u/Septembrino

Hi. I have been working for Varsity since 2017 and my income is usually pretty unpredictable. I guess that most of you can understand this. I am almost half retired because I have a pension, but my income is nowhere near the required amount.

Now, I have to declare my anual income to be able to sponsor my son-in-law's green card. Last year I made a bit over 5K, but this year I am already a bit over 7.

The questions are:

How much do you think I should declare? Would 10K be a credible amount if last year was about 50% of that? How would you calculate an amount for such a variable source of income?

Over the last few years, I made between 9 and 12, except for 2025. And they require, and pay attention, to the latest tax return, but the allow you to present up to 3 years, which I will do to make my case more credible.

Thank you for helping we with this family issue

We know exactly by what we can divide 1•3^n - 1. So, I compared the divisors of k*3^n - 1, where k are Fermat numbers (2^m + 1), to k = 1.

In this post, the word divisor refers to 2ᵐ, m > 0, and the expression is divisible by say 2⁵, means that the maximum divisor is 32. In some cases, we want to make sure that the quotient is odd. We can also say that the k•3ⁿ - 1 is congruent to 32 mod 64.

The first difference between k = 1 and k = 2^m + 1 happens at 2ᵐ⁻².

Let's consider the case where k = 257. Here m = 8, 2ᵐ⁻² = 64

257•3ⁿ - 1 = 256•3ⁿ + (3ⁿ - 1).

For n < 64, the expression will have the same divisors as k = 1. So, we can predict all of them. If n = 4, 257•3⁴ - 1 = 2⁸•3⁴ + (3⁴ - 1). The second term is divisible by 2⁶, while the first can be divided by 2⁸. So, the sum's divisor is the lesser one, 2⁶. Min {2⁶, 2⁸} = 2⁶.

For n ≥ 64, there are 2 cases:

1. 3ⁿ - 1 is divisible at least by 2⁹. In this case, the 1st term is the one that determines the divisor. The expression can be divided by the lesser power, which is 2⁸.
2. If both terms are divisible by 2⁸, both quotients are odd. So the sum is even. It's hard to predict by what the expression will be divisible. In general, we only know that the divisor is at least 2⁹. On the other hand, these unpredictable cases allow for divisors really huge.

I began developing conjectures that might help in the cases where the prediction fails. I also proved a couple of them, and I will keep working on the rest.

The pattern observed for 257 is common to all Fermat numbers, beginning at k = 17. The previous Fermat numbers are of no interest since 3 and 9 are multiples of 3 and 5 is a regular k. 5•3ⁿ - 1 can only be divided by 2 or by 4.

For 17, the first exception happens at n = 2², for 33, at n = 2³, for 65, at n = 2⁴, etc. These show up every 2ᵐ⁻¹. So 2ᵐ⁻² is a factor always present.

Summary for 17

p is a non negative integer

To clarify this table: For n = 20, we consider 4(4p+1). 20/4 = 5, and 4p+1 produces a solution (p=1). Because the divisor foe the 2nd line is 32, we know that 17•2²⁰ - 1 is divisible by 2⁵.

For 33, the top of the table is similar, but repeating factor is now 8

https://preview.redd.it/vfatd7ymijhh1.png?width=540&format=png&auto=webp&s=4ac1ac378aa4f37513dee01d54ed61088e6e3325

Observe that the table for 17 contains 16p + 11 and the one for 33 contains 32p+11. The divisor in the first case will repeat every 4•16 numbers, in the second case, every 8•32.

Comparing both tables, there are coincidences up to certain point, but then the pattern change.

That keeps happening for other Fermat numbers. There is a table showing that in my previous post. Notice that, in the table, n was divided by 2ᵐ⁻² to make the differences and similarities clearer.

As usual, ask away if something is unclear or you want more information. I will keep you posted when I notice more interesting facts that might be useful for people working on the Collatz conjecture.

