

A056911


Odd squarefree numbers.


53



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
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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 (onetoone 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 nth even squarefree number is very nearly p_i = 2 times the nth odd squarefree number (which means that the nth even squarefree number is very nearly (p_i + 1) = 3 times the nth squarefree number while the nth odd squarefree number is very nearly (p_i + 1)/ p_i = 3/2 the nth squarefree number.
For any prime p_i, the nth 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)
Sum(n>=1, a(n)/n^s) =((2^s)* zeta(s))/((1+2^s)*zeta(2*s)).  Enrique Pérez Herrero, Sep 15 2012


LINKS

Zak Seidov, Table of n, a(n) for n = 1..12000
G. J. O. Jameson, Even and odd squarefree numbers, Math. Gazette 94 (2010), 123127


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)^2 = 12/Pi^2.  Amiram Eldar, May 22 2020


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 !! (n1)
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 A036537.
A039956/2. Cf. A005117.
Cf. A238711 (subsequence).
Sequence in context: A247424 A305635 A334420 * A152955 A235866 A334141
Adjacent sequences: A056908 A056909 A056910 * A056912 A056913 A056914


KEYWORD

easy,nonn


AUTHOR

James A. Sellers, Jul 07 2000


STATUS

approved



