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Old 2022-09-27, 22:51   #1
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Default C000885 The Athanasii Kircheri triangle filled with non-negative integers.

C000885 The Athanasii Kircheri triangle filled with non-negative integers.

The original triangle used by Athanasii Kircheri in Mundus Subterraneus can be found on page 24. In the OEIS it is the sequence 071797.

Now, when we fill this triangle with non-negative integers, we get the following figure:



The symmetry line of this triangle

The central column of this triangle is the sequence 002378 oblong numbers = 2 * 000217 the triangular numbers.

Notice that, in terms of values, the line of symmetry is to the right of the central column:

1+2=3

4+5+6=7+8

9+10+11+12=13+14+15


...and so on.

We have prime numbers on both sides of the central column.

All rows start on the left with a 000290 square number, go through an 002378 oblong number and end with a number 005563 (square minus 1).

Then, we have:

\(square<oblong<oblong+\frac{1}{4}<\left( (next \ square) \ minus \ 1 \right)<next \ square\)

Or,

\(n^{2} = square<oblong<oblong+\frac{1}{4}<\left( (next \ square) \ minus \ 1 \right)<next \ square = \left( n+1 \right)^{2}\)

Or,

\(n^{2} < n^{2}+n < n^{2}+n+\frac{1}{4} < \left( \left( n+1 \right)-1 \right)\left( \left( n+1 \right)+1 \right) < \left( n+1 \right)^{2}\)

Or,

\(n^{2} < n^{2}+n < n^{2}+n+\frac{1}{4} < \left( n+1 \right)^{2} - 1 < \left( n+1 \right)^{2}\)

Or,

\(n^{2} < n^{2}+n < n^{2}+n+\frac{1}{4} < n^{2}+2n < n^{2}+2n+1\)


Now, when we take the square root of each element type, we get:

\(n < \sqrt{n^{2}+n} < n+\frac{1}{2} < \sqrt{n^{2}+2n} < n+1\)

Or,

\(n=\sqrt{square} < \sqrt{oblong(n)} < n+\frac{1}{2} < \sqrt{(next \ square) \ minus \ 1} < \sqrt{next \ square}=n+1\)

The square root of a perfect square number always results in an integer. Every integer is a rational number in the form of (even/2).

If we imagine that every square root of a square number results in a rational number, then since the square root of any number (oblong+1/4) results in a (odd/2)= rational number, we will consider (oblong+1/4) to be an "imperfect" square.

Because of that, we can imagine the symmetry line of the C000885 The Athanasii Kircheri triangle filled with non-negative integers as being an empty column with (oblong+1/4) elements in green as per figure below:



We will call the triangle above C000893 The Athanasii Kircheri triangle filled with non-negative integers showing the symmetry line.

See what happens when we take the square root of all its elements:



We will call the triangle above C000894 The square root of the Athanasii Kircheri triangle filled with non-negative integers showing the symmetry line.

Note that only perfect and "imperfect" squares have their rational square root. All other elements have their irrational square root.

The elements to the left of the green symmetry line have the first decimal place with a digit from 0 to 4.

The elements to the right of the green symmetry line have the first decimal place with digit from 5 to 9.

See the resulting table from those triangles:



See that between the square numbers in yellow of n and m columns there is a zigzag between the two sides of the table. Also there is a zigzag with the oblong numbers in red-dark color between the two sides of the table. There is an offset between both zigzags like sin and cos.

The same occur with the (square minus 1) numbers and the (odd/2) squared numbers. This occur in a such way that \(\sqrt{m} = \sqrt{n} - \frac{1}{2}\).

The prime number distribution

Because of that we can say:

*The first vertical column to the right of the central column is the sequence 002061 central polygonal numbers \( y^{2} \pm y +1 \).