In this subreddit, there is a part 1, where I studied k = 257. I kept studying several Fermat numbers, the ones of the form 2^m +1 , and I observed a few regularities.

Some are in the following formulas:

[(2^m + 1)•3^((2p)(2^(m-2)) - 1]  / 2^m is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((4p+1)(2^(m-2)) - 1]  / 2^(m+1) is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((8p+7)(2^(m-2)) - 1]  / 2^(m+2) is odd, p a non negative integer, m ≥ 4

[(2^m + 1)•3^((16p+3)(2^(m-2)) - 1]  / 2^(m+3) is odd, p a non negative integer, m ≥ 5

[(2^m + 1)•3^((32p+27)(2^(m-2)) - 1]  / 2^(m+4) is odd, p a non negative integer, m ≥ 6

...

The exponents have 2 factors, one is of the kind ap+b. If making a table, there is clearly a variable part and a constant part. The boundary between these regions in a diagonal line. The other factor is a power of 2 and depends on m. The first factor seems to be constant from certain m on.

The difference between 2 consecutive b's are powers of 2 in increasing order, or their negative version, or even powers of 2 multiplied by 3.

There is a part 1, where I studied k = 257 (link in comments). I kept studying several Fermat numbers, and I observed a few regularities. Some are in the following formulas and tables.

I will be extending the tables and posting them here probably in a few days.

Some of the formulas I got are:

[(2^m + 1)•3^((2p)(2^(m-2)) - 1]  / 2^m is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((4p+1)(2^(m-2)) - 1]  / 2^(m+1) is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((8p+7)(2^(m-2)) - 1]  / 2^(m+2) is odd, p a non negative integer, m ≥ 4

The exponents have 2 factors, one is of the kind ap+b, and the b's are in the simplified table below. The complete (ap +b)'s are in the table 2.

Table 1, simplified version of the table 2. It contains only the b's for exponents of the form (ap + b)•2^(m-2), m≥3, where a is a power of 2, p a non negative integer that provides the repetition of the same divisor.

Table 2, containing a and b for the exponents and the divisors for different m's, m&gt;3

There is clearly a variable part and a constant part, marked by the horizontal lines. The boundary between these regions in a diagonal line. The difference between 2 consecutive b's are powers of 2 in increasing order, or their negative version, or even powers of 2 multiplied by 3. You can see them in the tble 3.

Sometimes the differences are negative (in green) and in other cases the triple that they should be (in orange). Not sure yet why exactly that happens.

We know the divisors of 1•3^n - 1 from the p-adic valuation. We also can determine the divisors of k•3^n - 1, where k = 3, 9, etc. since these are shifts of k = 1.

NOTE: When I type "divisors" in this post, I mean the greatest divisors, the one that produce odd quotients.

We also know that k = 5 and 7 mod 8 produce a predictable pattern. The divisors are 4 for n = 0 mod 2 and 2 for n = 1 mod 2 (in the case of k = 5 mod 8), or viceversa (k = 7 mod 8).

k = 5

k = 7

Other than this, it seems that the divisors can be partially predicted by replying a few questions (like: by what do we divide k-1, 3k - 1, 9k-1). Once we locate the first divisors, there is a lot we can deduce, but, sadly, the rest needs to be studied on a one-by-one base.

I noticed, though, that for k = 257 = 2^6 + 1, the divisors are the same as the ones where k = 1 except in the case of n = 64 mod 128. It's still nice to be able to find similarities between 2 different k's.

k = 1. Compare the divisors row between k = 1 and k = 257

k = 127

If we consider 257*3^n - 1 = 256*3^n + (3^n - 1), there are 2 cases:

a) The divisor of 3^n - 1 is not 256, in this case, whichever is lesser divides the whole expression. The divisors are the same as for k = 1

Examples: 256*3^4 + (3^4 - 1). Since 3^4 - 1 is divisible by 16, then the whole is divisible by 16.

