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:


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.

  • $\begingroup$ $R(359)=539$ 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, 2018 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, 2018 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, 2018 at 22:58

1 Answer 1


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.

  • $\begingroup$ 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, 2018 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, 2018 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.