*The sequence of the (square minus 1) numbers 005563 forms the right edge of the triangle.
** The right side of this triangle are the numbers beetween the 002061 central polygonal numbers and the 005563 (square minus 1) numbers including them.
** The right side of the triangle is the sequence 063657. They are the integer numbers where the first decimal digit of its square root is 5, 6, 7, 8, or 9.
** The prime numbers of the right side of the triangle is the sequence 334163. They are the prime numbers where the first decimal digit of its square root is 5, 6, 7, 8, or 9.

*The sequence of the square numbers 000290 forms the left edge of the triangle.
** The left side of this triangle are the numbers beetween the A000290 square numbers and the 002378 oblong numbers including them.
** The left side of the triangle is the sequence 063656. They are the integer numbers where the first decimal digit of its square root is 0, 1, 2, 3, or 4.
** The prime numbers of the left side of the triangle is the sequence 307508. They are the prime numbers where the first decimal digit of its square root is 0, 1, 2, 3, or 4.

See the "square root sieve" of prime numbers table. The prime numbers from 307508 in the left and from 334163 in the right:



-----------------------------------------

To do the density and qty studies of:

Sequences 063656 and 063657.

Sequence \(\left\lfloor \left( \sqrt{A063656} - \left\lfloor \sqrt{A063656} \right\rfloor \right) *10 \right\rfloor\).

Sequence \(\left\lfloor \left( \sqrt{A063657} - \left\lfloor \sqrt{A063657} \right\rfloor \right) *10 \right\rfloor\).

Sequence \(\left\lfloor \left( \sqrt{A307508} - \left\lfloor \sqrt{A307508} \right\rfloor \right) *10 \right\rfloor\).

Sequence \(\left\lfloor \left( \sqrt{A334163} - \left\lfloor \sqrt{A334163} \right\rfloor \right) *10 \right\rfloor\).

-----------------------------------------

The sum of the rows
  1. The first vertical column to the left of the central column is 165900, the values of Fibonacci polynomial \( y^{2} \pm y -1 \). Also, 165900(y) \( = y*(y-1) \pm (y+(y-1)) \) is the sum or difference of the product of two consecutive integers and the sum of those two consecutive integers. 055112 is the product, 165900 is the sum or difference.
  2. The row sum is 055112 \( \pm 2 y^{3}-3 y^{2} \pm y \equiv y(y-1)(2y-1) = oblong*odd = y(y-1)(y+(y-1)) \) is the product of two consecutive integers multiplied by the sum of those two consecutive integers.
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Old 2022-09-28, 12:04   #2
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Default C000886 The Athanasii Kircheri triangle filled with positive integers

C000886 The Athanasii Kircheri triangle filled with positive integers.

When we fill the Athanasii Kircheri triangle with the positive integers, we get the following figure:



This triangle is also known as Klauber's triangle. He called this triangle the "triangle of primes". In the OEIS this triangle is mentioned in sequence 000027 in Kival Ngaokrajang's study.

The row sum is 005898 \(\pm 2y^{3}-3y^{2} \pm 3y-1 = (2y-1)(y^2-y+1) = \)central column * row length = (oblong+1) * odd = central polygonal numbers * odd \(= (2y^3-3y^2+y)+(2y-1) =\) the product of two consecutive numbers multiplied by the sum of the two consecutive numbers plus the odds \(= y^3+(y-1)^3 =\) the sum of two consecutive cubes.

The partial sum of 005898 is 037270 = \(\frac{\left( y^{2} \right) ^{2} + y^{2}}{2}\)

1st partial difference of 005898 is 005897 = 2 * 056107.

2nd partial difference of 005898 is 017593 = 2 * 016945 = 6 * 005408.

3rd partial difference of 005898 is 010851 = 6 * 007395.
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Old 2022-09-28, 13:08   #3
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Default C000884 The original Athanasii Kircheri triangle

C000884 The original Athanasii Kircheri triangle

In Pascal's triangle, the first row has 1 element, the second row has 2 elements, the third row has 3 elements, and so on. The number of elements in a row is equal to the row number.

