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%I #15 Apr 06 2014 22:30:42
%S 0,1,2,1,2,3,3,1,5,2,4,4,8,3,7,4,4,4,2,3,7,3,10,4,12,7,7,15,7,9,8,5,8,
%T 9,11,8,8,10,8,4,10,10,10,11,7,10,8,11,8,8,9,9,8,11,7,8,13,10,8,14,13,
%U 4,14,8,11,12,14,12,8,10,16,12,16,12,14,19,11,14,8,9
%N a(n) = |{0 < k < n: the two intervals (k*n, (k+1)*n) and ((k+1)*n, (k+2)*n) contain the same number of primes}|.
%C Conjecture: (i) a(n) > 0 for all n > 1. Moreover, if n > 1 is not equal to 8, then there is a positive integer k < n with 2*k + 1 prime such that the two intervals ((k-1)*n, k*n) and (k*n, (k+1)*n) contain the same number of primes.
%C (ii) For any integer n > 4, there is a positive integer k < prime(n) such that all the three intervals (k*n, (k+1)*n), ((k+1)*n, (k+2)*n), ((k+2)*n, (k+3)*n) contain the same number of primes, i.e., pi(k*n), pi((k+1)*n), pi((k+2)*n), pi((k+3)*n) form a 4-term arithmetic progression.
%H Zhi-Wei Sun, <a href="/A238281/b238281.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 each of the two intervals (7*8, 8*8) and (8*8, 9*8) contains exactly two primes.
%t d[k_,n_]:=PrimePi[(k+1)*n]-PrimePi[k*n]
%t a[n_]:=Sum[If[d[k,n]==d[k+1,n],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,80}]
%Y Cf. A000040, A000720, A237578, A238224, A238277, A238278.
%K nonn
%O 1,3
%A _Zhi-Wei Sun_, Feb 22 2014
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