Boring Math (11.0)

Based on

(-7, 5, -1)/4

I'm thinking:

0 (mod 7): n/7
1: 4n + 2
2: 4n + 3
3: 4n + 5
4: 4n + 7
5: (n+2)/7
6: (n+1)/7

Boring Math (11.0)

My current idea for mod 7 is approximately

0 (mod 7): n/7
1: 2n + 1
2: 2n + 2
3: 2n + 3
4: 4n + 4
5: (n+2)/7
6: 2n + 6

Boring Math (11.0)

I finally realized why my variant mod 4 Collatz tree is so predictable: odd numbers never become odd again.

Mod 6 * might * have a different property.

Boring Math (11.0)

If you look at it like this

{-3, 2, 7, 19, 47, 79, 199} - {1, 4+1-2, 7, 18+1, 47, 76+1+2, 199}

-->

{-4, -1, 0, 0, 0, 0, 0}

The first two elements relate to the half integers and possibly incorporates some concept of quadratic remainders.

Boring Math (11.0)

The error terms can be summarized by

{-3, 2, 7, 19, 47, 79, 199} - {1, 4+1, 7, 18+1, 47, 76+1, 199}

Boring Math (11.0)

It really feels like the Lucas numbers at indices

1
3
4
6
8
9
11

Are relevant. There's a little bit of "error", but this was qualitative to begin with.

Boring Math (11.0)

The indices

4
6
8

Are basically exact with the +1 to every 3rd element in the lucas numbers, shifting 18 --> 19.

Boring Math (11.0)

I'm looking at

-7 + {1+3,2+7,3+11,7+19,11+43,19+67,43+163}

-->

{-3, 2, 7, 19, 47, 79, 199}

Again and comparing it to oeis.org/A000204/list

Boring Math (11.0)

I'm so exhausted and behind on work for my day job to do anything major.

Boring Math (11.0)

sqrt(1367+x)/sqrt(139+x) = pi
--> x ~= -0.54963...

sqrt(1367+x)/sqrt(139-x) = pi
--> x ~= 0.44849...

Which is surprisingly close to +/- 1/2.

Boring Math (11.0)

sqrt(1367)/sqrt(139) ~= 2^3*5^-3*7^2

Boring Math (11.0)

For clarity, the sequence is sort of like:

1
0
-1
-2
-139

Where notably the 139 essentially shows up in the 2879 as well.

Boring Math (11.0)

Looking at:

{-2917, -40, -19, 2, 23}*2 - 6

Made me realize there's a difference of 21 between most of the elements.

The elements seem to always be congruent to 2 mod 21.

Boring Math (11.0)

I think this means I care about:

factor {-2879, -2, 19, 61}

Boring Math (11.0)

Python code:

def hailstone(n):
target = n+1
for iterations in range(2000):
residue = n % 3
if residue == 0:
n = 7*n + 3
elif residue == 1:
n = (7*n + 2)//3
else: # residue == 2:
n = (n - 2)//3
if n == target:
print((n-1))

Which finds items similar to oeis.org/A276260

Specifically:

-2917 = -(54^2 + 1) = -prime(422)
-40
-19
2 (can probably be ignored)
23

Boring Math (11.0)

I think "Matthews and Watts (1984)" is already correct. ( mathworld.wolfram.com/CollatzP )

If you look at "(1 + sqrt(7/3))/2" - oeis.org/A005186

The center point may be between 0 and -1 (so -1/2).

I need to check if negative cycles exist.

Boring Math (11.0)

Hypothetically, if I get the "mod 4" behavior and the "mod 7" behavior, it links to the (-7/4) in the 28=18+5+3+2 cycles in the Collatz conjecture and the Mandelbrot set.

Boring Math (11.0)

I think I am missing a mod 7 routine with (49?) stable points (possibly not integers).

I have zero idea where to look, beyond, perhaps, mathworld.wolfram.com/CollatzP

Which has "Matthews and Watts (1984)" and their mod 3 conjecture.

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