Quoting "turtle":

If by "our polygonal number counting function" you mean your equation(s)...

Of course that is what I mean! Let's be perfectly clear about this.

Here is the

**polygonal number counting function**:

[math] \varpi(x)\approx\left(\left(\sqrt{\left(\left(1-\frac{1}{\left(\alpha*\pi*e+e\right)}\right)*x\right)}-\frac{1}{4}\right)^{2}-\frac{1}{16}\right)*\left(1-\frac{\alpha}{\left(6*\pi^{5}-\pi^{2}\right)}\right) [/math]

where:

[math] \alpha=\left(\left(A^{-1}*\pi*e+e\right)*\left(\pi^{e}+e^{\left(\frac{-\pi}{2}\right)}\right)-\frac{\left(\left(\pi^e+e^{\frac{-\pi}{2}}+4+\frac{5}{16}\right)*\left(\ln\left(x\right)\right)^{-1}+1\right)}{\left(6*\pi^{5}*e^{2}-2*e^{2}\right)}\right)^{-1} [/math]

and here is how it compares to the

**empirical data** (observed results) which we have designated [math]\varpi(x)[/math]:

[math]x[/math]_______________________[math]\varpi(x)[/math]_________________ [math] B(x_{F\alpha}) [/math]_________Difference

10_______________________3______________________5___________________2

100______________________57_____________________60__________________3

1,000____________________622____________________628_________________6

10,000___________________6,357__________________6,364________________7

100,000__________________63,889_________________63,910_______________21

1,000,000________________639,946________________639,963______________17

10,000,000_______________6,402,325______________6,402,362_____________37

100,000,000______________64,032,121_____________64,032,273____________152

1,000,000,000____________640,349,979____________640,350,090____________111

10,000,000,000___________6,403,587,409__________6,403,587,408__________-1

100,000,000,000__________64,036,148,166_________64,036,147,620_________-546

1,000,000,000,000________640,362,343,980________640,362,340,975________-3005

10,000,000,000,000_______6,403,626,146,905______6,403,626,142,352_______-4554

100,000,000,000,000______64,036,270,046,655_____64,036,270,047,131_______476

200,000,000,000,000______128,072,542,422,652____128,072,542,422,781______129

300,000,000,000,000______192,108,815,175,881____192,108,815,178,717______2836

400,000,000,000,000______256,145,088,132,145____256,145,088,130,891_____-1254

500,000,000,000,000______320,181,361,209,667____320,181,361,208,163_____-1504

600,000,000,000,000______384,217,634,373,721____384,217,634,374,108______387

700,000,000,000,000______448,253,907,613,837____448,253,907,607,119_____-6718

800,000,000,000,000______512,290,180,895,369____512,290,180,893,137_____-2232

900,000,000,000,000______576,326,454,221,727____576,326,454,222,404______677

1,000,000,000,000,000____640,362,727,589,917____640,362,727,587,828_____-2089

In all of mathematics, this is the

*only* counting function for

**polygonal numbers of order greater than 2** or

**regular figurative numbers** as they are otherwise known, that actually

*converges* with the

**empirical data**.

Note that I didn't even bother to calculate the relative error because it is now virtually non-existant and

quickly approaching zero. Also note that this incredible convergence between the counting function and

the empirical data is

**not** possible without the

**fine structure constant** which, in turn,

involves the prime number generating

**Blazys constant** [math]A=2.5665438321713888444675291...[/math].

Quoting "turtle":

Without someone else actually counting how many polygonals in a range, your equations are meaningless.

Well, for one thing, people

*are* counting polygonals. As a matter of fact, just last August,

the great mathematician

**Lars Blomberg** counted them all the way to [math]10^{15}[/math].

**That's the world record!!! **Moreover, I am getting e-mails all the time, asking me for "code", to which I invariably respond that

I am not a programmer or "coder", and that I don't even own a computer. A lot of them "crashed" their

computers, and a lot of them gave up because their computers are too slow, but to my knowledge,

there are still

*several* ongoing attempts to break Lar's record. From this I can only assume that

counting

**order >2 polygonals** is an interesting programming challenge in and of itself, and that

efforts to penetrate the secrets of this incredibly fascinating erratic sequence will continue.

It doesn't matter much to me because the counting function is now so accurate that any further adjustments

would be minor indeed. The

**form** is definitely correct, and that, to me, is by far the most important thing.

The rest are just details.

I did my job, and delivered what I promised.

A super accurate counting function for

**polygonal numbers of order greater than 2**.

Quoting "turtle",

If by if by "our polygonal number counting function" you mean to imply you had a hand in conceiving of taking

the natural density of polygonal/nonpolygonal numbers, then again that was my conception.

The first case is just a silly claim; the second i consider theft. stop taking credit for my work Don.

"Taking the natural density of polygonal numbers" occurs the moment someone puts them in one to one correspondence

with the natural numbers. In other words, when somebody "numbers them" as follows:

1____6

2____9

3____10

4____12

5____15

6____16

7____18

8____21

9____22

10___24

then it is easy to see

**at a glance** that there are 3 order >2 polygonals less than or equal to 10.

I certainly don't want to "steal" any "credit" for that "achievment" which rightfully, does indeed belong to

both you and the ancient Greeks.

Don