A056911 Odd squarefree numbers.

1, 3, 5, 7, 11, 13, 15, 17, 19, 21, 23, 29, 31, 33, 35, 37, 39, 41, 43, 47, 51, 53, 55, 57, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 93, 95, 97, 101, 103, 105, 107, 109, 111, 113, 115, 119, 123, 127, 129, 131, 133, 137, 139, 141, 143, 145, 149, 151

Offset

1,2

Comments

From Daniel Forgues, May 27 2009: (Start)

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).

E.g. there are as many even squarefree numbers as there are odd squarefree numbers.

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.

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.

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)

Links

Zak Seidov, Table of n, a(n) for n = 1..12000

G. J. O. Jameson, Even and odd square-free numbers, Math. Gazette 94 (2010), 123-127; Author's copy.

Formula

A123314(A100112(a(n))) > 0. - Reinhard Zumkeller, Sep 25 2006

a(n) = n * (3/2) * zeta(2) + O(n^(1/2)) = n * (Pi^2 / 4) + O(n^(1/2)). - Daniel Forgues, May 27 2009

A008474(a(n)) * A000035(a(n)) = 1. - Reinhard Zumkeller, Aug 27 2011

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]

Example

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.

Mathematica

Select[Range[1, 151, 2], SquareFreeQ] (* Ant King, Mar 17 2013 *)

Prog

(Magma) [n: n in [1..151 by 2] | IsSquarefree(n)]; // Bruno Berselli, Mar 03 2011
(Haskell)
a056911 n = a056911_list !! (n-1)
a056911_list = filter ((== 1) . a008966) [1, 3..]
-- Reinhard Zumkeller, Aug 27 2011
(PARI) is(n)=n%2 && issquarefree(n) \\ Charles R Greathouse IV, Mar 26 2013

Crossrefs

Subsequence of A005117 and A036537.

A039956/2.

Cf. A238711 (subsequence).

Sequence in context: A334420 A342144 A357014 * A152955 A235866 A334141

Adjacent sequences: A056908 A056909 A056910 * A056912 A056913 A056914

Keyword

easy,nonn

Author

James A. Sellers, Jul 07 2000

Status

approved