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A182662
Number of ordered ways to write n = p + q with q > 0 such that p, 3*(p + prime(q)) - 1 and 3*(p + prime(q)) + 1 are all prime.
5
0, 0, 1, 0, 1, 0, 1, 2, 1, 1, 1, 0, 3, 2, 1, 1, 4, 3, 1, 1, 3, 3, 2, 3, 3, 1, 2, 3, 4, 2, 1, 6, 4, 4, 1, 4, 2, 1, 5, 4, 2, 1, 2, 4, 2, 2, 3, 3, 3, 4, 2, 3, 3, 2, 3, 1, 5, 2, 3, 1, 5, 6, 4, 5, 3, 3, 1, 4, 3, 2, 3, 5, 3, 3, 7, 4, 3, 1, 4, 5, 4, 3, 2, 4, 2, 5, 5, 4, 2, 2, 6, 8, 2, 2, 4, 2, 6, 1, 3, 2
OFFSET
1,8
COMMENTS
Conjecture: a(n) > 0 if n is not a divisor of 12.
Clearly, this implies the twin prime conjecture.
LINKS
EXAMPLE
a(11) = 1 since 11 = 7 + 4 with 7, 3*(7 + prime(4)) - 1 = 3*14 - 1 = 41 and 3*(7 + prime(4)) + 1 = 3*14 + 1 = 43 all prime.
a(210) = 1 since 210 = 97 + 113 with 97, 3*(97 + prime(113)) - 1 = 3*(97 + 617) - 1 = 2141 and 3*(97 + prime(113)) + 1 = 3*(97 + 617) + 1 = 2143 all prime.
MATHEMATICA
p[n_, m_]:=PrimeQ[3(m+Prime[n-m])-1]&&PrimeQ[3(m+Prime[n-m])+1]
a[n_]:=Sum[If[p[n, Prime[k]], 1, 0], {k, 1, PrimePi[n-1]}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 31 2014
STATUS
approved