Continuing with only one element at the top, the next possible triangle has 1 element in the first row, 3 elements in the second row, 5 elements in the third row, 7 elements in the fourth row, and so on. The number of elements in a row is an odd number equal to \((2*row \ number-1)\).

By sequentially numbering each cell in each row, we get the following triangle:



This last triangle has the shape shown in Kircheri, P. Athanasii. (1664) Mundus Subterraneus, pg. 24.

We will call this triangle shape the Kircheri triangle.

In OEIS, this triangle is the sequence 071797.

The sum of its y-th row is equal to \(2y^2 \pm y\), which yields the n-th hexagonal number, 000384 The square numbers plus oblong numbers.

The partial sum of the square numbers plus oblong numbers is 002412 the hexagonal pyramidal numbers, or greengrocer's numbers.

The partial differences of the square numbers plus oblong numbers is \(\pm 4y+1\) 016813.

The partial differences of \(\pm 4y+1\) is 010709 constant sequence: the all 4's sequence.
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Old 2022-09-29, 00:29   #4
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Default Kircheri triangle filled with non-negative integers and the OEIS sequence A055112

C000885 The Athanasii Kircheri triangle filled with non-negative integers is the example in the sequence 055112.

In this triangle, the central column is 002378 (oblong numbers) or 2 * 000217 (triangular numbers). - Bruce J. Nicholson, Aug 31 2017

The first vertical column to the left of the central column is 165900, Values of Fibonacci polynomial n^2±n–1. Also, 165900 = n(n–1)±(n+(n–1)) = oblong ± odd = 002378 ± 005408 is the sum or difference of the product of two consecutive integers and the sum of those two consecutive integers.

Almost similar, the sequence 055112 = n(n-1)*(n+(n-1)) = central column * row length = oblong * odd = 002378 * 005408 is the product of two consecutive integers multiplied by the sum of those two consecutive integers.

The sequence of square numbers 000290 forms the left edge of the triangle. The prime numbers of the sequence 307508 are located between the left edge 000290 and the central column 002378.

The sequence of (square minus 1) numbers 005563 forms the right edge of the triangle. The prime numbers of the sequence 334163 are located between the central column 002378 and the right edge 005563.
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Old 2022-10-01, 23:46   #5
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Default Suggested idea

I believe it is an excellent potential result if we use the same idea presented in https://www.youtube.com/watch?v=6SzZ_jAHasE



to do with all the possible triangles presented above...

The same with Pascal's triangle filled with (1) Kircheri filling; (2) non-negative integers; (3) positive integers...

Last fiddled with by Charles Kusniec on 2022-10-01 at 23:52
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Old 2022-10-27, 15:09   #6
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Default C001122 The Takumi Sato´s classification of natural numbers in Kircheri triangle.

In 2012 Takumi Sato classified the non-negative integers into 4 numerical sequences as follows:

See the distribution of these sequences in C000885 The Athanasii Kircheri triangle filled with non-negative integers:

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Old 2022-10-27, 21:20   #7
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Default The primes of the Takumi Sato´s 4 numerical sequences

The primes of the sequence https://oeis.org/A005563 is 3, only. It is the sequence https://oeis.org/A010701. (Today it is the best sequence to describe number 3).

The primes of the sequence https://oeis.org/A217571 are 5, 11, 19, 29, 41, 71, 89, 109, 131, 181, 239, 271, 379, ... . It is the sequence https://oeis.org/A002327.

The primes of the sequence https://oeis.org/A217575 are 2, 7, 13, 23, 31, 43, 47, 59, 61, 73, 79, 97, 113, 137, ... . It is the sequence https://oeis.org/A334163 with the prime number 2 in place of 3.

The primes of the sequence https://oeis.org/A217570 are 17, 37, 53, 67, 83, 101, 103, 107, 109, 127, 149, 151, ... . It is the sequence https://oeis.org/A307508 without the prime numbers of the sequence https://oeis.org/A217571 or without the prime numbers of the sequence https://oeis.org/A002327.
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Old 2022-10-27, 23:45   #8
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Quote:
Originally Posted by Charles Kusniec View Post
The primes of the sequence...
Charles... May I please reach out to you et al seriously?

