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 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 (list; graph; refs; listen; history; text; internal format)
 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*(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). LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 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]^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}] CROSSREFS Cf. A000040, A049002, A062326, A076297, A218829, A230502, A233204, A237348. Sequence in context: A135715 A089326 A359514 * A022923 A276858 A247656 Adjacent sequences: A237364 A237365 A237366 * A237368 A237369 A237370 KEYWORD nonn AUTHOR Zhi-Wei Sun, Feb 07 2014 STATUS approved

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