Here is an analysis of the (x+n) grid
Hey VQC. Huge fan
Should I keep doing the grid stuff I'm doing?
I can look further into the factoring method if we should be doing that. I think there is a lot to be learned from these grids and if its as versatile as you say then it could be useful. On the otherhand the first discovery might be so mindblowing that anything else we do to utilize the grid would be trivial. I don't want to be wasting time.
https://www.m ath.hmc.edu/funfacts/ffiles/20008.5.shtml
>Which whole numbers are expressible as sums of two (integer) squares?
>Here's a theorem that completely answers the question, due to Fermat:
>A number N is expressible as a sum of 2 squares if and only if in the prime factorization of N, every prime of the form (4k+3) occurs an even number of times!
>Examples: 245 = 577. The only prime of the form 4k+3 is 7, and it appears twice. So it should be possible to write 245 as a sum of 2 squares (in fact, try the squares of 14 and 7). But because 7 appears only once in 42=237, it is impossible to write 42 as the sum of two squares.
>A corollary of this fact is that every prime of the form (4k+1) can be written as the sum of two squares.
So since 145 is a sum of 2 squares so 145 = 4k+1 for k=36 = 1212 + 11
I think I learned in number theory way back when a way to get these squares for a number. Not sure though.
We know for a fact that it is coprime, so there are two factors. If one is equal to 4k+3, then the other must also be equal to 4k+3 for different k's because it can be written as a sum of two squares. If one factor is equal to 4k+1, then the other must also, because 0 is for this sake an even number.
So if a number is a sum of two squares, it is either
c = (4k+1)(4j+1) or (4k+3)(4j+3) for some j and some k
Lets look at the formula (d+n) = (b+a)/2 and (x+n) = (b-a)/2,
Suppose a=4k+1 b=4j+1,
d+n = (4j+1 + (4k+1))/2 = (4j + 4k + 2)/2 = 2(j+k) + 1
x+n = (4j+1 - (4k+1))/2 = (4j+1 - 4k - 1)/2 = 2(j-k)
for c=145 = 4(36) + 1
a = 5 = 4(1) + 1 ==k=1
b = 29 = 4(7) + 1 ==j=7
(d+n) = 2(j+k)+1 = 2(7+1)+1 =17
(x+n) = 2(j-k) = 2(7-1) = 12
Now suppose a=4k+3 and b=4j+3
d+n = (4j+3 + (4k+3))/2 = (4(j+k) + 6) / 2 = 2(j+k) + 3
x+n = (4j+3 - (4k+3))/2 = (4(j-k))/2 = 2(j-k)
For c = 253 = 4(63) + 1
a = 11 = 4(2) + 3 ==k = 2
b = 23 = 4(5) + 3 ==j = 5
(d+n) = 2(j+k) + 3 = 2(2+5) + 3 = 17
(x+n) = 2(j-k) = 2(5-2) = 6
This is super cool because every number that is the sum of two squares has (x+n) = 2(j-k).
Its almost as if it maps all the numbers from 2k+1 to 4k+1 in such a way.
Like for the 145 example,
a = 4(1) + 1
b = 4(7) + 1
Call the 1 and 7 the core components of the number. You can add and subtract these, and when you do you generate an odd and an even number
For these situations d+n is the odd component. I'd imagine when c=4k+3 that is the reverse but same type of stuff.
(d+n) = 2(7+1) + 1 = 17
(x+n) = 2(7-1) = 12
The same type of thing holds for c=253, but the leaf (odd number) is 3 instead of 1
a = 4(2) + 3
b = 4(5) + 3
(d+n) = 2(5+2) + 3 = 17
(x+n) = 2(5-2) = 6
Getting anywhere??
If we look in here
Pic 1 can show you a few things. The core element of b is always the sum of the core for d+n and x+n. Also the core for a is always the difference of d+n and x+n.
Whats interesting is that for these 4 numbers (a,b,d+n,x+n), the pattern always holds no matter what base you're in. Also another thing is that the leaf numbers of d+n and x+n always add up to the leaf number of b.
If a number c is a sum of two squares, then it is of the form 4k+1. Here, the 4 is the multiplier, the k would be the core and the 1 would be the leaf. I'm proposing that the multiplier can be any power of 2 and the k can be any number and then the leaf would then have to be a number less than the multiplier.
If a square c is of the form 4k+1, then according to the link I posted it's prime factorization must have primes of the form 4k+3 occurring an even number of times [here, 4 would be multiplier, k core and 3 leaf]. Since we know c is coprime (2 prime factors) and that they are odd, the only two types of numbers they can be are 4k+1 and 4k+3. Since the factors of the type 4k+3 must occur an even number of times and there are only 2 factors, the even numbers are either 2 or 0, so either both factors are 4k+3 for different k's or they are both 4k+1 for different k's. Keep in mind, this is only for c = 4k+1.
