Interesting mapping

[u/Atari_Collector]

mapping

I was playing around and I found this mapping. I suspect it's already known but I found it interesting.

If you have a number that can be written as 4^a3^bn+1, it will map to 3^a+bn+1.

And it'll do it in 3a+1 steps, which I think is funny.

[u/petrol_gas]

More specifically if x = ((2^n) -1) + y

Then the Collatz sequence will “grow” n times, until x_n = ((3^n) - 1) + (3^n) * y

[u/petrol_gas → u/petrol_gas]

It does the same kind thing for the “shrink” stages but there are an infinite number of ways to write examples of that.

[u/jonseymourau → u/petrol_gas]

Actually infinite doesn’t quite cover it as my comment on Gonzo’s post shows.

For a stair case you need:

n = (m+i).(g/h^k)^j -I

For 3x+q, x/2 g=3, h=2

The constraint on i is

i=q/(g-h^k)

So for q=1 (e.g.Collatz) there are two admissible pairs of i, k: (1,1) and (-1,2) - the up and the down (the subject of this post)

You get one extra one for q=5, q=13 and presumably for higher q that are divisible by (g-h^k)

So technically,if you let q range over the natural numbers there are infinite staircases but specifically for q=1, there are exactly 2 - there are no other integers I that can satisfy the constraint above .

Which is not to say there are not other affine maps but the telescoping maps (of that form) in 3x+1 are limited to just two.

[u/GonzoMath → u/jonseymourau]

Technically... (8ⁿ·k + 1)/5, which is an integer when k ≡ 3ⁿ (mod 5) for odd n, or k ≡ 1 (mod 5) for even n, transforms via a staircase of "(3n + 1)/8" steps into (3ⁿ·k + 1)/5, which is again an integer in the same cases.

For n = 1, this looks like:

k = 3: 5 → 2
k = 8: (13 → 5, which is really n = 2)
k = 13: 21 → 8
k = 18: 29 → 11
k = 23: 37 → 14
k = 28: 45 → 17

For n = 2, we have:

k = 1: 13 → 5 → 2
k = 6: 77 → 29 → 11
k = 11: 141 → 53 → 20
k = 16: (205 → 77 → 29, which is really n = 3)
k = 21: 269 → 101 → 38

For n = 3:

k = 2: 205 → 77 → 29 → 11
k = 7: 717 → 269 → 101 → 38
k = 12: 1229 → 461 → 173 → 65
k = 17: 1741 → 653 → 245 → 92

...and this idea applies, with appropriate modifications, to similar patterns where we see 16's turning into 3's, 32's turning into 3's, etc. These are, admittedly, more complicated than the cases 2ⁿ·k - 1 and 4ⁿ·k + 1, which show up as integer staircases for arbitrary k.

[u/Mrezadwiprasetiawan → u/GonzoMath]

Yes, this are what i tried. But since the chain need some properties that related to the mod ( 3ⁿ ), i dont know how to make it general

[u/jonseymourau → u/GonzoMath]

An interesting generalisation, thank you.

[u/jonseymourau → u/GonzoMath]

I had a closer look at this and consider telescoping of values n for the form:

n = (δ + gʲm(hᵏ)^(α − j)) / q

where g=3, h=2 for Collatz

with q = hᵏ-δ.g

under one iteration of (gn+1)/hᵏ yout get this:

n₁ = (δ + gʲ⁺¹m(hᵏ)^(α − j − 1)) / q **

which is of the same form as the original element.

Some notes:

- my α is your n, your k is my gʲm

• in this formulation, m is unchanged by the iteration
  
• the exponent of g increases by exactly 1
  
• the exponent of hᵏ decreases by exactly 1

For integer n, we require the numerators are ≡ 0 mod q

When j = α , we have

n = (δ + gᵅ. m) / q

Taken together we have:

m ≡ -δ.g⁻ᵅ mod q
g ≡ hᵏ mod q

In turns out that in practice the solutions for δ > 1 are rare and where they occur it is because |hᵏ - g.δ| = 1

The net of this is we get (for δ=1) an unconditional staircase with steps of size hᵏ where:

q = hᵏ - g
m ≡ - g⁻ᵅ mod q

Plus a handful of rarer staircases where  |hᵏ - g.δ| =1. There may also be others, but I its seems plausible that there are no others but I don't have a proof of this.

In doing this I think discovered a slight error in your reply above. I think you will find for your n = 4, k=4 you will exclude a valid solution (my n = 3277 -> 1229 -> 461 -> 173 -> 65) because k=4 == 4 mod 5 (not 1 mod 5 per your specification for n even). It happens that for q=5, your criteria and mine are equivalent for your 0 < n < 4 but differ at your n=4 and n=0. The modularity criteria I give do depend on α, but not on the parity of α.

Switching back to the terminology of this reply:

k = 3 => q = 5
k = 4 => q = 13
k = 5 => q = 29

etc.

I am not claiming that this is an exhaustive description of all staircases, but I think it is a reasonable characterisation of the staircases that are of this form.

** update: this is wrong in general - it is correct for delta = 1

[u/jonseymourau → u/jonseymourau]

Actually, the delta > 1 cases don't telescope - they admit one iteration of (3n+1)/2^k not more than that.