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a(n) = |{0 < k < n: pi(n + k^2) is prime}|, where pi(.) is given by A000720.
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%I #14 Apr 06 2014 22:18:22

%S 0,1,1,1,1,1,2,3,2,3,4,1,2,2,3,6,6,5,5,5,5,6,7,7,6,5,6,5,6,7,8,9,8,10,

%T 9,8,6,6,6,6,7,9,9,10,11,11,13,11,9,9,10,10,8,6,6,5,4,8,9,10,12,11,14,

%U 15,15,15,12,14,15,17,16,13,11,11,13,16,18,24,25,20

%N a(n) = |{0 < k < n: pi(n + k^2) is prime}|, where pi(.) is given by A000720.

%C Conjecture: (i) For each a = 2, 3, ... there is a positive integer N(a) such that for any integer n > N(a) there is a positive integer k < n with pi(n + k^a) prime. In particular, we may take (N(2), N(3), ..., N(9)) = (1, 1, 9, 26, 8, 9, 18, 1).

%C (ii) If n > 6, then pi(n^2 + k^2) is prime for some 0 < k < n. If n > 27, then pi(n^3 + k^3) is prime for some 0 < k < n. In general, for each a = 2, 3, ..., if n is sufficiently large then pi(n^a + k^a) is prime for some 0 < k < n.

%C For any integer n > 1, it is easy to show that pi(n + k) is prime for some 0 < k < n.

%H Zhi-Wei Sun, <a href="/A237582/b237582.txt">Table of n, a(n) for n = 1..2000</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(5) = 1 since pi(5 + 1^2) = 3 is prime.

%e a(6) = 1 since pi(6 + 5^2) = pi(31) = 11 is prime.

%e a(9) = 2 since pi(9 + 3^2) = pi(18) = 7 and pi(9 + 5^2) = pi(34) = 11 are both prime.

%e a(12) = 1 since pi(12 + 10^2) = pi(112) = 29 is prime.

%t p[n_]:=PrimeQ[PrimePi[n]]

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

%t Table[a[n],{n,1,80}]

%Y Cf. A000040, A000720, A185636, A237453, A237496, A237497, A237578.

%K nonn

%O 1,7

%A _Zhi-Wei Sun_, Feb 09 2014