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A266952 Least prime p such that p-2 and 6n-p and 6n+2-p are also prime, or 0 if no such prime exists.

%I A266952 #19 Jan 09 2016 14:40:56
%S A266952 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,
%T A266952 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,
%U A266952 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
%N A266952 Least prime p such that p-2 and 6n-p and 6n+2-p are also prime, or 0 if no such prime exists.
%C A266952 If a(n) > 0, then the triple {6n-2, 6n, 6n+2} of consecutive even numbers allows a "simultaneous Goldbach decomposition" using two pairs of twin primes, 6n-2 = p-2 + 6n-p ; 6n = p + 6n-p ; 6n+2 = p + 6n+2-p.
%C A266952 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.)
%C A266952 This seems equivalent to a conjecture Zwillinger made in 1978, see reference in LINKS.
%C A266952 See A266953 for another variant with a slightly relaxed condition (instead of 6n+2-p one can also have 6n+4-p prime, but this affects only n=2 and n=67), and A266948 for another variant with less restrictive conditions (only p-2 and 6n-p have to be prime).
%H A266952 Harvey Dubner, <a href="/A007534/a007534.pdf">Twin Prime Conjectures</a>, Journal of Recreational Mathematics, Vol. 30 (3), 1999-2000.
%H A266952 Dan Zwillinger, <a href="http://dx.doi.org/10.1090/S0025-5718-1979-0528060-5">A Goldbach Conjecture Using Twin Primes</a>, Math. Comp. 33, No.147 (1979), p.1071.
%o A266952 (PARI) A266952(n)=my(GP(n, p=2)=forprime(p=p, n+1, isprime(n*2-p)&&return(p))); for(p=1, 3*n, isprime(-2+p=GP(3*n, p))+!p&&(!p||isprime(6*n+2-p))&&return(p))
%Y A266952 Cf. A007534, A266948, A266953.
%K A266952 nonn
%O A266952 0,3
%A A266952 _M. F. Hasler_, Jan 06 2016