>>6185

Welcome back!

>staircase

You made me think of polite numbers…

Polite numbers are numbers that can be expressed as the sum of consecutive integers.

The set of impolite numbers are numbers that cannot.

The impolite numbers are the powers of 2.

The number of ways you can define a given polite number as a sum of consecutive numbers is equal to the number of odd factors (including 1 and itself).

Triangle numbers are obviously a subset of the polite numbers (whereby the sequence starts at 1), and a square is the sum of 2 consecutive triangle numbers.

An odd square is the 8 triangles + 1.

An even square can be reduced by dividing by 4.

Therefore the problem can be reduced down to differences of triangle numbers where one triangle is off by 1, and this difference of triangles can be expressed as polite numbers…

Am I at all close?

Hey everyone,

Good to see you all - VQC, Topol, VA, AA, PMA, MM, CA/GA (congrats!), Isee, Hobo, 3dA, PA - and all the other anons!

I've been reading all the posts and following along, even though I haven't been posting.

The reason why I haven't been posting is that I have had nothing new to add!

But today I rewatched Chris' youtube vid, and I discovered something that I think hasn't been mentioned here before:

• Goto to cell (c, 1)

• Find the row where the x is equal to a

• Factor the a value at that record

• Result should be a*(d+n)

• Goto to cell (c, 1)

• Find the row where the x is equal to b

• Factor the a value at that record

• Result should be b*(d+n)

• Goto to cell (-c, 1)

• Find the row where the x is equal to a

• Factor the a value at that record

• Result should be a*(x+n) <- this is not perfect, but always a * a factor of (x+n)

• Goto to cell (-c, 1)

• Find the row where the x is equal to b

• Factor the a value at that record

• Result should be b*(x+n) <- this is not perfect, but always b * a factor of (x+n)

Moreso, it seems that these 4 records are the only records whereby the x value is a factor of the a value.

This all seems important since we may be able to derive a better n value from these columns - not just na/a.

>>6671

Thanks VA!

Sorry for being so silent.

Great progress all round!

>>6670

And a little more…

Because at any record d=x+a:

• at (c,1) where x = our original a

• the a value in that record = a*(d+n)

• the d value in that record = a*(d+n+1)

Then go back 1 record, so that x = a-2:

• the d value in that record = a*(d+n-1)

Therefore, there are 2 records in (c,1) next to each other, where d is divisible by the higher x value.

I believe this is only true where x is a divisor of c, so 4 times per (c,1) (1, a, b, c) for our semiprimes.

I'll follow up with some equations to show how

I think this difference of 1, again something vqc has been referencing, can help in finding factors of c.

>>6674

Thanks!

Super interesting suggestion!

I am finishing up one (or two) more posts to explain my current direction, but right after that, I'll look into (e,c) && (c, e)… I think there's something here relative to that, because the columns (e, 1) , (f, 1), (c, 1) and (-c, 1) are definitely related.

Equations coming v soon…

For odd e, each (e, 1) cell can be produced via:

x[t] = 2t - 1

a[t] = (x^2+e)/2

= ((2t-1)^2 + e)/2

= (4t^2 -4t +e+1)/2

d[t] = a+x

= (4t^2 -4t +e+1)/2 + (2t-1)

= (4t^2 -4t +e+1)/2 + (4t-2)/2

= (4t^2 -4t +e+1 +4t-2)/2

= (4t^2 + e-1)/2

In (e, 1), the row directly before a given row is:

x[t-1] = 2(t-1) - 1

= 2t - 3

a[t-1] = (x^2+e)/2

= ((2t-3)^2 + e)/2

= (4t^2 -12t +e+9)/2

d[t-1] = a+x

= (4t^2 -12t +e+9)/2 + (2t-3)

= (4t^2 -12t +e+9)/2 + (4t-6)/2

= (4t^2 -12t +e+9 +4t -6)/2

= (4t^2 -8t +e+3)/2

So according to the rules in my previous message:

