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%I #66 Sep 26 2023 02:09:46
%S 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 59,61,65,67,69,71,73,77,79,83,85,87,89,91,93,95,97,101,103,105,107,
%U 109,111,113,115,119,123,127,129,131,133,137,139,141,143,145,149,151
%N Odd squarefree numbers.
%C From _Daniel Forgues_, May 27 2009: (Start)
%C 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 E.g. there are as many even squarefree numbers as there are odd squarefree numbers.
%C 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 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 For any prime p_i, the n-th squarefree number odd to p_i (not divisible by p_i) is: 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 Zak Seidov, <a href="/A056911/b056911.txt">Table of n, a(n) for n = 1..12000</a>
%H G. J. O. Jameson, <a href="https://www.jstor.org/stable/27821897">Even and odd square-free numbers</a>, Math. Gazette 94 (2010), 123-127; <a href="http://www.maths.lancs.ac.uk/~jameson/sqf.pdf">Author's copy</a>.
%F A123314(A100112(a(n))) > 0. - _Reinhard Zumkeller_, Sep 25 2006
%F a(n) = n * (3/2) * zeta(2) + O(n^(1/2)) = n * (Pi^2 / 4) + O(n^(1/2)). - _Daniel Forgues_, May 27 2009
%F A008474(a(n)) * A000035(a(n)) = 1. - _Reinhard Zumkeller_, Aug 27 2011
%F Sum_{n>=1} 1/a(n)^s = ((2^s)* zeta(s))/((1+2^s)*zeta(2*s)). - _Enrique Pérez Herrero_, Sep 15 2012 [corrected by _Amiram Eldar_, Sep 26 2023]
%e 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.
%t Select[Range[1,151,2],SquareFreeQ] (* _Ant King_, Mar 17 2013 *)
%o (Magma) [n: n in [1..151 by 2] | IsSquarefree(n)]; // _Bruno Berselli_, Mar 03 2011
%o (Haskell)
%o a056911 n = a056911_list !! (n-1)
%o a056911_list = filter ((== 1) . a008966) [1,3..]
%o -- _Reinhard Zumkeller_, Aug 27 2011
%o (PARI) is(n)=n%2 && issquarefree(n) \\ _Charles R Greathouse IV_, Mar 26 2013
%Y Subsequence of A005117 and A036537.
%Y A039956/2.
%Y Cf. A238711 (subsequence).
%K easy,nonn
%O 1,2
%A _James A. Sellers_, Jul 07 2000
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