%I #10 Mar 03 2014 11:27:53
%S 0,0,1,1,2,1,3,3,1,3,4,0,4,2,1,3,5,0,4,4,1,4,5,0,3,4,0,3,6,0,5,4,1,6,
%T 6,0,7,4,1,5,4,0,7,6,0,8,5,0,8,7,1,6,7,0,9,9,1,9,8,0,6,7,0,7,12,0,9,7,
%U 1,11,10,0,6,8,0,7,9,0,7,12
%N a(n) = |{0 < k < n: floor(k*n/3) is prime}|.
%C Conjecture: If n > m > 0 with n not divisible by m, then floor(k*n/m) is prime for some 0 < k < n.
%H Zhi-Wei Sun, <a href="/A238703/b238703.txt">Table of n, a(n) for n = 1..6000</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(4) = 1 since floor(2*4/3) = 2 is prime.
%e If p is a prime, then a(3*p) = 1 since floor(k*3p/3) = k*p is prime only for k = 1. If m > 1 is composite, then a(3*m) = 0 since floor(k*3m/3) = k*m is composite for all k > 0.
%t p[n_,k_]:=PrimeQ[Floor[k*n/3]]
%t a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,80}]
%Y Cf. A000040, A237578, A237615, A238645, A238701.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Mar 03 2014