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Note: I will use the abbreviation RCF for the Reduced Collatz Function.

The arrangement of certain specific digits produce a particular pattern on the next iteration of the Reduced Collatz Function. Before dwelving into the pattern, lets have a recap of the function.

The RCF applies only to odd number, and is defined as: $$R(n)=\frac{3n+1}{2^r}$$ where $r$ denotes the largest exponent of the largest power of two that divides $3n+1$. Denote $i$ as the $i$'th iteration of the function. Define also $r$ and $n\in\mathbb{Z_0^+}$.

Lets do some examples (the initial numbers are explicitly chosen):

\begin{array}\hline R(23)\to35\to53\to5\to1 \\ R(27)\to...\to395\to593\to... \\ R(359)\to539\to809\to607\to... \\ R(3559)\to5339\to8009\to6007\to... \\ R(35559)\to53339\to80009\to60007\to... \\ \end{array}

Note these are just observations of mine. You see the patterns? Digits $3$, $5$ and $9$ are noticeable. $5$ and $3$ swapped places. $9$ is at the same digit-position. However, in $R^i(n)$, the digits $9$ and $5$ can't be the first digit or else $R^{i+1}(n)$ will not produce the desired effect.

Q: Im sure these patterns have been researched on, any literature about these specifics?

When $R^i(n) = 539$ then $n=359$ for $i-1$. This particular iteration of the function looks like it has an inverse. There might be examples that I am not aware of that work for other numbers but I have not found any. I've seen these numbers in other sequences, namely prime numbers.

The following formula gives primes where $n$ is defined by a sequence that can be found here:

$$\frac{320\cdot10^n+31}{9}$$

Set $n=7$

This results in the number: $355555559$. Whatever $n$ we set, the count of $5$'s between $3$ and $9$ will be same as $n$.

If we put this into the RCF as input? Well we get $R^i(355555559)\to R^{i+1}(533333339)$.

There does not seem to be a relation between the number of digits between $5$ and $9$ and the primes in the collatz function, but the most significant and least significant digit might have?

Q: Are there any core math on this subject of the Primes and Collatz? I am open to listen.

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  • $\begingroup$ $R(359)=539$ as you say, but $R(1437)=R(5749)=539$ too so I am not sure about the point you are making $\endgroup$
    – Henry
    Oct 9 '18 at 22:36
  • $\begingroup$ Read $i+1$ next iteration. Im not talking about the numbers infront or beyond those two, even though that might be good research too. Although they might be connected some way. But I want to keep it simpler than all this as groundwork. $\endgroup$ Oct 9 '18 at 22:52
  • $\begingroup$ Three threes is nine. Three times a sequence of threes gives a sequence of nines. Follow those threes by a five, then three times 5 plus one is sixteen. The one ten carries to the sequence of nines and by induction turns them into a product of a power of ten. If they're preceded by a 5, the highest 10 carries to turn 3x5 into 16. Dividing by two turns the six of the sixteen into a 3 and 16 timesthe power of ten into 8 times a power of 10. $\endgroup$ Oct 12 '18 at 22:58
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By applying successively $R(n)=\frac{3}{2^r}\cdot n + \frac{1}{2^r}$ to $n_0, n_1,...$

$n_1 = \frac{3}{2^{r_1}}\cdot n_0 + \frac{1}{2^{r_1}}$

$n_2 = \frac{3}{2^{r_2}}\cdot n_1 + \frac{1}{2^{r_2}} = \frac{3^2}{2^{r_1+r_2}}\cdot n_0+\frac{3^1}{2^{r_1+r_2}}+\frac{3^0}{2^{r_1}}$

we find $n_i$ which is $i$ step away from $n_0$

$$n_i = \frac{3^i}{2^{r_1+r_2+...+r_i}}\cdot n_0+\frac{3^{i-1}}{2^{r_1+r_2+...+r_i}}+\frac{3^{i-2}}{2^{r_1+r_2+...+r_{i-1}}}+...+\frac{3^0}{2^{r_1}}$$

