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A166737 Number of primes in (n^2*log(n)..(n+1)^2*log(n+1)] semi-open intervals, n >= 1.

%I A166737 #7 Mar 31 2012 14:01:22
%S A166737 1,3,4,4,6,6,8,8,10,11,10,13,13,14,16,14,17,20,18,21,21,22,21,24,22,
%T A166737 30,22,31,28,25,34,32,32,33,33,34,36,38,41,35,41,40,41,45,41,41,48,49,
%U A166737 48,49,48,48,48,54,56,54,51,56,56,61,62,57,60,62,63,59,65,66,64,65,77,67
%N A166737 Number of primes in (n^2*log(n)..(n+1)^2*log(n+1)] semi-open intervals, n >= 1.
%C A166737 Number of primes in (n*(n*log(n))..(n+1)*((n+1)*log(n+1))] semi-open intervals, n >= 1.
%C A166737 The semi-open intervals form a partition of the real line for x > 0, thus each prime appears in a unique interval.
%C A166737 The n-th interval length is:
%C A166737  (n+1/2)*[2*log(n+1/2)+1]
%C A166737  2*n*log(n) as n goes to infinity
%C A166737 The n-th interval prime density is:
%C A166737  1/[2*log(n+1/2)+log(log(n+1/2))]
%C A166737  1/(2*log(n)) as n goes to infinity
%C A166737 The expected number of primes for n-th interval is:
%C A166737  (n+1/2)*[2*log(n+1/2)+1]/[2*log(n+1/2)+log(log(n+1/2))]
%C A166737  n as n goes to infinity
%C A166737 The actual number of primes for n-th interval seems to be (from graph): a(n) = n + O(n^(1/2))
%C A166737 The partial sums of this sequence give:
%C A166737 pi((n+1)^2*log(n+1)) = Sum_{i=1}^n {a(i)} ~ Sum_{i=1}^n {i} = t_n = n*(n+1)/2
%H A166737 Daniel Forgues, <a href="/A166737/b166737.txt">Table of n, a(n) for n=1..141</a>
%F A166737 a(n) = pi((n+1)^2*log(n+1)) - pi(n^2*log(n)) since the intervals are semi-open properly.
%Y A166737 Cf. A166712 (for intervals containing an asymptotic average of one prime.)
%Y A166737 Cf. A014085 (for primes between successive squares.)
%Y A166737 Cf. A166332, A166363.
%Y A166737 Cf. A000720.
%K A166737 nonn
%O A166737 1,2
%A A166737 _Daniel Forgues_, Oct 21 2009
%E A166737 Corrected and edited by _Daniel Forgues_, Oct 23 2009