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A056911 Odd squarefree numbers.

%I A056911 #75 Jul 02 2025 16:02:00
%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 squarefree numbers.
%C A056911 From _Daniel Forgues_, May 27 2009: (Start)
%C A056911 For any prime p, there are as many squarefree numbers having p as a factor as squarefree numbers not having p as a factor amongst all the squarefree numbers (one-to-one correspondence, 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, the density of squarefree numbers having p as a factor is 1/p of the density of squarefree numbers not having p as a factor.
%C A056911 E.g. the density of even squarefree numbers is 1/p = 1/2 of the density of odd squarefree numbers (which means that 1/(p + 1) = 1/3 of the squarefree numbers are even and p/(p + 1) = 2/3 are odd). As a consequence the n-th even squarefree number is very nearly p = 2 times the n-th odd squarefree number (which means that the n-th even squarefree number is very nearly (p + 1) = 3 times the n-th squarefree number while the n-th odd squarefree number is very nearly (p + 1)/p = 3/2 the n-th squarefree number).
%C A056911 For any prime p, the n-th squarefree number not divisible by p is: n * (1 + 1/p) * zeta(2) + O(n^(1/2)) = n * (1 + 1/p) * (Pi^2 / 6) + O(n^(1/2)) (End)
%H A056911 Zak Seidov, <a href="/A056911/b056911.txt">Table of n, a(n) for n = 1..12000</a>
%H A056911 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 A056911 A123314(A100112(a(n))) > 0. - _Reinhard Zumkeller_, Sep 25 2006
%F A056911 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 A056911 A008474(a(n)) * A000035(a(n)) = 1. - _Reinhard Zumkeller_, Aug 27 2011
%F A056911 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 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.
%t A056911 Select[Range[1,151,2],SquareFreeQ] (* _Ant King_, Mar 17 2013 *)
%o A056911 (Magma) [n: n in [1..151 by 2] | IsSquarefree(n)]; // _Bruno Berselli_, Mar 03 2011
%o A056911 (Haskell)
%o A056911 a056911 n = a056911_list !! (n-1)
%o A056911 a056911_list = filter ((== 1) . a008966) [1,3..]
%o A056911 -- _Reinhard Zumkeller_, Aug 27 2011
%o A056911 (PARI) is(n)=n%2 && issquarefree(n) \\ _Charles R Greathouse IV_, Mar 26 2013
%o A056911 (PARI) list(lim)=my(v=List()); forsquarefree(k=1,lim\1, if(k[1]%2, listput(v,k[1]))); Vec(v) \\ _Charles R Greathouse IV_, Jan 14 2025
%Y A056911 Subsequence of A005117 and A036537.
%Y A056911 Equals A039956/2.
%Y A056911 Cf. A238711 (subsequence).
%K A056911 easy,nonn
%O A056911 1,2
%A A056911 _James Sellers_, Jul 07 2000