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a(n) = |{0 < k <= n/2: prime(k) + prime(n-k) - 1 is prime}|
7

%I #13 Apr 04 2014 18:45:58

%S 0,1,0,1,1,0,2,1,3,1,3,1,3,1,3,4,5,2,5,2,5,5,4,5,4,5,6,8,2,8,9,11,4,6,

%T 1,3,6,8,8,7,3,11,9,8,8,9,12,8,10,10,10,8,6,3,8,11,13,14,13,15,8,15,

%U 15,14,8,18,11,14,5,10,7,10,15,12,10,5,10,11,12,16,21,15,16,14,8,15,19,14,16,18,13,10,28,21,14,20,18,24,20,19

%N a(n) = |{0 < k <= n/2: prime(k) + prime(n-k) - 1 is prime}|

%C Conjecture: (i) a(n) > 0 except for n = 1, 3, 6. Also, a(n) = 1 only for n = 2, 4, 5, 8, 10, 12, 14, 35.

%C (ii) For each integer n > 7, there is a positive integer k < n/2 with (prime(n-k) - prime(k))/2 prime. Also, for any positive integer n not among 1, 3, 5, 9, 21, (prime(k) + prime(n-k))/2 is prime for some 0 < k < n.

%C (iii) For any integer n > 6, prime(k)^2 + prime(n-k)^2 - 1 is prime for some 0 < k < n. Also, for any integer n > 4 not equal to 14, (prime(k)^2 + prime(n-k)^2)/2 is prime for some 0 < k < n.

%C (iv) For any integer n > 3, (prime(k) - 1)^2 + prime(n-k)^2 is prime for some 0 < k < n. Also, if n > 4 then (prime(k) + 1)^2 + prime(n-k)^2 is prime for some 0 < k < n.

%H Zhi-Wei Sun, <a href="/A232465/b232465.txt">Table of n, a(n) for n = 1..5000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014

%e a(8) = 1 since prime(4) + prime(4) - 1 = 13 is prime.

%e a(10) = 1 since prime(4) + prime(6) - 1 = 7 + 13 - 1 = 19 is prime.

%e a(14) = 1 since prime(6) + prime(8) - 1 = 13 + 19 - 1 = 31 is prime.

%e a(35) = 1 since prime(2) + prime(33) - 1 = 3 + 137 - 1 = 139 is prime.

%t a[n_]:=Sum[If[PrimeQ[Prime[k]+Prime[n-k]-1],1,0],{k,1,n/2}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000040, A232443, A232463.

%K nonn

%O 1,7

%A _Zhi-Wei Sun_, Nov 24 2013