Particular base-10 digit patterns in Collatz

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.

• $R(359)=539$ as you say, but $R(1437)=R(5749)=539$ too so I am not sure about the point you are making Oct 9, 2018 at 22:36
• 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. Oct 9, 2018 at 22:52
• 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. Oct 12, 2018 at 22:58

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.

• 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. Oct 10, 2018 at 22:01
• 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. Oct 11, 2018 at 8:19