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User:Bogart B. Strauss

This is my approach to the elementary factorization of integers:

Given an integer ${\displaystyle Q=(2^{b})*(3^{c})*(6v+(-1)^{r})*(-1)^{a}}$ with a, b, c, v, and r equal to the positive integers and zero.

Since we can find factors of 2 and 3 with a glance and the sign of the integer is of no importance for factorization, they are considered trivial and thus we are left with ${\displaystyle 6v+(-1)^{r}}$ which combines two cases (r-even and r-odd). If this set of numbers equals R, then R*R is an element of R.

${\displaystyle 6v+(-1)^{r}}$ is the sequence ${\displaystyle 1,5,7,11,13,17,19,23,25,29,31,35,37,41,43}$.....sequence A007310 of the OEIS.

Considering that this can be formulated with a single variable, let's define a new sequence z.

${\displaystyle z=(1/2)*(6x-(-1)^{x}+3)}$ with x equal to the positive integers or zero.

Using a sieving method (if you are curious, email me), we can then derive something to equate ${\displaystyle 6v+(-1)^{r}}$ with.

${\displaystyle 6v+(-1)^{r}=z^{2}+z*(6y+3-(-1)^{x})}$ with y equal to the positive integers or zero.

Thanks to Fermat realizing that a difference of squares with integer bases turns into a product of integers, we can then add that methodology to what we already have.

${\displaystyle 6v+(-1)^{r}=z^{2}+z*(6y+3-(-1)^{x})=n^{2}-m^{2}}$ where n and m are positive integers and m can equal zero with ${\displaystyle n>m}$.

Considering that the factors will be found in divisor pairs, we can state that ${\displaystyle 6v+(-1)^{r}=z*f}$ .....with the set of all f being equal to the set of all z.

Notable relations:

${\displaystyle n-m=z}$

${\displaystyle n+m=f}$

${\displaystyle n=3k+(1/4)*(1+(-1)^{r})*(3-(-1)^{x})}$ where k is a positive integer or zero

${\displaystyle m=3y+(1/4)*(1-(-1)^{r})*(3+(-1)^{x})}$

Things start getting strange when we look at the resulting patterns from n and m as they relate to ${\displaystyle 6v+(-1)^{r}}$. There is a clear split into 6 cases (3 for r-even, 3 for r-odd) and a monolith of a difference of squares formula:

${\displaystyle 6v+(-1)^{r}}$ can be split in three: ${\displaystyle 18p+(-1)^{r}}$, ${\displaystyle 18p+7*(-1)^{r}}$, and ${\displaystyle 18p+13*(-1)^{r}}$.

6v+(-1)^r=18p+(6q+1)*(-1)^r=( ((1/4)*(18d+18+9*(-1)^r+(6*(-1)^q-1-q*(1+(-1)^q))*(-1)^(d+r)))^2 - ((1/2)*(3y+3k+(3y-3k)*(-1)^r))^2 )*(-1)^r where p and d are positive integers or zero while q={0,1,2}. I wouldn't read too much into this formula without figuring a desired q and r first.

I could not derive any more obvious patterns from this point and had to mess around with, well, anything. That anything is 3 times the number to be factored as another difference of squares.

${\displaystyle 3*(6v+(-1)^{r})=g^{2}-h^{2}}$ where g is a positive integer not divisible by 3 and h is an integer not divisible by 3 with ${\displaystyle g>h}$.

Notable resulting relations:

${\displaystyle g=2n-m}$

${\displaystyle h=n-2m}$

${\displaystyle 6v+(-1)^{r}=gm+nh=(1/2)*(gn+mh)}$

From that last formula, we realize that either ${\displaystyle gn}$ or ${\displaystyle mh}$ has only one possible factor of two or "critical 2". The reason is that ${\displaystyle 6v+(-1)^{r}}$ is always an odd number and thus it is always a sum of an odd number and an even number.

If we list the first 15 numbers in the sequence ${\displaystyle 6v+(-1)^{r}}$, we can identify a pattern of critical 2's. Every unique divisor pair has a unique combination of h,m,n,g.

1*1____h=1, m=0, n=1, g=2

1*5____h=-1, m=2, n=3, g=4

1*7____h=-2, m=3, n=4, g=5

1*11___h=-4, m=5, n=6, g=7

1*13___h=-5, m=6, n=7, g=8

1*17___h=-7, m=8, n=9, g=10

1*19___h=-8, m=9, n=10, g=11

1*23___h=-10, m=11, n=12, g=13

1*25___h=-11, m=12, n=13, g=14

5*5____h=5, m=0, n=5, g=10

1*29___h=-13, m=14, n=15, g=16

1*31___h=-14, m=15, n=16, g=17

1*35___h=-16, m=17, n=18, g=19

5*7____h=4, m=1, n=6, g=11

1*37___h=-17, m=18, n=19, g=20

1*41___h=-19, m=20, n=21, g=22

1*43___h=-20, m=21, n=22, g=23

The pattern of critical 2's is ${\displaystyle g,m,h,n,m,g,n,h}$ repeating. Also notice that the critical 2's are unique to the number to be factored and thus relates each divisor pair of the number. This splits things up into 8 different cases (four for r-even and four for r-odd).

I can no longer reasonably subject anyone to a formula as we now have a total of 48 unique cases to deal with. Here's an example of just one case with B equaling positive integers and zero:

${\displaystyle 6v+(-1)^{r}=18p+1}$ (from the three-split)${\displaystyle =24B+1}$ (from the four-split)${\displaystyle =72B+1}$ (from the forty eight split)

m is an element of ${\displaystyle 12B}$

n is an element of ${\displaystyle 18B+or-1}$

h is an element of ${\displaystyle +or-z}$

g is an element of ${\displaystyle 2*z}$

Recall that z is defined earlier. You can see here that the critical 2 is g for all numbers equal to ${\displaystyle 72B+1}$.

So ${\displaystyle 145=17^{2}-12^{2}=(17+12)*(17-12)=29*5}$. While ${\displaystyle 3*145=22^{2}-(-7)^{2}=(22-7)*(22+7)=15*29}$.

That's as far as I've gotten with the elementary factoring of integers and I encourage anyone to try and dig deeper, I certainly am. Patterns exist, we merely have yet to find them. But here it seems that there will always be a "next" pattern.

I will be transferring more summaries of my work here soon. The next piece will be an N choose K type of formula for figuring out the number of divisor pairs with a given prime factorization. This one will be a while as I am trying for a better notation for the extension of Sigma summation I have to use.

This is no place for philosophical statements or meandering through things I may perhaps misunderstand. For the things I may be mistaken about, I have yet to understand the proofs otherwise. I don't believe, I think. It does not escape me that this is a very precious responsibility.