Knowing this, I algebraically looked into the d+n and x+n values in this format through the formulas:
d+n = (b+a)/2
x+n = (b-a)/2
I found that if
a = (4)(k) + L
b = (4)(j) + L
For L=1 or 3 and since b>a, j>k
Then you can find the d+n and x+n values like this
d+n = (2)(j+k) + L
x+n = 2(j-k)
So instead of doing all the calculations you could look at it like this:
>Divide the multiplier by two
>Take sum and difference of core elements
>Decrement L by one for the difference
I'm thinking we might be able to take this formula and do other things with it, but I am still looking. For instance what if we increase all the powers? Then does the function map to a different calculation? If you increase the multiplier to greater than the initial number, then the entire number would just be a leaf because the core would be 0. Obviously every single number can be written as another number.
15 = 16(0) + 15 = 8(1) + 7 = 4(3) + 3 = 2(7) + 1 = 1*(15) + 0
Look here, the leaf goes 15->7->3->1->0
and the core goes 0->1->3->7->15
They reverse but the power decreases.
I think we should look into this
Another thing we can use this type of stuff for is Legendre Symbols.
The legendre symbol (I'll denote it as (a//p)) is:
0 if a is divisible by p
1 if a is a quadratic residue mod p
-1 if a is not a quadratic residue mod p
If a IS a quadratic residue mod p, then there exists a number M where MM - a = kp for some k.
Or the square of M is a units more than a multiple of p.
We could make a a negative number, which would switch the equation to MM + a = kp.
Then its a square plus a number equals a product of 2 numbers (one of which is necessarily prime), which is our entire problem.
If you look further into these numbers, you'll see things where this type of math is used.
If you let a=-1, then the legendre symbol (a//p) can be calculated directly through this calculation:
(a//p) = 1 if p = 4k+1
(a//p) = 3 if p = 4k+3
If you let a=-2, then
(a//p) = 1 if p = 8k + 1 or 8k + 3
(a//p) = -1 if p = 8k + 5 or 8k + 7
If you let a=-3, then
(a//p) = 1 if p = 12k + 1 or 12k + 7
(a//p) = -1 if p = 12k + 5 or 12k + 11 (3 or 9 dont seem to show up)
a = -4
(a//p) = 1 if p = 8k + 1 or 8k + 5
(a//p) = -1 if p = 8k + 3 or 8k + 7
a = -5
(a//p) = 1 if p = 20k + {1, 3, 5, 7, 9}
(a//p) = -1 if p = 20k + {11, 13, 17, 19} (15 doesnt seem to show up)
a = -6
(a//p) = 1 if p = 24k + {1, 5, 7, 11}
(a//p) = -1 if p = 24k + {13, 17, 19, 23}
Basically it looks like you can check divisibility without actually calculating anything except the mod of 4 * the number. The -4 case is unique though I'll look more into that. There may be numbers missing from these but generally speaking the ideas is right. I'll write a script later to more easily calculate this stuff and verify it.
>We admitted we were powerless over our addiction - that our lives had become unmanageable.
>Came to believe that a [Power] [greater than] ourselves [c]ould restore us to sanity.
>Made a decision to turn our will and our lives over to the care of [God] as we understood [God].
>Made a searching and fearless moral inventory of [ourselves].
>Admitted to [God], to ourselves and to another human being the exact nature of our [wrongs]
>Were entirely ready to have G[o]d remove all these [defects] of [c]haracter.
>Humbly asked God to [remove] our [shortcomings].
>Made a list of all [persons] we had harmed, and became willing to make [amends] to them all.
>Made direct amends to such [people] wherever possible, except when to do so would [injure] them or others.
>[Continued] to take personal inventory and when we were wrong promptly admitted it.
>Sought through prayer and meditation to improve our conscious contact with God as we understood God, praying only for knowledge of God's will for us and the power to carry that out.
>Having had a spiritual awakening as the result of these steps, we tried to carry this message to other addicts, and to practice these principles in all our affairs.
>The Twelve Steps are the solution for bEtter meN.
EN is the obvious clue.
These steps above could refer to this puzzle?
Lets say 'we' are the number c.
>Came to believe that a Power greater than ourselves could restore us to sanity.
Power greater than c so
(d+1)^2 dd+e
c^2 c
Either of these could potentially be leads. I think its c^2 cuz VQC said that before it was important
>Made a decision to turn our will and our lives over to the care of God as we understood God.