Some of us spend more time than we should drilling down on ideas. Even ideas that turn out to be less than optimal. Often thrown up as a distraction by those who don't actually want us to get our jobs done.

You have thrown out thousands of words (and even more numbers) without really explaining your argument. Heuristics suggests this means you aren't serious.

Care to change our preception?

Believe it or not, we are all ears for serious (a waste of time otherwise)...
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Old 2022-10-28, 01:03   #9
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Quote:
Originally Posted by chalsall View Post
Charles... May I please reach out to you et al seriously?

Some of us spend more time than we should drilling down on ideas. Even ideas that turn out to be less than optimal. Often thrown up as a distraction by those who don't actually want us to get our jobs done.

You have thrown out thousands of words (and even more numbers) without really explaining your argument. Heuristics suggests this means you aren't serious.

Care to change our preception?

Believe it or not, we are all ears for serious (a waste of time otherwise)...
Dear Chris Halsall,

I humbly ask for all possible forgiveness for giving the impression that I am not "serious and wasting your time".

That was never my intention and if that is happening, I would certainly like to change your precept.

1. I do not participate in any other forum.

2. Everything I write I do alone, without help or revision, much less exchange of ideas.

3. I am finding it very difficult to quickly show the big picture as I have several studies in several directions that tie together. Writing everything out takes time and my productivity is low.

4. For example this last post you used to consider your fatigue of my words and numbers is directly related to the new class of numbers I wrote at https://www.mersenneforum.org/showthread.php?t=27328. At the same time, everything I wrote in "How and why Legendre's conjecture works" I have other studies extending that to prove Goldbach's conjecture and Landau's problem at the same time.

5. Many things and solutions that I write are those that I cannot find in the literature available on the Internet. Many things are new and many I don't know if they are new. One example was my post on "Divisibility by 7". Since I couldn't find the sequences in OEIS, I thought it was something new. Maybe that is why you might have a bad impression of my posts.

6. On the other hand, writing all this takes time. Once, to figure out a table, I had to generate a dozen other tables, each with over 20 columns.
... but I never posted anything that didn't make mathematical sense.

CONCLUSION: If this were a soccer game, I would say that I am just a player. The field, the ball, the judges, the fans are all yours. The decision is yours.

Please, let me know if you prefer me to continue or stop sending the ideas and studies here.

Thank you.
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Old 2022-10-28, 01:28   #10
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Quote:
Originally Posted by Charles Kusniec View Post
CONCLUSION: If this were a soccer game, I would say that I am just a player. The field, the ball, the judges, the fans are all yours. The decision is yours.

Please, let me know if you prefer me to continue or stop sending the ideas and studies here.

Thank you.
Charles... That was a well thought out answer to my questions. To share, I hated sports as a child...

I now know you are not a 'bot. Sorry, we actually have to manage that kind of "attack surface" around these here parts. I'm only half joking... 9-)

Please do keep "radiating". But please do so understanding the whole SnR thing...

Not all "get" the humor. But those that do tend to also get the deeper stuff.
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Old 2022-10-28, 21:36   #11
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Default Missing information in the 1st post.

Dear Chris Halsall,

Your remark "...without really explaining your argument," intrigued me.

Then, I realized that I neglected to write an important paragraph in the 3rd line of the first post at https://www.mersenneforum.org/showpo...43&postcount=1. (I guess I'm getting old and forgetful...)

Below are the 3 triangles that should be placed as information after the 3rd line, when filling Kircheri's triangle with the non-negative integers.

I am now taking the opportunity to add a spoiler on each of the 3 triangle shapes to show a little more where each thing fits in.

Lastly, if you have the power to please transfer this missing information to the first post and delete this current one, I thank you.

In fact, please, you can delete it too:
https://www.mersenneforum.org/showpo...54&postcount=8
https://www.mersenneforum.org/showpo...58&postcount=9
https://www.mersenneforum.org/showpo...1&postcount=10

Thank you very much for the remarks.

