%I A056911
%S A056911 1,3,5,7,11,13,15,17,19,21,23,29,31,33,35,37,39,41,43,47,51,53,55,57,
%T A056911 59,61,65,67,69,71,73,77,79,83,85,87,89,91,93,95,97,101,103,105,107,
%U A056911 109,111,113,115,119,123,127,129,131,133,137,139,141,143,145,149,151
%N A056911 Odd square-free numbers.
%C A056911 Contribution from Daniel Forgues (squid(AT)zensearch.com), May 27 2009: 
               (Start)
%C A056911 For any prime p_i, there are as many squarefree numbers having p_i as 
               a factor as squarefree numbers not having p_i as a factor amongst 
               all the squarefree numbers (one-to-one correspondance, both cardinality 
               aleph_0).
%C A056911 E.g. there are as many even squarefree numbers as there are odd squarefree 
               numbers.
%C A056911 For any prime p_i, the density of squarefree numbers having p_i as a 
               factor is 1/p_i of the density of squarefree numbers not having p_i 
               as a factor.
%C A056911 E.g. the density of even squarefree numbers is 1/p_i = 1/2 of the density 
               of odd squarefree numbers (which means that 1/(p_i + 1) = 1/3 of 
               the squarefree numbers are even and p_i/(p_i + 1) = 2/3 are odd) 
               and as a consequence the n_th even squarefree number is very nearly 
               p_i = 2 times the n_th odd squarefree number (which means that the 
               n_th even squarefree number is very nearly (p_i + 1) = 3 times the 
               n_th squarefree number while the n_th odd squarefree number is very 
               nearly (p_i + 1)/ p_i = 3/2 the n_th squarefree number.
%C A056911 a(n) = n * (3/2) * zeta(2) + O(n^(1/2)) = n * (3/2) * (pi^2 / 6) + O(n^(1/
               2))
%C A056911 For any prime p_i, the n_th squarefree number odd to p_i (not divisible 
               by p_i) is:
%C A056911 n * ((p_i + 1)/p_i) * zeta(2) + O(n^(1/2)) = n * (p_i + 1)/p_i) * (pi^2 
               / 6) + O(n^(1/2)) (End)
%H A056911 Zak Seidov, <a href="b056911.txt">Table of n, a(n) for n = 1..12000.</
               a>
%F A056911 A123314(A100112(a(n))) > 0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Sep 25 2006
%e A056911 The exponents in the prime factorization of 15 are all equal to 1, so 
               15 appears here. The number 75 does not appear in this sequence, 
               as it is divisible by the square number 25.
%Y A056911 A039956/2. Cf. A005117.
%Y A056911 Sequence in context: A070087 A100933 A088828 this_sequence A152955 A155113 
               A103796
%Y A056911 Adjacent sequences: A056908 A056909 A056910 this_sequence A056912 A056913 
               A056914
%K A056911 easy,nonn
%O A056911 1,2
%A A056911 James A. Sellers (sellersj(AT)math.psu.edu), Jul 07 2000