[seqfan] Re: Natural Density of Odd Squares ( Euler)

[Gordon Speagle]

Thanks everyone, I think this sequence https://oeis.org/A217739 and  the
comments both explain my question and provided an answer :.

format <https://oeis.org/A217739/internal>)
OFFSET

0,1
COMMENTS

This is the probability that a randomly chosen singly even number is
squarefree. (The probability that any randomly chosen integer is squarefree
is 6/Pi^2).

This number also arises in the study of the Fourier series for a triangle
wave. By Equation 6 given by Weisstein, this number is b_1, since b_n =
8/(Pi^2 n^2) for odd n. Springer labels this a_1.

This is also the probability that the greatest common divisor of two
randomly chosen positive integers will be a power of 2. Generally, the
probability that the greatest common divisor of two random integers will be
a power of p, a prime, is (6/Pi^2)/(1-1/p^2). Here we are considering the
integer 1 to be a power of p. - Geoffrey Critzer
<https://oeis.org/wiki/User:Geoffrey_Critzer>, Jan 13 2015

The probability that two randomly chosen odd numbers will be coprime
(Nymann,

On Mon, Dec 7, 2020, 21:37 Neil Sloane <njasloane at gmail.com> wrote:

> Knopp's book on Theory and Applications of Infinite Series is what you
> should read.
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Mon, Dec 7, 2020 at 10:33 PM Allan Wechsler <acwacw at gmail.com> wrote:
>
> > Gordon, perhaps you have misunderstood what it means to be a "square-free
> > integer". It's not a matter of taking all the integers and discarding the
> > squares. Rather, an integer is said to be "square-free" if it is not
> > divisible by any square greater than 1.
> >
> > On Mon, Dec 7, 2020 at 1:15 PM jean-paul allouche <
> > jean-paul.allouche at imj-prg.fr> wrote:
> >
> > > Hi
> > >
> > > I am not sure of what you mean by "1/( 2k-1)": is there a misprint?
> > > I guess you might have meant 1/(2k-1)^2?
> > > Anyway the density of squares is zero -- a fortiori the density of
> > > odd or even squares is 0 as well
> > >
> > > best wishes
> > > jean-paul
> > >
> > >
> > > Le 07/12/2020 à 17:10, Gordon Speagle a écrit :
> > > > Hello Everyone,
> > > >
> > > > This is my inaugural message to the group.  My name is Gordon
> > > >
> > > > I am not an academic mathematician, but in my investigations into
> > > numbers I
> > > > will sometime encounter something  that is interesting to me and I
> find
> > > > myself unsure of its role in the larger context of professional
> > > > mathematics.
> > > >
> > > > After revisiting Euler's solution to the Basel problem and his
> linking
> > of
> > > > pi to the distribution of square free integers, using his same logic,
> > it
> > > > can be shown that the series of 1/( 2k-1) converges to  (pi2)/ 8.
> > > >
> > > > Heuristically, it seems that this would mean that the naturally
> density
> > > of
> > > > non odd square integers is 8/pi2.
> > > >
> > > > Would the density of the set of the remaining odd square  integers be
> > > >
> > > > 1- 8/pi2 ?
> > > >
> > > > It seems that follows from an Erdos theorem on natural density.
> > > >
> > > > I am unsure if my reasoning is sound and would like to discuss among
> > > those
> > > > with similar interests. Which is what precipitated my post.
> > > >
> > > > Thanks
> > > >
> > > > Gordon
> > > >
> > > > --
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> > >
> > >
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