, 5, 0, 5, 3, 3, 11, 7, 0, 2, 7, 0, 5, 109
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OFFSET

1,1


COMMENTS

Numbers n such that a(n) = 0: 13, 16, 19, 25, 28, 31, 40, 43, 46, 52, 55, 58, 61, 67, 73, 76, ...
a(n) = 0 if n = 1 mod 3 and none of the pairs {n1, 3n+1}, {2n1, 4n+1}, {n+1, n+2} have both members prime. On Dickson's conjecture "if" can be replaced with "if and only if".  Charles R Greathouse IV, May 07 2013
Smallest k > 1 such that n^k  n  1 and n^k + n + 1 are both prime, or 0 if no such k exists: 0, 3, 2, 0, 2, 2, 0, 3, 3, 0, 4, 2, 0, 2, 3, 0, 2, 3, 0, 2, 2, 0,...  JuriStepan Gerasimov, May 09 2013


LINKS



EXAMPLE

a(1) = 5 because 1*5  1  1 = 3 and 1*5 + 1 + 1 = 7 are both prime,
a(2) = 5 because 2*5  2  1 = 7 and 2*5 + 2 + 1 = 13 are both prime,
a(3) = 3 because 3*3  3  1 = 5 and 3*3 + 3 + 1 = 13 are both prime.


MATHEMATICA

a[n_] := Block[{p = 2}, If[n < 5, {5, 5, 3, 2}[[n]], If[Mod[n, 3] == 1, If[PrimeQ[2*n1] && PrimeQ[4*n+1], 3, 0], While[! PrimeQ[n*(p  1) 1]  ! PrimeQ[n*(p + 1) +1], p = NextPrime@p]; p]]]; Array[a, 80] (* Giovanni Resta, May 05 2013 *)


PROG

(PARI) a(n)=forprime(p=2, 5, if(isprime(n*pn1) && isprime(n*p+n+1), return(p))); if(n%3==1, return(0)); forprime(p=7, , if(isprime(n*pn1) && isprime(n*p+n+1), return(p))) \\ Charles R Greathouse IV, May 07 2013


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



