Is there any known finite sequence of positive integers $a_i$ such that 2ドル^{a_k}-3^k$ is a positive proper divisor of $\sum_{i=0}^k 3^i2^{a_k-a_i}$? [closed]

Question 1

Is there any known finite sequence of positive integers $(a_i)_{i=0}^k$ such that 2ドル^{a_k}-3^k$ is a positive proper divisor of $\sum_{i=0}^k 3^i2^{a_k-a_i}$?

Any nontrivial loop in the Collatz algorithm generates a solution (and indeed an infinite family of solutions); but it's unclear to me if the converse is true. So I wonder if any known solutions exist.

Question 2

This question is similar to: If $\sum\limits_{i=0}^{k-1} 2^{a_i} 3^i$ is divisible by 2ドル^m - 3^k,ドル then the quotient is 1ドル$ or 2ドル$ (Existence of cycles in the Collatz-problem). If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.

Question 3

If $k=2$ , the problem is already solved no cycles are of length 2ドル$ . As of 2025ドル,ドル the best known bound on cycle length is 217ドル$ 976ドル$ 794ドル$ 617ドル$

Question 4

@Lhachimi Look carefully at the equation: in the example by DejanGovc, 2ドル^{a_k}-3^k = 2^4-3^2=16-9=7$.

Question 5

the only known sequence that satisfies your diophantine equation is $S_i$

which will come from $T^k(n)=n$ , i.e a collatz cycle which is still an open problem :(

$$T^{k}\left(n\right)=\displaystyle \frac{3^{k}n+\displaystyle \sum_{i=1}^{k}3^{k-i}2^{S_{i-1}}}{2^{S_{k}}}=n$$

So,

$$n=\displaystyle\frac{\displaystyle\sum_{i=1}^{k}3^{k-i}2^{S_{i-1}}}{2^{S_k}-3^k}$$

However , you still can break the only possible cycle : $T^k(1)=1$

like by allowing the denominator to be negative , i.e : 2ドル^{S_k}-3^k<0$

Here's an example (for $k=3$):

$$-1=\frac{2^{1+1}+3\times2^{1+1}+9}{2^{1+1+1}-3^3}$$

However your specific question asks about series of this form $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$:

let us for instance explore them at $k=2$ ,

we want : 2ドル^{a_2}-9 \mid 2^{a_2-a_0}+3\times2^{a_2-a_1}+9$

that is 2ドル^{a_2}-9 \mid 2^{a_0+a_1} \times (2^{a_2-a_0}+3\times2^{a_2-a_1}+9)=2^{a_2+a_1}+3\times2^{a_2+a_0}+9\times2^{a_0+a_1}$

set $N=2^{a_2}-9$ , so we have 2ドル^{a_2}\equiv 9 \pmod N$

$\implies 2^{a_2}-9\mid 9\times2^{a_1}+3^3\times2^{a_0}+9\times2^{a_0+a_1}=9\times(2^{a_1}+3\times 2^{a_0}+2^{a_0+a_1})$

So $ 2^{a_2}-9 \mid 2^{a_1}+3\times 2^{a_0}+2^{a_0+a_1} $

$\implies 2^{a_2}-9 \mid 2^{a_0}+3\times2^{a_0-a_1}+1$ (if $a_0 \ge a_1$ )

$\implies 2^{a_2}-9 \mid 2^{a_1}+2^{a_1-a_0}+3$ (if $a_0 \le a_1$)

In both cases , we may have many potential solutions !

As $a_2$ is independent of the choices of $a_1$ and $a_0$

that is you can fix $a_2=3$ and the two divisibilities will always hold :)

So $(a_0,a_1,a_2)=(u,v,3)$ for arbitrarily $u,v \in \mathbb{N^2}$ is valid where $a_2\ge a_0,a_1$

or fix $a_k$ at any integer and simply solve for the $a_i$ remaining !