If I had to guess what 'God' is in this context I'd say it is the number 1. Thats what a lot of people say it is anyway
Maybe thats the (e,1) row and use what we know about that row to do stuff with the record
>Made a searching and fearless moral inventory of ourselves.
This could be some tree generation with the dd+e trees.
>Admitted to God, to ourselves and to another human being the exact nature of our wrongs.
I'd say that since we have dd+e e has always seemed to be a little extra piece that we don't need. I think maybe then we could collect the extra parts from each division. The sum of those could be the 'defects of character' in that they aren't perfect squares.
>Were entirely ready to have God remove all these defects of character.
>Humbly asked God to remove our shortcomings.
c - (sum of all minor e values) = something
>Made a list of all persons we had harmed, and became willing to make amends to them all.
So somehow we are going to be using numbers along the line, I think for this it would mean add a certain amount to these values you've seen along the way. I'd say its the numbers in the tree.
>Made direct amends to such people wherever possible, except when to do so would injure them or others.
Add the numbers to the cells unless it would invalidate a cell
>Continued to take personal inventory and when we were wrong promptly admitted it.
This implies some sort of while loop I think
>Sought through prayer and meditation to improve our conscious contact with God as we understood God, praying only for knowledge of God's will for us and the power to carry that out.
Something something I don't know maybe trust row n=1
>Having had a spiritual awakening as the result of these steps, we tried to carry this message to other addicts, and to practice these principles in all our affairs.
Use this algorithm to do everything because its super cool
If we look at any number base, the 1's place is 1, the next place is b, the next is bb, then bb*b etc. Obviously, you can't really have non-integers as bases and e=2.718 is a decimal.
In a way, though, the fibonacci numbers are like base e. because F(n) * phi ~= F(n+1), which is similar to how, in base 10 for example, if you have 7 * 10 = 70.
Sexagesimal again obvious pointer to fibonacci numbers. (great site btw)
https://www.mathsisfun.com/numbers/fibonacci-sequence.html
If you look at the last digit of fibonacci numbers it repeats every 60. The tens repeats every 300, etc.
>Full circle
Implication we need to use a circle.
>It's perfect
Perfect circle
>I think base 60 + Babylonian fractions would round this out nicely
>round this out
Either another reference to the roundness of a circle or has to do with rounding the numbers.
>Absolutely positive its the key
Absolute value of something
I gotta go back to work, but I thikn we could do something with this notation
If we have an number like:
'100001000' in fibonacci notation,
we could rewrite it as
'100000110'
or
'011000110'
'002100110'
'000320110'
'000053110'
etc.
Like any number can be turned into the two next to it.
I think that if we can make this a constant digit then it must be divisible by that digit at the very least.
https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem
Heres some examples for shrinking the fibonacci notation down. First is for full notations.
Next is if you keep it as 1's and 0's. I think that is the better route to take but idk why I say that. It seems that the most sparse it will be is always 101010101 etc and the most dense is 11111111. Basically the rule of this when you collapse it is that it is 1s broken up by 0s and there are never two zeroes touching, because if there were then you could populate them with the next higher digit. If you have a streak of 1's (aka a sum of consecutive fibonacci numbers) then it is as follows
>https://math.stackexchange.com/questions/833231/the-sum-of-n-consecutive-fibonacci-numbers
Sum of (F(1) .. F(k)) = F(k+2) - 1
So the sum from j to k would be:
[F(k+2) - 1] - [F(j+2) - 1] = F(k+2) - F(j+2)
I don't know where you'd fit in the negatives. Maybe make a second number. Basically I'm thinking we collapse it until we can't anymore then trim out the streaks by removing the negative element, then keep collapsing until its just a string of 1's then do something with the negative
>Admitting powerlessness over the addiction and our lives had become unmanageable
Powerless AKA not a difference of squares. AKA 1 dimensional.
If our lives were manageable then we'd have two factors for it
>Believing that a higher power (in whatever form) can help
I think this higher power is phi and in specific this big spiral.
>Deciding to turn control over to the higher power
I think this would be to represent ones self as a sum of fibonacci numbers (none consecutive)
For example 3*5 = 15 would be 13 + 2 = 1000100 = (13)(0)(0)(0)(2)(0)(0)
>Taking a personal inventory
[Interesting how 145 = 144 + 1 and 144 and 1 are both fibonacci numbers]
I think this is to look at your own bits within the number.
