What strange pattern emerges from the Pell Numbers?

[tejolson]

Is this of interest to the mathematics community?

((sqrt(2*x^2-1))^4+(sqrt(2*x^2-1))^2*2+1)/(x^4*4)=1

((sqrt(2*x^2-1))^6+(sqrt(2*x^2-1))^4*3+(sqrt(2*x^2-1))^2*3+1)/(x^6*8)=1

This is just a warped Pell Number. But it turns out it is true for all numbers, not just Pell Numbers. I don't think it's supposed to prove anything worthwhile.

I also have a Pell Number of 2,373,210 digits. Which means I have the 2,373,210th perfect square that is also a triangular number. It also means I have the 2,373,210th leg/leg primitive Pythagorean triangle. And it means I have 2 triangular numbers that square to make another Primitive Triangle. I don't know how many decimals this square root of two will go to but I'm sure the other methods are better than this one. I guess I can keep going and stop it at a number that is relatively prime, but I doubt that would be of any use. I'm sure there is more that I don't know about.

 

[DonAntonio]

Is this of interest to the mathematics community?

((sqrt(2*x^2-1))^4+(sqrt(2*x^2-1))^2*2+1)/(x^4*4)=1



You seem to have written $\frac{\left(\sqrt{2x^2-1}\right)^4+\left(\sqrt{2x^2-1}\right)^2+1}{4x^4}=1$ , which of course is true but pretty trivial since the numerators is just


a perfect square by Newton's binomial theorem:$\left(\sqrt{2x^2-1}\right)^4+\left(\sqrt{2x^2-1}\right)^2+1=(2x^2-1+1)^2=(2x^2)^2=4x^4$ ...


((sqrt(2*x^2-1))^6+(sqrt(2*x^2-1))^4*3+(sqrt(2*x^2-1))^2*3+1)/(x^6*8)=1



Just as above, but this time a perfect cube: $\left(\sqrt{2x^2-1}\right)^6+3\left(\sqrt{2x^2-1}\right)^4+3\left(\sqrt{2x^2-1}\right)^2+1=(2x^2-1+1)^3=(2x^2)^2=8x^6$


This is just a warped Pell Number. But it turns out it is true for all numbers, not just Pell Numbers. I don't think it's supposed to prove anything worthwhile.


Well, if you reached the above by yourself then it is very nice, and of course it proves some really interesting relations between coefficients

of binomials squared, cubed, etc., but it's been with us for centuries.



I also have a Pell Number of 2,373,210 digits. Which means I have the 2,373,210th perfect square that is also a triangular number. It also means I have the 2,373,210th leg/leg primitive Pythagorean triangle. And it means I have 2 triangular numbers that square to make another Primitive Triangle. I don't know how many decimals this square root of two will go to but I'm sure the other methods are better than this one. I guess I can keep going and stop it at a number that is relatively prime, but I doubt that would be of any use. I'm sure there is more that I don't know about.

If you're interested in this stuff you may want to try to study Analytic Number Theory, though it'd be a good idea to have first some

background in Analysis (preferably also complex but at least real) and abstract Algebra, so that you'd enjoy and make the most of it.

DonAntonio

 

[tejolson]

Actually, the binomial thing had occurred to me. I don't know the details about binomial expansion but I thought it involved triangular numbers. Now that I'm looking at it more closely, I'm seeing it all. I have a bag full of tricks. I would appreciate it if you all dismantled them as well.

Can I reuse this thread?

 

[tejolson]

I guess I'll get told if it's not allowed. So my current mission is a set of leg/hyp twin pythagorean triangles that hit 0 degrees without the use of angles (no pi). I'm using a 3 4 5 triangle right now because it is the easiest.

A really good candidate is the 177th triangle. After simplification the mathematics ends up being 59/17. There are 17 cycles. This is using Microsoft Expression, a drawing program. I would like to prove with 100% accuracy that there is a consistent 0 for all pythagorean triangles. The radius is 4.5 (4+5)/2. The length of each for the perimeter is 3.
(3*177)/(4+5)17
(3*3*59)/(9)17
59/17

What is the purpose in this? I want to make a really big leg/hyp twin pathagorean triangle and use it to measure pi. I would like to point out that this is not a perfect method for measuring pi because this is still a perimeter method that has straight sides.

I guess this next part is a little bit out there but I couldn't help but notice the 3' and 5' on DNA. Maybe there is no 0 degree. Maybe it is just very close, close enough for DNA to use and close enough to confuse us when we try and assign numbers to it.

 

[tejolson]

So I did a goolge search for "pell number." Then I seen this on the sixth search. I thought, what kind of idiot would call their thread "Pell Number Craziness." Sure enough it was me.

What do you think of this? If you split the Pell Number into odd and even sets and remove the even you'll notice something with the last number. It starts out as 1, 5, 9, then after it has a pattern. 1, 5, 9, 9, 5, 1, 1, 5, 9, 9, 5, 1, 1, 5, 9, 9... Anyway, the numbers keep doing this no matter what. It doesn't change when they cycle back.
6900001st ...00001
6900003rd ...00005
6900005th ...00029
6900007th ...00169
6900009th ...00985
I don't know what the 6899999th Pell Number is. It would take about 2 hours to find out. I have my computers tied up right now. But I'm willing to bet it has a 1 (judging from 1345, 2760, and others). That being said, the -1th Pell Number would also be a 1. I also noticed there are no threes. For an odd type of equation, I find this unusual.

I also found a way to turn these Pell Numbers into a cool integer triangle. Look at the file attachement.
3^2+4^2=5^2
(3*3)^2+(4*3)^2=(5*3)^2
9^2+12^2=15^2
5^2+12^2=13^2

20^2+21^2=29^2
(20*20)^2+(21*20)^2=(29*20)^2
400^2+420^2=580^2
29^2+420^2=421^2