Work in progress:

Definition of F(n)

Consider all possible values of

(2^a₁ - 2^a₂ - 3*2^a₃ – 3²*2^a₄ - ... – 3^(m-2)*2^a_m – 3^(m-1))/3^m  for a₂ > 0 a₁-a₂ ≠ 2

(2^a₁ - 1)/3 for a₂ = 0

where a_k > a_k+1 and a_k is a positive integer.

I call expressions of this form the Collatz representation of a number. If the Collatz conjecture is true then every odd positive integer will have a unique Collatz representation. The stopping time for a number with a Collatz representation is a₁.

For brevity I will refer to the above number only by the powers of two i.e. x = (a₁,a₂,…,a_m)

Proof of uniqueness:

Let x have two distinct Collatz representations (a₁,a₂,…,a_m) and (b₁,b₂,…,b_n). Multiply both by 3 and add 1. If a_m=b_n divide both sides by a_m and continue. Since the representations are distinct, at some point a_m-i ≠ b_n-i. Without loss of generality assume a_m-i > b_n-i. Now multiply both sides by 3 and add 1 then divide both sides by b_n-i. The numerator of (b₁,b₂,…,b_n) has been reduced to an odd number but the numerator of (a₁,a₂,…,a_m) is even after applying the same operations to both sides. Therefore (a₁,a₂,…,a_m) ≠ (b₁,b₂,…,b_n).

Let F(n) = the number of integer solutions for a₁=n.

For a₁ = n there are 2^(n-1) - 2^(n-3) possible equations and the number of equations with m powers of 2 will have binomial distribution over n-1 minus the equations where a₁ – a₂ = 2 which will have m powers of 2 binomially distributed over n-3.

e.g. a₁ = 1

(1) = (2¹ - 1) / 3 = 1/3

so F(1) = 0

e.g. a₁ = 2

(2,1) = (2² – 2¹ - 3) / 3² = -1/9

(2) = (2² – 2⁰) / 3 = 1

so F(2) = 1

e.g. a₁ = 3

(3,2,1) = (2³ – 2² – 3¹*2¹ – 3²) / 3³ = -11/27

(3,2) = (2³ – 2² – 3¹) / 3² = 1/9

(3,1) = (2³ – 2¹ – 3¹) / 3² = 3/9

(3) = (2³ – 2⁰ ) / 3 = 7/3

None are integers so F(3) = 0

Note that (3,1) is the same as (1) since a₁ – a₂ = 2 so this equation would not be included.

Some Observations

What other restrictions can we place on this expression to ensure it is an integer?

1. Clearly a₁ ≡ a₂ mod 2 otherwise 3 does not divide 2^a₁ - 2^a₂

If we ignore a₁ then what restrictions can we place on the remaining powers of 2? The answer is none. To see this first note that 2 is a primitive root of 3ⁿ so the order of 2 mod 3ⁿ is 2*3^n-1

Now  2^a₁ - 2^a₂ - 3*2^a₃ – 3²*2^a₄ - ... – 3^(m-2)*2^a_m – 3^(m-1) = 2^a₁ - 2^a₂ - 3*x

And since 2 is a primitive root of 3ⁿ

2^a₁ - 2^a₂ - 3*x ≡ 2^a₁ - 2^β mod 3ⁿ for some β between 1 and 2*3^n-1

So 2^a₁ - 2^β ≡ 0 mod 3ⁿ has the solution a₁ = β+q*2*3^n-1 q=0,1,2,….

For example consider (a₁,1,2) what is the solution for a₁?

2^a₁ – 2¹ - 3*2² – 3² = 2^a₁ – 23 and 2¹¹ ≡ 23 mod 3³

So a₁ = 11 +q*18. Letting q=0 gives (2¹¹ – 2¹ - 3*2² – 3²)/3³ = 75

But the Collatz representation for 75 is (11,3,1). So (11,3,1) = (11,1,2). What happened to our uniqueness for the Collatz representation? The answer is we relaxed the condition that a_i is monotonically decreasing. This raises some questions.

1. How many different ways can we represent a number if the a_i are not monotonically decreasing?

2. What is the minimum number of terms needed to represent a number if we relax the monotonic condition?

Transformations of x

Consider the term a_m. This can tell us something about the number. Let x = (a₁,a₂,…,a_m) Since x is odd it is 1,3,5 or 7 mod 8.

(8n +1) * 3 + 1 = 24n + 4 = 2² * ( 6n +1) so the value of a_m must be 2.