256*3^128 + (3^128 - 1). Since 3^128 - 1 is divisible by 512, then the whole is divisible by 256.

b) The divisor of 3^n - 1 is 256. Then there is no easy way I know to predict the divisors of the whole. 2 fractions can add up to an integer.

That might a blessing in disguise, though. This is the only way of obtaining divisors greater than 256.

For now I located up to 16384.

Summary of divisors that differ from k = 1 and their fractal-like structure

I am also planning to observed other 2^m + 1. BTW, I saw online that these are called Fermat numbers.

k = 129

I will keep you posted.

We know the divisors of 1•3^n - 1 from the p-adic valuation. We also can determine the divisors of k•3^n - 1, where k = 3, 9, etc. since these are shifts of k = 1.

We also know that k = 5 and 7 mod 8 produce a predictable pattern. The divisors are 4 for n = 0 mod 2 and 2 for n = 1 mod 2 (in the case of k = 5 mod 8), or viceversa (k = 7 mod 8).

Other than this, it seems that the divisors can be partially predicted by replying a few questions (like: by what do we divide k-1, 3k - 1, 9k-1). Once we locate the first divisors, there is a lot we can deduce, but, sadly, the rest needs to be studied on a one-by-one base.

I noticed, though, that for k = 257 = 2^6 + 1, the divisors are the same as the ones where k = 1 except in the case of n = 64 mod 128. It's still nice to be able to find similarities between 2 different k's.

If we consider 257*3^n - 1 = 256*3^n + (3^n - 1), there are 2 cases:

a) The divisor of 3^n - 1 is not 256, in this case, whichever is lesser divides the whole expression. The divisors are the same as for k = 1

Examples: 256*3^4 + (3^4 - 1). Since 3^4 - 1 is divisible by 16, then the whole is divisible by 16.

256*3^128 + (3^128 - 1). Since 3^128 - 1 is divisible by 512, then the whole is divisible by 256.

b) The divisor of 3^n - 1 is 256. Then there is no easy way I know to predict the divisors of the whole. 2 fractions can add up to an integer.

That might a blessing in disguise, though. This is the only way of obtaining divisors greater than 256.

For now I located up to 16384. I am also planning to observed other 2^m + 1. BTW, I saw online that these are called Fermat numbers.

I will keep you posted.

I was working with 11*3^n - 1 and trying to find by which power of 2 they are divisible to produce odd integers. Since there is a 2 every other power, I removed these, getting the 2nd level, then I removed the 8's, getting the 3rd level, then the 16’s, etc.

For each level, except the 1st, the base line, code 0 corresponds to 2^n, and 2^(n+i), was assigned code i. Example: if the base line is 8, then 8 was assigned to code 0, 16 to code 1, 32 to code 2, etc.

For the 1st level, the base line was 2, code 0. But, since there are no divisors 4, 8 was assigned to code 1, 16, to code 2, etc.

The codes in bold fond correspond to higher order powers, enclosing "a period". These powers are part of higher order structures.  For each level, I studied a random section of at least 2 consecutive periods. I studied till level 14, observing a similar behavior all along.

Tables available under request.

Levels 1 to 4. Observe the orange codes. The blue codes are actually part of higher levels

Levels 5 to 8. Observe the orange codes. The blue codes are actually part of higher levels

I suspect that 17*3^n - 1, and other coefficients of the exponential part, behave in a similar way.

I have been working on expressions of the kind k 3^n - 1. By what power of 2 can we divide a number we obtain at the top of the matrix? I noticed that, in some cases, we can predict the power. If k = 5 or 7 mod 8, what I call regular k's, we get only divisors 2 and 4.

Interestingly enough the ones that are not regular don't present any 4's, but there is a 2 every other divisor. So, we can also predict the 2's and maybe some of the small powers in some cases.

I thought of discussing some diophantine equations that can be used to predict these divisors, mostly when we get a hight power of 2.

