OFFSET
1,1
COMMENTS
Conjecture: There are infinitely many primes p(k) such that p(k)+2 and p(k+m)-2 are both primes for all m > 1.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
prime(13) = 41, and both prime(13)+2 = 43 and prime(13+5)-2 = 59 are primes, so 41 is in the sequence.
MATHEMATICA
For[n = 1, n < 500, n++, If[PrimeQ[Prime[n] + 2], If[PrimeQ[Prime[n + 5] - 2], Print[Prime[n]]]]] (* Stefan Steinerberger, Feb 07 2006 *)
PROG
(PARI) pnpk(n, m=5, k=2) = { local(x, v1, v2); for(x=1, n, v1 = prime(x)+ k; v2 = prime(x+m)-k; if(isprime(v1)&isprime(v2), print1(prime(x), ", ") ) ) ; } \\ corrected by Michel Marcus, Sep 14 2015
(PARI) lista(pmax) = {my(k = 1, p = primes(6)); forprime(p1 = p[#p], pmax, k++; p[#p] = p1; if(p[2]- p[1] == 2 && p[6] - p[5] == 2, print1(p[1], ", ")); for(i = 1, #p-1, p[i] = p[i+1])); } \\ Amiram Eldar, Oct 04 2024
(Magma) [NthPrime(n): n in [1..1500] | IsPrime(NthPrime(n)+2) and IsPrime(NthPrime(n+5)-2)]; // Vincenzo Librandi, Sep 14 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Cino Hilliard, May 02 2005
STATUS
approved