For instance, take $a_2=4$

$\implies 2^4-3^2=7$ so you gotta solve 2ドル^{a_1}+2^{a_1-a_0} \equiv -3 \pmod 7$

$\implies 2^{a_1-a_0}\times(2^{a_0}+1)\equiv 4 \pmod 7$

As all our goal is to find valid solutions , we can take $a_1-a_0=1$

$\implies 2^{a_0}+1 \equiv 2 \pmod 7 \implies 2^{a_0} \equiv 1 \pmod 7$

and hence this couple is also valid : $(3\lambda,3\lambda+1,4)$ for any $\lambda \in \mathbb{N}$ where $a_2\ge a_0,a_1$

$a_2 - a_0\ge 0$ and $a_2-a_1\ge 0$ guarantees we get a positive exponent at the end !

that is we will always generate each couple with the condition : $a_i \le a_k$

The case $\lambda=1$ , gives the example given by @Dejan Govc !

To conclude , same think can be done for the $k$-th case , we can generate infinitely many contradictory couples !

I adivse you to try to take a look about the attempts made to attack the cycles :

like :

The 3x+1 problem: new lower bounds on nontrivial cycle lengths Author links open overlay panel

The Collatz conjecture, Littlewood-Offord theory, and powers of 2 and 3

Question 6

Hmm. You should perhaps note, that you introduce the property from the Collatz-problem, that the sequence of $a_i,ドル $i=0..k$ follows the inequality 0ドル<=a_0<a_1<... a_i ... < a_k$ while in the OP such a relation is not at all considered. In fact we read only of some sequence in the $a_i$ - and no further property on the sizes of the $a_i$ are given - this seems to allow a lot of solutions, as given in the two examples in answer&comments at @DejanGovc 's answer

Question 7

the $a_k$ I used to derive a lot of solutions have only to obey this condition $a_k \ge a_i$ to make the exponents in $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$ positive , and I didn't consider any relation or other constraint from collatz

Question 8

Yes, you can take $k=2$ and $(a_0,a_1,a_2)=(3,4,4)$. Then $2ドル^{a_k}-3^k=7\qquad\text{and}\qquad\sum_{i=0}^k 3^i2^{a_k-a_i} = 2+3+9=14.$$ Since 7ドル$ is a proper divisor of 14ドル$ this gives a solution to your problem.

Question 9

Can you clarify which $x$ and $y$ you're talking about ? because for 2ドル$ length cycles we have $n=\displaystyle \frac{2^x+3}{2^{x+y}-3^2}$

Question 10

@Lhachimi: I am simply giving an example of a finite sequence as asked in the question.

Question 11

@Lhachimi That may be a question to the author of the question - maybe there is another different connection to the Collatz problem. Note that the question just asks about a finite sequence with a certain property, and does not ask for connecting it to the Collatz problem.

Question 12

Well, here is another solution: take $k=3$ and $\left(a_{0},a_{1},a_{2},a_{3}\right)=\left(2,3,4,5\right)$; then $2ドル^{a_{k}}-3^{k}=2^{5}-3^{3}=5\text{ and }\sum_{i=0}^{k}{3^{i}2^{a_{k}-a_{i}}}=8+12+18+27=65$$ and 5 is a proper divisor of 65.

Question 13

to solve 2ドル^{a_k}-3^k$ divides $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$ , Firstly I started by multiplying $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$ by 2ドル^{a_0+a_1+\cdots+a_{k-1}}$ , so that the we cancel 2ドル^{a_k}$ from $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$ by some sample manipulations (considering $N=2^{a_k}-3^k$ and thus 2ドル^{a_k} \equiv 3^k \pmod {N}$) then I can fix $a_k$ at some value and solve for the remaining $a_i$ , respecting the condition $a_k\ge a_i$ so that the exponents in the initial sum stay positive as well

Lhachimi 6021 gold badge2 silver badges19 bronze badges · Accepted Answer · 2025-10-09 20:53:19Z

the only known sequence that satisfies your diophantine equation is $S_i$

which will come from $T^k(n)=n$ , i.e a collatz cycle which is still an open problem :(

$$T^{k}\left(n\right)=\displaystyle \frac{3^{k}n+\displaystyle \sum_{i=1}^{k}3^{k-i}2^{S_{i-1}}}{2^{S_{k}}}=n$$

So,

$$n=\displaystyle\frac{\displaystyle\sum_{i=1}^{k}3^{k-i}2^{S_{i-1}}}{2^{S_k}-3^k}$$