(8n + 3) * 3 + 1 = 24n + 10 = 2¹ * (12n + 5) so the value of a_m must be 1.

(8n + 5) * 3 + 1 = 24n + 16 = 2³ * (3n + 2) so a_m >= 3 since 3n+2 is even if n is even.

(8n + 7) * 3 + 1 = 24n + 22 = 2¹ * (12n + 11) so a_m is 1.

Let’s take a look at some transformations of x. It’s easy to show that if x = (a₁,a₂,…,a_m) then (a₁+2,a₂+2,…,a_m+2) = 4x + 1 and 4x + 1 is 5 mod 8. If we think about the numbers we can create for a₁ = n we can show a one to one correspondence between the a₁ = n where x ≡5 mod 8 and all the numbers for a₁ = n-2. a₁ = n inherits all it’s numbers 5 mod 8 by simply “inflating” the numbers a₁ = n-2 by adding 2 to all the powers of 2. We now have some of the answer to the question what is F(n)?

F(n) = F(n-2) + the numbers a₁=n where a_m is 1 or 2.

Now take a look at a_m=2. Clearly (3x + 1)/4 : (a₁,a₂,…,a_m-1,2) → (a₁-2,a₂-2,…,a_m-1-2) and since x ≡ 1 mod 8, (3x +1)/4 ≡ 1 mod 6.

Conversely if x = (a₁-2,a₂-2,…,a_m-1-2) ≡ 1 mod 6 then (4x – 1)/3 : (a₁-2,a₂-2,…,a_m-1-2) → (a₁,a₂,…,a_m-1,2) ≡ 1 mod 8. So we have shown a 1 to 1 correspondence between the numbers x ≡ 1 mod 8 for a₁ and x ≡ 1 mod 6 for a₁-2.

Last we look at a_m=1. Again (3x + 1)/2 : (a₁,a₂,…a_m-1,1) → (a₁-1,a₂-1,…,a_m-1-1) and since x is 3 or 7 mod 8, (3x+1)/2 ≡ 5 mod 6.

Conversely if x = (a₁-1,a₂-1,…,a_m-1-1) ≡ 5 mod 6 then (2x – 1)/3 : (a₁-1,a₂-1,…,a_m-1-1) → (a₁,a₂,…,a_m-1,1) ≡ 3 or 7 mod 8. So we have shown a 1 to 1 correspondence between the numbers x ≡ 3 or 7 mod 8 for a₁ and x ≡ 5 mod 6 for a₁-1.

We now have a full definition for F(n).

                                      F(n) = F(n-2) + F(n-2)_{x≡1 mod 6} + F(n-1)_{x≡5 mod 6}  Except when n=4 since (4x-1)/3 maps 1 to itself.

This also gives us a definition for the number of integers a₁=n with a given value of m since

                                      F(n)_m=b = F(n-2)_m=b + F(n-2)_{x≡1mod6 m=b-1} +F(n-1)_{x≡5mod6 m=b-1}

If we define S(n) as the sum of the integers that comprise F(n) then

                                      S(n) = ∑_a1=n-2 (4x+1) + ∑_{a1=n-2 x≡1mod6} (4x-1)/3  + ∑ _{a1=n-1 x≡5mod6} (2x-1)/3

If we assume an even distribution of x over mod 6 then we can interpret F(n) as 1/3 F(n-1) +4/3 F(n-2). Solving this difference equation with initial conditions F(0) = 0, F(1) = 1, F(2) = 0 gives F(n) = 9/28 (4/3)ⁿ – 4/7(-1)ⁿ and lim n→∞ F(n)/F(n-1) = 4/3.

Questions:

Yes.

If n ≡ 0 mod 2 then (2ⁿ – 1) / 3 is an integer

If n ≡ 1 mod 2 then consider the 3 cases n ≡ 1 mod 6, n ≡ 3 mod 6 and n ≡ 5 mod 6

If n ≡ 1 mod 6 then (2ⁿ – 2³ – 3) / 3² is an integer

If n ≡ 3 mod 6 then (2ⁿ – 2⁵ – 3) / 3² is an integer

If n ≡ 5 mod 6 then (2ⁿ – 2¹ – 3) / 3² is an integer

2. Is the sequence F(n) monotonic for n > 3 ?

Yes if the distribution of the numbers mod 6 for a₁=n is uniform

3. Does the sequence F(n)/F(n-1) converge ?

Yes if the distribution of the numbers mod 6 for a₁=n is uniform then F(n)/F(n-1) → 4/3 as n→ ∞.

4. Can we show the distribution of x mod 6 becomes uniform?