%I #19 Sep 08 2022 08:45:45
%S 1,1,2,3,1,4,3,4,3,5,4,1,5,2,6,7,5,1,6,1,1,7,2,9,8,7,8,9,1,4,10,9,10,
%T 9,10,1,3,12,11,12,11,3,12,11,13,10,13,3,10,11,15,4,12,13,11,12,17,13,
%U 1,16,13,17,15,7,16,1,15,17,13,7,1,15,1,17,9,11,7,18,23,13,20,19,20,17,16
%N Numerator of (b(n+1) - b(n))/(b(n+2) - b(n)), where b(n) = A038107(n) is the number of primes up to n^2.
%C If the limit of R(n) exists as n->oo it is 1/2, but existence of the limit is conjectural. R(n) generalizes to R_k(n) by substituting PrimePi_k for PrimePi(n), where PrimePi_k(n) is the number of numbers with k prime factors (including repetitions) <= n. Convergence of {R(n)} to 1/2 implies Legendre's conjecture. For discussion of the order of the number of prime factors of a number n see reference [1], below. The PNT and reference [1] suggest but offer no proof that R_k(n)-> 1/2 as n -> oo. The corresponding sequence for near-primes would be {R_2(n)} = {1/3, 2/3, 1/2, ...}.
%D S. Ramanujan, The Normal Number of Prime Factors of a Number n, reprinted at Chapter 35, Collected Papers (Hardy et al., ed), AMS Chelsea Publishing, 2000.
%H Harvey P. Dale, <a href="/A161621/b161621.txt">Table of n, a(n) for n = 1..1000</a>
%e R(3) = (PrimePi(4^2)-PrimePi(3^2)) / (PrimePi(5^2)-PrimePi(3^2)) = (PrimePi(16)-PrimePi(9)) / (PrimePi(25)-PrimePi(9)) = (6-4)/(9-4) = 2/5. Hence a(3) = 2. - _Klaus Brockhaus_, Jun 15 2009
%t Numerator[(#[[2]]-#[[1]])/(#[[3]]-#[[1]])&/@Partition[PrimePi[ Range[ 90]^2],3,1]] (* _Harvey P. Dale_, Jan 06 2017 *)
%o (Magma) [ Numerator((#PrimesUpTo((n+1)^2)-a) / (#PrimesUpTo((n+2)^2)-a)) where a is #PrimesUpTo(n^2): n in [1..85] ]; // _Klaus Brockhaus_, Jun 15 2009
%Y Cf. A014085
%Y Cf. A161622 (denominators). - _Klaus Brockhaus_, Jun 15 2009
%K nonn,frac
%O 1,3
%A _Daniel Tisdale_, Jun 14 2009
%E a(1) inserted and extended beyond a(13) by _Klaus Brockhaus_, Jun 15 2009
%E Simplified title by _John W. Nicholson_, Dec 13 2013