(this is just my idea for now of what this entails)
1000100
0110011
0101111
This streak of four 1's is 1+1+2+3 = 7
The 1 on the left is 8
7+8 = 15
Sum of consecutive fibs = F(1) + … + F(k) = F(k+2) - 1
15 = 0101111 = 0100000 + 0001111
and 0001111 = 0100000 - 1
15 = 0100000 + 0100000 - 1
15 = 0111000 - 1
15 = 0110110 - 1
15 = 0110110 - 1 = 0110000 + 0000110 - 0000001
0000110 = 0010000 - 0001000
0110000 = 10000000 - 1000000
so
15 = 10000000 + 0010000 - [ 0001000 + 1000000 - 0000001 ]
15 = 10010000 - 1001001
15 = 10001100 - 0110111
15 = 10001011 - 0110111
etc..
>Admitting to the higher power, oneself, and another person the wrongs done
Something something get the negative amount and give it to the fibonacci
15 = 1000100 = 1111111 - 0111011 Maybe this value is the negative.
>Being ready to have the higher power correct any shortcomings in one’s character
Correction of shorcomings would be the result of this calculation whatever that may be
>Asking the higher power to remove those shortcomings
Subtract that amount from you
Anybody want some grids? More incoming
In order that is E by A
E by C
E by D
E by (d+n)
E by F
I still lurk. I got sidetracked during the holidays. Still do work on this though. Noticed something recently. thanks to
I was looking at (1,1).
In (1,1) the values for x and a are tied to a pythagorean triple,
(x, a-1, a) so that aa = xx + (a-1)(a-1). Then I couldn't really follow what d was getting to, but I think this might be a way to tie together unrelated squares, because xx, aa, (a-1)(a-1), d*d, are all related to eachother in some way in these equations. Granted there isn't much but forgive me.
Another thing I saw is that if you look at the records in row 1, all of their 'ones' place values repeat over time, no matter what base you are using. For instance, if we have a base equal to 45, then all of the values in column A % 45 will repeat over time. For instance, if we use base 60, then for every even e, the a's repeat every 30, then if we use an odd e, it repeats every 15. Hovever, if our base is odd, it will always repeat at that interval. Also within the repeating the same pattern occurs backwards so theres half as many options.
For base divisible by 4:
Even e:
A[t] % base = A[t+base/4] % base
Odd e:
A[t] % base = A[t+base/2] % base
Base divisible by 2:
A[t] % base = A[t+base/2] % base
Other base:
A[t] % base = A[t+base] % base
We could then iterate through the first t records in (e,1) because those would have the correct a%t values,
then maybe we could use modular arithmetic to solve it from there. Maybe mix in the -f entry to get two equations. Maybe deliberately use a base that is divisible by 4 so that d will be even and -f will be odd, and you'll have two different bases to use in the modular equation.
Lets use base 60 for c=145
We know that A[t] = na for some value in (1,1), and that the values A[t]%60 for this are pic related:
(1, 5, 13, 25, 41, 1, 25, 53, 25, 1, 41, 25,13,5,1,1,5,13,…)
We calculated all of these and there are a bunch of repeats so we filter them out
now our options are na == 1, 5, 13, 25, 41, 53 (mod 60)
Second pic related is -24.
If we look at (-24, 1) = (-f, 1), then there is an A[t] = n(a-1).
Now all the options are :
(6, 20, 38, 00, 26, 56, 30, 8, 50, 36, 26, 20, 18, 20, 26, 36, 50, 08, …
which can be reduced to n(a-1) == 6, 8, 18, 20, 26, 36, 38, 50, 56
We can check every permutation which would be 6*9 attempts = 54 iterations, and each time you'd do this.
Lets say I'm testing n(a-1) 18, and na 41 (mod 60)
na - n(a-1) == 41 - 18 (mod 60)
na - na + n == 23 (mod 60)
n == 23 (mod 60)
So I guess it would always turn out to be just getting the difference of all these numbers, which isn't too intense of a calculation. Of course once we get n (mod 60) it's not like we have n or anything. We might be able to narrow it down further with other equations and known variables but idk. Just thought this was neat and wasn't sure if anyone else had posted it.
That was a tkinter app I made that would do it for any records. Never found much from it. Then again I never really used it because I saw the flaws and I made a better app. I think we could use the coloring to highlight these cells but then view them from a different perspective. I'm stil working on a program that will let us view the cells from any angle. AKA originally the axes are (e,n) but I'm trying to make something that will enable use to view from any axes and highlight any cells.
>using only c d e f
I think that this could be the key. He probably chose those letters for a certain reason. Maybe theres a bunch of steps. c gives you d, d and c give you e, d and e give you f, f and e give you g, g and f give you h, …. up until 'n' I'd guess.
c d e f g h i j k l m n -finally -> a or b
>VQC: Twelve steps are the key
c -n is 12 steps if we follow this pattern. Now obviously I have no idea what these steps are in specific, but if this is the case then vqc has given us up to f. This is 1/3 of the way already.