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a(n) = |{0 < k < n: k^2 + k - 1 and pi(k*n) are both prime}|, where pi(.) is given by A000720.
5

%I #10 Apr 17 2016 09:25:40

%S 0,0,1,1,0,2,2,1,2,1,3,2,1,4,1,3,4,4,2,4,3,6,2,2,2,3,7,4,3,4,5,6,1,3,

%T 2,3,9,3,3,4,7,5,8,5,2,2,5,5,4,5,6,4,5,6,10,6,6,10,9,9,10,12,2,8,7,3,

%U 6,6,4,6

%N a(n) = |{0 < k < n: k^2 + k - 1 and pi(k*n) are both prime}|, where pi(.) is given by A000720.

%C Conjecture: (i) a(n) > 0 for all n > 5.

%C (ii) For each n = 4, 5, ..., there is a positive integer k < n with k^2 + k - 1 and pi(k*n) + 1 both prime. Also, for any integer n > 6, there is a positive integer k < n with k^2 + k - 1 and pi(k*n) - 1 both prime.

%C (iii) For every integer n > 15, there is a positive integer k < n such that pi(k) - 1 and pi(k*n) are both prime.

%C Note that part (i) is a refinement of the first assertion in the comments in A237578.

%H Zhi-Wei Sun, <a href="/A237615/b237615.txt">Table of n, a(n) for n = 1..5000</a>

%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;9ffe75c.1402">A combinatorial conjecture on primes</a>, a message to Number Theory List, Feb. 9, 2014.

%e a(8) = 1 since 4^2 + 4 - 1 = 19 and pi(4*8) = 11 are both prime.

%e a(33) = 1 since 28^2 + 28 - 1 = 811 and pi(28*33) = 157 are both prime.

%t p[k_,n_]:=PrimeQ[k^2+k-1]&&PrimeQ[PrimePi[k*n]]

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

%t Table[a[n],{n,1,70}]

%Y Cf. A000040, A000720, A002327, A045546, A237578, A237597, A237598.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Feb 10 2014