• a[t] is divisible by x[t] (d+n times)

• d[t] is divisible by x[t] (d+n+1 times)

• d[t-1] is divisible by x[t] (d+n-1 times)

• therefore d[t] - d[t-1] = 2x[t]

Proof:

d[t] - d[t-1] = 2x[t]

= (4t^2 + e-1)/2 - (4t^2 -8t +e+3)/2

= ((4t^2 +e -1) - (4t^2 -8t +e +3))/2

= (4t^2 +e -1 -4t^2 +8t -e -3) /2

= (4t^2-4t^2 +e-e +8t -3-1) /2

= (8t-4) /2

= 2(4t-2)/2

= 2(2t-1)

= 2x[t]

Still trying to figure out how to use this…

d[t] - d[t-1] = 2x[t]

This is true in all (e, 1), before y'all jump on me.

>>6678

I missed that one, thank you very much.

I'll experiment with that next.

Right now I'm researching relationships between (e,1) and (c,1).

>>6680

I have it open in another tab, haven't read it yet.

>>6681

Thanks PMA - I've been lurking, great job so far man.

It all just gets too deep for me sometimes and my real world responsibilities creep back up, so i need breaks.

While I'm posting a bit, here's something else I've been toying with regarding recursive factoring of d & e:

We know c = ab = (d+n)^2 - (x+n)^2 = dd+e.

The part to focus on is:

dd+e = (d+n)^2 - (x+n)^2

Assuming we know d's factors, we can express d in the same form ie the difference of 2 squares.

Same for e.

So if we express d and e as the difference of 2 squares, for this sake, Big (B and b) & Small (S and s):

(B^2-S^2)(B^2-S^2) + (b^2 - s^2)

So there are 2 main operations that we have to figure out how to translate: squaring d, and adding e.

Ok, so to square the difference of 2 squares, we can perform the following math:

c = (B^2 - S^2)^2

= ((B-S)(B+S))^2

= (BB-2BS+BB)(BB+2BS+SS)

= (BB+SS -2BS)(BB+SS +2BS)

= (BB+SS)^2 - (2BS)^2

Here we're expressed it back in terms of the difference of 2 squares.

Therefore, back in our original notation:

((d+n)^2 - (x+n)^2)^2 = ((d+n)^2+(x+n)^2)^2 - (2(d+n)(x+n))^2

The more difficult operation is the addition of dd & e, I'm still trying to figure this one out, but if we can, we can solve the problem recursively easily.

So one potential path is to figure out how to express the addition of 2 difference of 2 squares, in terms of the difference of 2 squares:

(B^2 - S^2) + (b^2 - s^2)

>>6685

Yep, thats the direction I'm talking.

Use 12=2*6 and you'll be able to express it as the difference of 2 squares.

>>6685

Here's my (unfinished) progress so far on addition of 2 diff of squares:

Uppercase and lowercase represent the 2 terms, in our case dd & e.

((D+N)^2 - (X+N)^2) + ((d+n)^2 - (x+n)^2)

= DD+2ND +NN -XX -2XN -NN + dd+2nd +nn -xx -2xn -nn

= (DD+dd+NN+nn) - (XX+xx+NN+nn) +2(ND-XN+nd-xn)

= (DD+dd) - (XX+xx) +2(ND-XN+nd-xn)

= (DD+dd) - (XX+xx) +2(ND-XN+nd-xn) +2Dd-2Dd +2Xx-2Xx

= (DD+dd+2Dd) - (XX+xx+2Xx) +2(ND+nd-XN-xn-Dd-Xx)

= (D+d)^2 - (X+x)^2 + 2(NA+na-Dd-Xx)

Some things I'm trying to substitute into the last term:

Dd = Da-Dx

= dA-dX

Xx = Xd-Xa

= xD-xA

Sorry if this is a little messy and unconventional variables (casing), its a wip.