Assuming that k 3^n - 1 can be divided by 128 (that's an example), then k 3^n - 1 = 128 x, and that generates this diophantine: k 3^n - 128 x = 1.

Restrictions: k can't be 5 or 7 mod 8, and both, k and x, have to be odd. So, we should also take that into account.

Predicting high divisors is not an easy task. If you have any tips to help here, feel free to share.

This is a follow up of this thread:

https://www.reddit.com/r/Collatz/comments/1sviw0y/working_on_prediction_of_collatz_trajectories/

In it, I explained how to get the same number in the column 1 (C1), in the colored row, as the one get if using k as a seed number. To do that, we can use the seed number 2k-1.

To get the same number in C2, we can do 4k-1.

These expressions get really complicated as n (in Cn) grows.

To get the same number in column 3, in the colored row, we need to do 64k - 7.

Example 1:

Matrix k = 3. Observe the 13 in column 3 (C3)

Matrix k = 185, observe the same 13 in C3. 185 = 3x64 - 7

To get the same number in C4 we have to use k*2^18-(2^18-1)/27 will provide the same number in C4 (column 4), colored row as k.

Example 2: k = 1 and k = 252435.

Matrix k - 1, showing a 13 in Column 4 (c4)

Matrx k = 252435, showing the same 13 in C4.

Note how large most numbers are. The divisor is also large, which makes sense since 13 is much smaller than the rest of the numbers,

To understand the matrices, we can generate them multiplying the previous number by 2 and adding 1. Going across, we multiply the previous number by 3 and we add 2. In a diagonal number we get pieces of the Syracuse function. More information on matrices in these threads:

https://www.reddit.com/r/Collatz/comments/1liaya4/collatz_matrices_base_on_the_p2p1_theorem/

https://www.reddit.com/r/Collatz/comments/1s1d02t/creating_collatz_matrices_using_a_spreadsheet/

The rows on top of the matrix body are explained here: https://www.reddit.com/r/Collatz/comments/1s5w71y/row_on_top_of_the_collatz_matrices/

This is related to the Collatz matrices. See link below. Please, take a look at these before reading this post. There are more links in my profile.

https://www.reddit.com/r/Collatz/comments/1s1d02t/creating_collatz_matrices_using_a_spreadsheet/

The numbers in the colored row below are part of the Collatz sequences.

I realised that I can get the exact same number in the column 1 (C1) if I use (2k-1) instead of k, the seed of a matrix.

Example: if k = 3, I get a 1 on top of the 3. If instead I use 3x2-1 = 5, I get the same 1 above the 5. The same happens if I use 5x2-1=9, 9x-1 = 17, etc. Going that way, I can get the exact same 1 on top of a large number. BTW, when I get a 1 in that row, it means that there is a Collatz sequence that got to 1.

k=3, 1 in C1 and in C2, 1 in C3

k=232769, 1 in C1 and 49153 in C2. Observe that the other numbers are large, but the 1 repeats

Of course, this is not magic. The equations predict that. See proof in comments.

Logically, these very low reduced top numbers in the colored row are possible since large powers of 2 are their divisors.

Similarly, for k = 7, there is a 3 on top of the 7. The equations (7x2-1 = 13, 13x2-1 = 25, etc.) predict that there will also be a 3 on top of large numbers like 12289.

k=7, 3 in C1 and 5 in C2

k = 12289, 3 in C1 and 18433 in C2

I am only posting 2 examples, but that works for any k.

I also discovered that, by doing 4k-1, I can predict the number in C2 (column 2). As you can see from the previous examples, that number is not the same for each pair of matrices.

From k = 3, and using 11 = 4x3-1, we observed that the 49 in the 3rd column repeats, for example, if k = 697 = 11x64-7

k = 11, 1 in C2, same as in K=3, and 49 in C3

k = 697, 49 in C3

Right now I am working on the proof of this statement. I will post it as soon as I am done writing it.