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%I #13 Mar 01 2014 09:52:43
%S 0,1,1,0,3,0,4,1,2,3,1,3,4,4,4,4,1,3,6,5,3,3,4,6,3,8,5,6,3,4,2,10,6,5,
%T 7,8,6,8,7,5,7,5,11,7,7,8,8,11,6,5,7,11,11,7,4,9,7,3,5,7,7,11,8,13,9,
%U 8,7,7,12,10,8,11,8,15,8,9,9,15,13,4
%N a(n) = |{0 < k <= n: prime(k*n) + 2 is prime}|.
%C Conjecture: (i) a(n) > 0 for all n > 6, and a(n) = 1 only for n = 2, 3, 8, 11, 17. Moreover, for any n > 0 there exists a positive integer k < 3*sqrt(n) + 6 such that prime(k*n) + 2 is prime.
%C (ii) For any integer n > 6, prime(k^2*n) + 2 is prime for some k = 1, ..., n.
%C (iii) If n > 5, then prime(k^2*(n-k)) + 2 is prime for some 0 < k < n.
%C Clearly, each of the three parts implies the twin prime conjecture.
%C We have verified part (i) of the conjecture for n up to 2*10^6.
%H Zhi-Wei Sun, <a href="/A238573/b238573.txt">Table of n, a(n) for n = 1..5000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e a(2) = 1 since prime(1*2) + 2 = 3 + 2 = 5 is prime.
%e a(3) = 1 since prime(1*3) + 2 = 5 + 2 = 7 is prime.
%e a(8) = 1 since prime(8*8) + 2 = 311 + 2 = 313 is prime.
%e a(11) = 1 since prime(3*11) + 2 = 137 + 2 = 139 is prime.
%e a(17) = 1 since prime(1*17) + 2 = 59 + 2 = 61 is prime.
%t p[k_,n_]:=PrimeQ[Prime[k*n]+2]
%t a[n_]:=Sum[If[p[k,n],1,0],{k,1,n}]
%t Table[a[n],{n,1,80}]
%Y Cf. A000040, A001359, A006512, A218829, A236531, A237578, A238576.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Mar 01 2014
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