>>6687

To work through your example:

c = dd+e = 145 = 12^2 + 1

12 = 4^2 - 2^2

1 = 1^2 - 0^2

Using the formula in >>6684

12 ^2 = (4(4) + 2(2))^2 - (2(4(2))^2

= (16+4)^2 - (16)^2

= 20^2 - 16^2

So now we have:

dd = 20^2 - 16^2

e = 1^2 - 0^2

so c = 20^2 - 16^2 + 1^2 - 0^2

Thats where the work in my post >>6688 comes in.

I'm still working out how to add 2 difference of squares and end in the difference of squares.

So I'm trying to find a way algebraically to define:

20^2 - 16^2 + 1^2 - 0^2 = 17^2 - 12^2

>>6688 is my progress so far.

>>6699

Thanks Hobo. Its good to be back and posting again - I missed you guys!

>>6703

Yeah, I've been thinking about that too…

This may be known, but if you take 2c and calculate e from it, the original c will always be in (e,1) as an A value.

We may be able to combine these relationships somehow using column 0. Just thinking out loud.

>>6704

>(4,10) -> (4, 1, 6, 2, 4, 10)

>(50,68) -> (36, 1, 58, 8, 50, 68)

>(148,178) -> (100, 1, 162, 14, 148, 178)

I might be stating the obvious but the pattern I see here is the e's are even squares… very interesting indeed!

For the pattern in (-f, 1), the other thing i see is that the e's are separated by (1)(72), then (2)(72).

Although 72 doesn't appear significant in c=13*41 as far as I can tell…

>>6697

Regarding a*a in (e, 1), I don't think it works for all numbers.

Take a = 101 b = 103.

a*a = 10201

I can't find it as a factor in (e, 1).

However, in (-f,1) its first found at t = a*a.

But this rule is not true for all c.

Here's some thoughts on xa…

b = a + 2(x+n)

c = ab

= a(a + 2(x+n))

= aa + 2xa + 2na

Therefore:

2na = c - aa - 2xa

We know c%8, d%8 and e%8.

We know a is odd for semi-prime c.

We know the parity of x from d-a.

We know that aa is an odd square.

We know that xx is either an even or odd square.

We know all odd squares are 8T+1.

We know all even squares are 4 times a smaller square.

xa is interesting because x and a are linearly related.

xa = (d-x)(d-a)

I cycled through the possible ax combinations for different d's.

As a increases -:

For odd d, even x:

xa % 8 = … 0,2,4,6, 0,2,4,6, ….

or

xa % 8 = … 6,4,2,0, 6,4,2,0, ….

For even d, odd x:

xa % 8 = … 1, 5, 1, 5, 1, 5, ….

or

xa % 8 = … 3, 3, 3, ….

or

xa % 8 = … 5, 1, 5, 1, 5, 1, ….

or

xa % 8 = … 7, 7, 7, ….

The selection of which pattern to choose just cycles as d increases.

When you multiply xa by 2 the pattern simplifies:

2xa % 8 = 0, 4, 0, 4, …

or

2xa % 8 = 2, 2, 2, 2, …

or

2xa % 8 = 4, 0, 4, 0, …

or

2xa % 8 = 6, 6, 6, 6, …

Again, regularly cycles as d increases.

This is another way of deriving the value of na % 8.

There's always a max xa when x = a or x = a+1.

Whats even more interesting is that when I divide xa by 8, and minux max_xa, you get another repeating pattern!

That pattern repeats as d increases too:

floor(max_xa/8) - floor(xa/8) =

0, 1, 3, 6, 9, 14, 20, 27, 34, 43, 53, 64, …

or 0, 1, 2, 5, 8, 13, 18, 25, 32, 41, 50, 61, …

or 0, 2, 4, 7, 11, 17, 23, 30, 38, 48, 58, 69, …

or 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, …

Notice that last pattern… triangles!!!

The other ones are very close to the triangles series too.

I know this is messy, but I'll drop by tomorrow and clarify anything.