

A266952


Least prime p such that p2 and 6np and 6n+2p are also prime, or 0 if no such prime exists.


4



0, 0, 7, 7, 7, 13, 7, 13, 7, 13, 19, 7, 13, 7, 13, 19, 0, 31, 7, 7, 13, 19, 31, 31, 7, 13, 7, 13, 19, 73, 31, 7, 13, 7, 7, 13, 19, 31, 31, 7, 13, 7, 13, 19, 73, 31, 7, 13, 7, 13, 19, 109, 31, 7, 13, 19, 109, 31, 109, 7, 13, 19, 61, 31, 73, 43, 199, 0, 61, 103, 73, 7, 13, 7, 13, 19, 109, 31, 7, 13, 19, 139, 31, 151, 43, 199, 0, 61, 7, 13, 19, 199, 31, 139, 43
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OFFSET

0,3


COMMENTS

If a(n) > 0, then the triple {6n2, 6n, 6n+2} of consecutive even numbers allows a "simultaneous Goldbach decomposition" using two pairs of twin primes, 6n2 = p2 + 6np ; 6n = p + 6np ; 6n+2 = p + 6n+2p.
Up to 10^5, the only indices for which a(n)=0 are {0, 1, 16, 67, 86, 131, 151, 186, 191, 211, 226, 541, 701}. I conjecture that this list is finite, and probably complete. Is it a coincidence that all odd numbers > 1 in this list are primes? (See also A144094.)
This seems equivalent to a conjecture Zwillinger made in 1978, see reference in LINKS.
See A266953 for another variant with a slightly relaxed condition (instead of 6n+2p one can also have 6n+4p prime, but this affects only n=2 and n=67), and A266948 for another variant with less restrictive conditions (only p2 and 6np have to be prime).


LINKS



PROG

(PARI) A266952(n)=my(GP(n, p=2)=forprime(p=p, n+1, isprime(n*2p)&&return(p))); for(p=1, 3*n, isprime(2+p=GP(3*n, p))+!p&&(!pisprime(6*n+2p))&&return(p))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



