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A235187
Number of ordered ways to write 2*n = p + q with p, q and prime(p) + q - 1 all prime.
3
0, 0, 1, 1, 2, 1, 1, 2, 2, 3, 2, 3, 2, 1, 2, 2, 2, 3, 1, 2, 5, 5, 2, 5, 3, 2, 5, 2, 1, 6, 2, 4, 4, 1, 5, 3, 4, 3, 6, 6, 3, 5, 5, 2, 9, 3, 3, 7, 2, 4, 7, 6, 3, 6, 7, 5, 4, 4, 4, 12, 3, 2, 5, 3, 3, 9, 3, 1, 7, 4, 2, 8, 6, 3, 8, 3, 4, 7, 6, 3, 10, 3, 3, 10, 8, 3, 11, 5, 3, 10, 6, 1, 9, 8, 2, 7, 4, 3, 9, 4
OFFSET
1,5
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 2.
(ii) Any integer n > 4 can be written as p + q with q > 0 such that p and p - 1 + prime(q) are both prime.
(iii) Each integer n > 7 can be written as p + q with q > 0 such that prime(p) + sigma(q) is prime, where sigma(q) denotes the sum of all positive divisors of q.
Clearly, part (i) is stronger than Goldbach's conjecture.
EXAMPLE
a(7) = 1 since 2*7 = 7 + 7 with 7 and prime(7) + 7 - 1 = 17 + 6 = 23 both prime.
a(14) = 1 since 2*14 = 11 + 17 with 11, 17 and prime(11) + 16 = 47 all prime.
a(92) = 1 since 2*92 = 47 + 137 with 47, 137 and prime(47) + 136 = 347 all prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[2n-Prime[k]]&&PrimeQ[Prime[Prime[k]]+2n-Prime[k]-1], 1, 0], {k, 1, PrimePi[2n-1]}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 04 2014
STATUS
approved