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.

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

Offset: 1

Author

Zhi-Wei Sun, Feb 07 2014

Keywords

• nonn

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*(k-1) and prime(m) + m*(m-1) 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).

Examples

			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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A049002, A062326, A076297, A218829, A230502, A233204, A237348.

Programs

Mathematica

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