

A237367


Number of ordered ways to write n = k + m with k > 0 and m > 0 such that 2*k  1, prime(k)^2  2 and prime(m)^2  2 are all prime.


3



0, 0, 1, 2, 3, 3, 3, 3, 2, 3, 2, 4, 3, 5, 2, 6, 3, 6, 2, 4, 3, 4, 2, 4, 3, 4, 4, 4, 3, 8, 3, 4, 5, 6, 6, 5, 6, 5, 5, 3, 4, 7, 5, 6, 3, 7, 3, 3, 5, 4, 5, 6, 5, 8, 10, 4, 5, 11, 6, 3, 6, 5, 5, 5, 6, 5, 8, 4, 3, 5, 6, 5, 1, 7, 6, 3, 3, 5, 6, 4
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OFFSET

1,4


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 73, 81, 534.
(ii) Any integer n > 2 can be written as k + m with k > 0 and m > 0 such that 2*k  1, prime(k) + k*(k1) and prime(m) + m*(m1) are all prime.
(iii) Every n = 9, 10, ... can be written as k + m with k > 0 and m > 0 such that 6*k  1, prime(k) + 2*k and prime(m) + 2*m are all prime.
Clearly, part (i) of this conjecture implies that there are infinitely many primes p with p^2  2 also prime. Similar comments apply to parts (ii) and (iii).


LINKS



EXAMPLE

a(3) = 1 since 3 = 2 + 1 with 2*2  1 = 3, prime(2)^2  2 = 3^2  2 = 7 and prime(1)^2  2 = 2^2  2 = 2 all prime.
a(73) = 1 since 73 = 55 + 18 with 2*55  1 = 109, prime(55)^2  2 = 257^2  2 = 66047 and prime(18)^2  2 = 61^2  2 = 3719 all prime.
a(81) = 1 since 81 = 34 + 47 with 2*34  1 = 67, prime(34)^2  2 = 139^2  2 = 19319 and prime(47)^2  2 = 211^2  2 = 44519 all prime.
a(534) = 1 since 534 = 100 + 434 with 2*100  1 = 199, prime(100)^2  2 = 541^2  2 = 292679 and prime(434)^2  2 = 3023^2  2 = 9138527 all prime.


MATHEMATICA

pq[k_]:=PrimeQ[Prime[k]^22]
a[n_]:=Sum[If[PrimeQ[2k1]&&pq[k]&&pq[nk], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 80}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



