Function for a new math pattern that emerged while working on the Collatz conjecture

Question

So, this is a follow up to my previous question on the same topic, and in this question, I used the same technique, only with a larger value. Here's the set below:

S no.	Resultant Value
1	227
2	14549
3	931157
4	59594069
5	3814020437
6	244097307989
7	15622227711317
8	999822573524309

I know this is a short set of values, but the calculations were getting large, and it was hard to generate more values. However, if they are needed for finding the function, do let me know and I will try my best to provide.

Please refer to the answer of my previous post for the general base formula of sorts for this one.

I tried a lot to get a function that generates the value for this, but I really couldn't find one, as I am also not as much of an expert at math, so I would really appreciate your help in finding the function for this.

Answer 1

$7+22\cdot 9\cdot 4^{z+1} + \cdot \sum\limits_{k=0}^{z+1} 22\cdot 4^k$

with $z=3n-1, n\geq 0$

But as said in the first post, there are many ways to write this sequence. Why do you want to use those weird sums and not the formulas proposed in the first post or by Gottfried?

EDIT:

Depends on what representation you prefer, but let's start with $a_{n+1}=64\cdot a_n+21$

I will assume you have some knowledge of the tree structure.

There are branches where the exponent of 2 are odd, in other words, when you multiply elements of that branch by 3 (+1), the number of division by 2 is odd:

$\begin{array}{c|lcr} \text{division by}&\text{branch element} \\ \hline 2^1&4n+3\\ 2^3&16n+13\\ 2^5&64n+53\\ ...&... \end{array}$ They all lead to $6n+5$

There are branches where the exponent of 2 are even, in other words, when you multiply elements of that branch by 3 (+1), the number of division by 2 is even:

$\begin{array}{c|lcr} \text{division by}&\text{branch element} \\ \hline 2^2&8n+1\\ 2^4&32n+5\\ 2^6&128n+21\\ ...&... \end{array}$ They all lead to $6n+1$

Now apparently you are only interested in branch values $x\equiv 2\mod (3)$ and since elements on the branches are cyclic$\mod (3)$, you have to pick an element every 3 elements on that branch, hence the multiplication by $64$ and addition of $21$ to get the third next element (you multiply by $4$ and add $1$ to get the next element).

since the first $x\equiv 2\mod (3)$ can be in position 1, 2 or 3 of a branch, the starting value $a_0$ of one of your sequences can be of one of those 6 forms (the above forms with the additional constraint that they are $\equiv 2 \mod(3)$):

$\begin{array}{c|lcr} \text{division by}&a_0\equiv 2\mod(3) \\ \hline 2^1&12n+11\\ 2^2&24n+17\\ 2^3&48n+29\\ 2^4&96n+5\\ 2^5&192n+53\\ 2^6&384n+149\\ \end{array}$

e.g. $a_0=227$ is of the form $12n+11$ and only have 1 division by 2, but any number of this forms is valid.

Note: if you are interested in $x\equiv 1\mod (3)$ or $x\equiv 0\mod (3)$, you can apply the same reasoning for calculating the $a_0$ admissible forms.

Now, if you are trying to find which form or which representation will show you a pattern to know how many odd steps away (number of multiplication by 3 (+1)) you are from 1 (e.g. $53,113,227$ are 2 odd steps away from 1), good luck. Nobody found anything yet (but there are other representations for that).

Answer 2

In general you can use this formula for $\quad O>1$

$$n = \frac{2^{E}-b}{3^{O}}\tag1$$

$$b = 3^{O-1} + \sum_{i=0}^{O-2} {3^{i} \cdot 2^{\left(\sum_{j=1}^{O-1-i} {a_{j}}\right)}} $$

which is obtained by considering the sequence of odd numbers

$n_{i}=\frac{3 \cdot n_{i-1} +1}{2^{a_i}}\quad$ with $\quad n_0=n$

with $O$ the number of odd elements and with $E$ the number of even elements in the sequence and $n_O=1$.

In your case $\quad O=2\quad $ and $\quad b = 3 + 2^{a_1} $

for $n=227=\frac{2^{11}-5}{3^2}\quad$ then $\quad E=11\quad$ and $\quad b=3 + 2^1 = 5$

$n=14549=\frac{2^{17}-131}{3^2}\quad$ then $ \quad E=17\quad$ and $\quad b=3 + 2^7 = 131$

$n=931157=\frac{2^{23}-8195}{3^2}\quad$ then $\quad E=23 \quad$ and $\quad b=3 + 2^{13} = 8195$

$\cdots$

$n=\frac{2^{11+6\cdot k}-(3+2^{1+6\cdot k})}{3^2}\quad$ with $\quad k\geq 0 \quad$then $\quad E=11+6\cdot k \quad$ and $\quad b=3 + 2^{1+6 \cdot k}$

Edit I add other ways to write the same result as a summation

Let's consider the Branch-1 in the previous post:

$$3 + \sum_{i=0}^{3\cdot k+1} 10 \cdot 4^i=3 +10\cdot \sum_{i=0}^{3\cdot k+1} 4^i$$

you can verify that

$$\sum_{i=0}^{3\cdot k+1} 4^i=\frac{16\cdot64^k-1}{3}$$

then for Branch-1:

$n=\frac{2^{9+6\cdot k}-(3+2^{5+6\cdot k})}{3^2}=\frac{32\cdot 64^k \cdot(16-1)-3}{3^2}= 3+10\cdot\frac{16\cdot 64^k -1}{3}=3 +10\cdot \sum\limits_{i=0}^{3\cdot k+1} 4^i$

for Branch-2:

$n=\frac{2^{10+6\cdot k}-(3+2^{2+6\cdot k})}{3^2}= 3+60\cdot 64^k + 10\cdot\frac{16\cdot 64^k -1}{3}=3 +60\cdot 64^k + 10\cdot \sum\limits_{i=0}^{3\cdot k+1} 4^i$

for Branch-3:

$n=\frac{2^{11+6\cdot k}-(3+2^{1+6\cdot k})}{3^2}= 3+174\cdot 64^k + 10\cdot\frac{16\cdot 64^k -1}{3}=3 +174\cdot 64^k + 10\cdot \sum\limits_{i=0}^{3\cdot k+1} 4^i$

Note that $113-53=$60 and $227-53=$174.