------------------



Here, the sequences of odd prime numbers can only appear diagonally.

This is because all verticals follow the sequence of square numbers that alternately have an even and an odd number.

See the highlighted 45° or 1:1 diagonal, showing a segment of Euler's famous sequence x=y^2-y+41. Further up the triangle, you can see part of all the other sequences shown at Euler's "Lucky" numbers.

It is quite visible that all quadratic sequences of prime numbers on a 1:1 diagonal, if they do not end earlier, will at most always end in a square number.

Spoiler:

We will prove later when we continue Kircheri's triangle by completing all rows in the XY plane, then all sequences of prime numbers in 1:1 always end at (or start from) a quadratic CG with a=1 in the form of x=y(y+-b) shown in Understanding A056737 and the quadratic sequences: not so "quibbles".

At the same time as all CG quadratic sequences with a=1 limit all quadratic prime number sequences with a=1, and because they cannot cover all integers, then all quadratic sequences have an infinite number of prime numbers.
Then we will extend to all other polynomials of any order (Bunyakovsky conjecture).



Here, the quadratic sequences with a=1 of odd prime numbers can only appear vertically and specifically in the odd columns. This is because all vertical columns follow the sequence of the oblong numbers which have only even numbers.

See that the 1:1 diagonal highlighted earlier in C000885.L (left), now appears in C000885.C (center) on the 1:2 slope. Also, when we extend this C000885.C (center) triangle toward the two infinite sides, this same sequence will also appear in column 41.

Klauber detected this by the first time in 1931.
(I take this opportunity to ask: if anyone has access to Klauber's original triangle material and can take a clear picture, especially on the upper part of the triangle, I would appreciate it in advance.
This is because I would like to know if he managed to get to the negative integers (passing through 0).
It is the 0's that generate the quadratic GCs that limit all prime number sequences.
I am very curious to know if he realized this too, but all the pictures on the internet that I could get are not sharp enough to know if he saw this
).

Spoiler:

See how here we have a huge hint of how to prove Goldbach's conjecture.

See what we get as a result, when we apply the method of common differences to the columns of the triangle:
  • Between the first row and the second row the difference between its elements in each column is always 2.
  • Between the second row and the third row the difference between its elements in each column is always 4.
  • Between the third row and the fourth row the difference between its elements in each column is always 6.
  • And so on, covering all positive even numbers.

So, to prove Goldbach's conjecture, it is enough to prove that between two subsequent rows we will always find at least one pair of prime numbers aligned vertically.

If we proved Legendre's conjecture, then we have at least one prime number in each subsector SO and OM.

So, we simply extend the rows of the triangle C000885.C (center) to the two infinities. When we extend the rows to the two infinities, the quadratic CGs cannot cover all the integers always forcing the appearance of infinite alignments of prime numbers vertically.

Since there are infinite alignments, then this will also occur for the infinitely many alignments of a positive prime number with another negative prime number in all rows. In this way, at the same time we prove Goldbach we prove Landau.



Here again the sequences of odd prime numbers can only appear diagonally.

This is because all verticals follow the sequence of the (square minus 1) numbers that also has alternately an even and an odd number.

Now, the 1:1 diagonal highlighted earlier at C000885.L (left) appears at C000885.C (center) on the 1:3 slope.

In fact, when we extend this triangle C000885.R (right) to the two infinite sides, the famous Euler sequence x=y^2-y+41 also appears on the 1:1 slope (seen by Klauber).
Just fit C000885.L (left) to the right of C000885.R (right) with an offset of one unit for the continuity of the row sequences.

Spoiler:

These behaviors show us how the polynomial sequences interact with each other where each of them appears at all offsets.
These observations were the inspiration for the pre-prints (which are outdated) in https://easychair.org/publications/preprint/FQgX e https://easychair.org/publications/preprint/Ps32, https://easychair.org/publications/preprint/l731.
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