1,2

Contribution from Daniel Forgues (squid(AT)zensearch.com), 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.

a(n) = n * (3/2) * zeta(2) + O(n^(1/2)) = n * (3/2) * (pi^2 / 6) + O(n^(1/2))

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)

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

A123314(A100112(a(n))) > 0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 25 2006

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.

A039956/2. Cf. A005117.

Sequence in context: A070087 A100933 A088828 this_sequence A152955 A155113 A103796

Adjacent sequences: A056908 A056909 A056910 this_sequence A056912 A056913 A056914

easy,nonn

James A. Sellers (sellersj(AT)math.psu.edu), Jul 07 2000