Equations for some glides in the Collatz Conjecture

By Erika Schwibs

I’ve been focusing on 5 and 16, as in 3(5)+1=16. 15/16 is the lowest combo that goes to 4, 2, 1.

This is a work in progress, and I didn’t major in math, so for right now, first notice the progressions, like numbers 1, 4, 7, increasing by 3. Another increases by 16. The result is equations for some glides, I believe they’re called.

First, 5A+B=C, where A begins with 1 and increases by 3, B begins with 0 and increases by 1, and C begins at 5 and increases by 16:

1×5+0=5

4×5+1=21

7×5+2=37

10×5+3=53

13×5+4=69

16×5+5=85

19×5+6=101

22×5+7=117

25×5+8=133

28×5+9=149

31×5+10=165

34×5+11=181

37×5+12=197

Etc.

Insert these products (5, 21, 37, etc.) as x into 3x+1, on the left side of the equation. The products on the right side, in turn, will increase by 48 and also be the products of 16 and another number, starting with 1 (as in 16=16×1), and that number will increase by 3:

3(5)+1= 16 =16×1

3(21)+1= 64 =16×4

3(37)+1= 112 =16×7

3(53)+1= 160 =16×10

3(69)+1= 208 =16×13

3(85)+1= 256 =16×16

3(101)+1= 304 =16×19

3(117)+1= 352 =16×22

3(133)+1= 400 =16×25

3(149)+1= 448 =16×28

3(165)+1= 496 =16×31

3(181)+1= 544 =16×34

3(197)+1= 591 =16×37

Etc.

Of course, if you have one side, you can solve for the other. I’ve put this into the form 3x+1=16y.

If y=10:

3x+1=16(10)

3x+1=160

3x=159

x=53

If x=53:

3(53)+1=16y

159+1=160=16y

y=10

And this, too. B begins with 1, and increases by 3. If b=37:

3(5a+b)+1=16a

3(5a+37)+1=16a

15a+111+1=16a

112=a

Meaning*:

(*From here on out, I’m using a period to represent multiplication).

3(5.112+37)+1=16.112

1791+1=1792

The number 1792 is divisible by 2 all the way down to the number 7.

Another example. If b=10:

3(5a+10)+1=16a

15a+30+1=16a

31=a

3(5.31+10)+1=16a

3(165)+1=16a

495+1=496=16a

31=a

496 is divisible by 2 down to the number 31.

And a third example. If b=9:

3(5a+9)+1=16a

15a+27+1=16a

28=a

So:

3(5.28+9)+1=16.28

3(140)+27+1=16.28

420+27+1=16.28

=448

448 is divisible by 2 down to the number 7.

If b=8:

3(5a+8)+1=16a

15a+24+1=16a

a=25

3(5.25+8)+1=16.25

3(125+8)+1=400

3(133)+1=400

399+1=400

400 is divisible by 2 down to the number 25.