We set $\delta$ (positive) being the parts which are not dependent on $n_i$ or $n_0$ to simplify:

$$\delta = \frac{3^{i-1}}{2^{r_1+r_2+...+r_i}}+\frac{3^{i-2}}{2^{r_1+r_2+...+r_{i-1}}}+...+\frac{3^0}{2^{r_1}}$$

So we have:

$$n_i = \frac{3^i}{2^{r_1+r_2+...+r_i}}\cdot n_0+\delta$$

and by setting $j=r_1+r_2+...+r_i$ (Note: $j\geq i$ since all $r_i\geq 1$)

$$n_i = \frac{3^i}{2^j}\cdot n_0+\delta$$

Since obviously $$k\cdot3^i=\frac{3^i}{2^j}\cdot k\cdot2^j$$

We add both LHS and both RHS and we get this formula $$(n_i+k\cdot3^i)=\frac{3^i}{2^j}\cdot (n_0+k\cdot2^j)+\delta$$ It means that if you successively apply $R(n)$ to $n_0$ till you get $n_i$, and if you apply the same $R(n)$ (or should I say the same successive $r_i$) to $(n_0+k\cdot2^j)$, you'll reach $(n_i+k\cdot3^i)$

Elements in the sub-sequence will look like this: $$\{n_0+k\cdot2^j, n_1+k\cdot3^1\cdot2^{j-r_i}, n_2+k\cdot3^2\cdot2^{j-r_i-r_{i-1}},...,n_i+k\cdot3^i\}$$ e.g: If you take $\{n_0=359, n_1=539, n_2=809,...,n_{i-1}=607, n_i=911\}$ and you set $k=10^2$ and $2^j=2^5$, the above sub-sequence will be:

$$\{359+3200, 539+4800, 809+7200,...,607+5400,911+8100\}$$ or $$\{3559, 5339, 8009,...,6007,9011\}$$ You can do the same with $n_0=3559...$ and $k=10^3$ (or if you start with $n_0=359$, you just set $k=10^2+10^3$). Note: setting $k=100, 1100, 11100, 111100, 1111100....$ to get your different sub-sequences is probably the main key to this "digit pattern" behaviour.

$$$$ By taking $k=10^n$ you get your extra digits, but to see a "digit pattern" you can't do that with any number. My guess is that $2^j$ must be 2 digits, the first digit of $n_0$ must be the same as the first digit of $2^j$ and the second must be the sum of the 2 digits of $2^j$.

e.g. take $2^j=2^4=16$ -> $n_0$ start with a $1$ like $2^j$, and the second digit must be $1+6=7$. Now if we set $n_0=179$ (or 173 or 175...), and you add $1600$ to it, you'll get $1779$. If you add $16000$ again you'll get $17779$...and if you look at the successors they exhibit the same "digit pattern" behaviour

e.g. take $2^j=32$ -> $n_0$ start with $3$ and second digit must be $3+2=5$ like $359$.

EDIT: It is not limited to 2 digits. If you take $n_0=1421$ and $2^j=128$ it works too (1421+12800=14221....142221.....1422221.....), but the logic to find working number is a bit different.

My guess is that if you find a working number $n_0$, $n_1$ will be transformed the same way as $k\cdot 2^j$ and exhibit the same "digit pattern", but I still need to check this part

Anyway, these are particular cases that exhibit properties of the Collatz function AND properties of the decimal base. I don't know if you can get anything from it or if there is any link to prime numbers. It might be interesting to investigate.

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  • $\begingroup$ Instead of raising to base $3$, have you tried base $8$ or $16$? I indirectly use those bases, but its much simpler when those bases are computed in binary. What is your $\delta$-symbol representing, and can it be negative? Also I can't get my head around $r^i$? It looks more complicated than $\frac{3^0}{2^1}+\frac{3^1}{2^1}\cdot35559 = 53339$ Even if I dont understand your definitions entirely I hope you can crunch it down to some simpler formula I can understand and I wish you good luck. $\endgroup$ Oct 10 '18 at 22:01
  • $\begingroup$ My intent was not to work in base 3, but indeed you could see it that way since there are only exponents of 2 and 3. I added some details to my post but I have not much time to investigate more deeper, sorry. $\endgroup$
    – Collag3n
    Oct 11 '18 at 8:19

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