However , you still can break the only possible cycle : $T^k(1)=1$

like by allowing the denominator to be negative , i.e : 2ドル^{S_k}-3^k<0$

Here's an example (for $k=3$):

$$-1=\frac{2^{1+1}+3\times2^{1+1}+9}{2^{1+1+1}-3^3}$$

However your specific question asks about series of this form $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$:

let us for instance explore them at $k=2$ ,

we want : 2ドル^{a_2}-9 \mid 2^{a_2-a_0}+3\times2^{a_2-a_1}+9$

that is 2ドル^{a_2}-9 \mid 2^{a_0+a_1} \times (2^{a_2-a_0}+3\times2^{a_2-a_1}+9)=2^{a_2+a_1}+3\times2^{a_2+a_0}+9\times2^{a_0+a_1}$

set $N=2^{a_2}-9$ , so we have 2ドル^{a_2}\equiv 9 \pmod N$

$\implies 2^{a_2}-9\mid 9\times2^{a_1}+3^3\times2^{a_0}+9\times2^{a_0+a_1}=9\times(2^{a_1}+3\times 2^{a_0}+2^{a_0+a_1})$

So $ 2^{a_2}-9 \mid 2^{a_1}+3\times 2^{a_0}+2^{a_0+a_1} $

$\implies 2^{a_2}-9 \mid 2^{a_0}+3\times2^{a_0-a_1}+1$ (if $a_0 \ge a_1$ )

$\implies 2^{a_2}-9 \mid 2^{a_1}+2^{a_1-a_0}+3$ (if $a_0 \le a_1$)

In both cases , we may have many potential solutions !

As $a_2$ is independent of the choices of $a_1$ and $a_0$

that is you can fix $a_2=3$ and the two divisibilities will always hold :)

So $(a_0,a_1,a_2)=(u,v,3)$ for arbitrarily $u,v \in \mathbb{N^2}$ is valid where $a_2\ge a_0,a_1$

or fix $a_k$ at any integer and simply solve for the $a_i$ remaining !

For instance, take $a_2=4$

$\implies 2^4-3^2=7$ so you gotta solve 2ドル^{a_1}+2^{a_1-a_0} \equiv -3 \pmod 7$

$\implies 2^{a_1-a_0}\times(2^{a_0}+1)\equiv 4 \pmod 7$

As all our goal is to find valid solutions , we can take $a_1-a_0=1$

$\implies 2^{a_0}+1 \equiv 2 \pmod 7 \implies 2^{a_0} \equiv 1 \pmod 7$

and hence this couple is also valid : $(3\lambda,3\lambda+1,4)$ for any $\lambda \in \mathbb{N}$ where $a_2\ge a_0,a_1$

$a_2 - a_0\ge 0$ and $a_2-a_1\ge 0$ guarantees we get a positive exponent at the end !

that is we will always generate each couple with the condition : $a_i \le a_k$

The case $\lambda=1$ , gives the example given by @Dejan Govc !

To conclude , same think can be done for the $k$-th case , we can generate infinitely many contradictory couples !

I adivse you to try to take a look about the attempts made to attack the cycles :

like :

The 3x+1 problem: new lower bounds on nontrivial cycle lengths Author links open overlay panel

The Collatz conjecture, Littlewood-Offord theory, and powers of 2 and 3

Hmm. You should perhaps note, that you introduce the property from the Collatz-problem, that the sequence of $a_i,ドル $i=0..k$ follows the inequality 0ドル<=a_0<a_1<... a_i ... < a_k$ while in the OP such a relation is not at all considered. In fact we read only of some sequence in the $a_i$ - and no further property on the sizes of the $a_i$ are given - this seems to allow a lot of solutions, as given in the two examples in answer&comments at @DejanGovc 's answer
the $a_k$ I used to derive a lot of solutions have only to obey this condition $a_k \ge a_i$ to make the exponents in $\displaystyle \sum_{i=0}^k 3^i2^{a_k-a_i}$ positive , and I didn't consider any relation or other constraint from collatz

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Is there any known finite sequence of positive integers $a_i$ such that 2ドル^{a_k}-3^k$ is a positive proper divisor of $\sum_{i=0}^k 3^i2^{a_k-a_i}$? [closed]

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Is there any known finite sequence of positive integers $a_i$ such that 2ドル^{a_k}-3^k$ is a positive proper divisor of $\sum_{i=0}^k 3^i2^{a_k-a_i}$? [closed]

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