OFFSET
0,3
COMMENTS
If a(n) > 0, then the triple {6n-2, 6n, 6n+2} of consecutive even numbers allows a "simultaneous Goldbach decomposition" using only 4 different primes, 6n-2 = p-2 + 6n-p ; 6n = p + 6n-p ; 6n+2 = p + 6n+2-p = p-2 + 6n+4-p.
See A266952 for the version which does not allow the second decomposition of the last member. See A266948 for a variant which does not require 6n+2-p to be prime.
Up to 10^5, the only indices for which a(n)=0 are {0, 1, 16, 86, 131, 151, 186, 191, 211, 226, 541, 701}. (Only 2 and 67 require the alternative primality of 6n+4-p and have thus A266952(n)=0.) I conjecture that this list is finite, and probably complete. Is it a coincidence that all odd numbers in this list are primes?
PROG
(PARI) A266953(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)||isprime(6*n+4-p))&&return(p))
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jan 06 2016
STATUS
approved