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I want to prove that, given this equation:

$$ \prod_{i=1}^N \left(3+\frac{1}{y_i}\right) = 2^k $$

any power of 2 can be reached by at most one unique multiset of positive integers $y_i$ (being all congruent to 1 or 5 (mod 6)). How do you approach this problem?

EDIT

Yes, this problem is connected to Collatz Conjecture.

What I do know:

  1. Under a variant of Collatz function, $f(y) = \frac{3y_i + 1}{2^{\nu_2(y_i)}}$, collatz cycles are formed by numbers $y_i$ congruent to 1 or 5 (mod 6).
  2. If a set of numbers $y_i$ for a cycle, then the product formula will produce a power of 2 within the interval $\left(3^N, 4^N \right]$.
  3. The only known solution is for $N=1$ and $y_i=1$, which maps to $4$.
  4. If a particular $y_i$ belongs to a cycle, it won't appear in any other cycle. Also, cycles admit just one instance of every element that belongs to it.
  5. If a set of numbers is a cycle, then it's a solution to the product formula, and if we allow multisets of $y_i$ numbers to be plugged into the formula (allowing repetition), then any "linear integer union" of solutions constitute a solution. Hence, a multiset of $1$s is a solution because the set $\{1\}$ is a solution.
  6. These multisets of $1$s map to all the powers of 4, or equivalently, to all the powers of 2 that are perfect squares.
  7. Fixing $N$, the only possibility for the numbers $y_i$ to reach $4^N$ is for all of them to be equal to 1 (the trivial multiset).
  8. The logic after the question: Because starting at some $N$ and above, the intervals $\left(3^N, 4^N \right]$ start to imbricate, to intersect, it might be the case that a power of two that is also a perfect square could be reached by other set of $y_i$ numbers from a greater interval overlapping with the previous one. Then, if the only way of reaching a perfect power of 2 is through the trivial multiset, such set of numbers doesn't exist.
  9. The same can be deduced with powers of 2 that are not a perfect power. Let's assume that a cycle exists that hits a power of 2 that is not a perfect power when plugged into the product formula. Then, the multiset composed by these numbers duplicated reaches a power of 2 that is a perfect square.
  10. If the only allowed multiset of numbers that can reach a power of 4 is the trivial one, then no other cycle should exist.

The flaw in the argument:

As @collag3n has shown, the product formula can be satisfied by $y_i$ values that don't belong to a cycle, and two counterexamples have been provided. The user @Gottfried Helms has pointed out that the product formula is not necessarily satisfied only by numbers $y_i$ belonging to cycles.

Carlos Toscano-Ochoa
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  • No, and I don't agree: the "duplicate" claims that, if the product is NOT restricted to be a power of 2, then it admits more than one possible solution. I already know that. Also, the "duplicate" only requires the chosen integers to be odd, whereas I rule out multiples of 3: the provided counterexample doesn't rule out the answer to my question. – Carlos Toscano-Ochoa Feb 22 '25 at 11:49
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    Alright, I will reopen the question and link the post here, which is related in any case. You should also add more context, why you approach this problem (Collatz?), and what you have tried so far. – Dietrich Burde Feb 22 '25 at 11:58
  • Related: https://math.stackexchange.com/questions/2249909/is-it-possible-to-find-general-formulas-that-can-express-all-collatz-numbers?noredirect=1 – Anand Feb 22 '25 at 12:21
  • Hmm, since for the case of the generalized Collatz-transformation using $m=181$ instead of $m=3$ in $x_{k+1}=(m\cdot x_k+1)/2^A$ (having $A$ such that $x_{k+1}$ is odd) such duplicity actually does occur I doubt, I could help with a proof... (For $N=2$. $m=181$ and $k=15$) we have duplicity by the two solutions $(y_1,y_2)=(27,611)$ and $(y_1,y_2)=(35,99)$. So I think, an argument against specifically the possibility of duplicity in the case of $m=3$ should not be simple... – Gottfried Helms Feb 22 '25 at 17:31
  • the 2 counterexamples also disprove uniqueness (your original question) – user489810 Feb 24 '25 at 09:35

2 Answers2

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$(3+\frac{1}{5})(3+\frac{1}{7})(3+\frac{1}{11})(3+\frac{1}{13})(3+\frac{1}{17})(3+\frac{1}{293})(3+\frac{1}{55})(3+\frac{1}{83})(3+\frac{1}{125})(3+\frac{1}{47})(3+\frac{1}{71})(3+\frac{1}{107})(3+\frac{1}{161})(3+\frac{1}{121})(3+\frac{1}{91})(3+\frac{1}{137})(3+\frac{1}{103})(3+\frac{1}{155})(3+\frac{1}{233})(3+\frac{1}{175})(3+\frac{1}{263})(3+\frac{1}{395})(3+\frac{1}{593})(3+\frac{1}{445})(3+\frac{1}{167})(3+\frac{1}{251})(3+\frac{1}{377})(3+\frac{1}{283})(3+\frac{1}{425})(3+\frac{1}{319})(3+\frac{1}{479})(3+\frac{1}{719})(3+\frac{1}{1079})(3+\frac{1}{1619})(3+\frac{1}{2429})(3+\frac{1}{911})(3+\frac{1}{1367})=2^{59}$

$(3+\frac{1}{5})(3+\frac{1}{7})(3+\frac{1}{11})(3+\frac{1}{13})(3+\frac{1}{17})(3+\frac{1}{347})(3+\frac{1}{521})(3+\frac{1}{391})(3+\frac{1}{587})(3+\frac{1}{881})(3+\frac{1}{661})(3+\frac{1}{31})(3+\frac{1}{47})(3+\frac{1}{71})(3+\frac{1}{107})(3+\frac{1}{161})(3+\frac{1}{121})(3+\frac{1}{91})(3+\frac{1}{137})(3+\frac{1}{103})(3+\frac{1}{155})(3+\frac{1}{233})(3+\frac{1}{175})(3+\frac{1}{263})(3+\frac{1}{395})(3+\frac{1}{593})(3+\frac{1}{445})(3+\frac{1}{167})(3+\frac{1}{251})(3+\frac{1}{377})(3+\frac{1}{283})(3+\frac{1}{425})(3+\frac{1}{319})(3+\frac{1}{479})(3+\frac{1}{719})(3+\frac{1}{1079})(3+\frac{1}{1619})=2^{59}$

Collag3n
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  • Both are equal to $2^{59}$ ? – Aurel-BG Feb 22 '25 at 20:45
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    @Aurel-BG, yes they are. To check you can just edit and copy/paste + remove all "\frac{", replace "}{" by "/", remove all remaining "}", replace ")(" by ")*(" and run it in PARI/GP or wolfram... – Collag3n Feb 22 '25 at 20:57
  • Nice (I upvoted) ! :)

    do you have any PARI/GP code for this please ?

     – Aurel-BG Feb 22 '25 at 21:02
  • Just paste it in PARI/GP. Apparently this is too long for wolfram. I will put both of them in comments bellow. – Collag3n Feb 22 '25 at 21:06
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    (3+1/5)(3+1/7)(3+1/11)(3+1/13)(3+1/17)(3+1/293)(3+1/55)(3+1/83)(3+1/125)(3+1/47)(3+1/71)(3+1/107)(3+1/161)(3+1/121)(3+1/91)(3+1/137)(3+1/103)(3+1/155)(3+1/233)(3+1/175)(3+1/263)(3+1/395)(3+1/593)(3+1/445)(3+1/167)(3+1/251)(3+1/377)(3+1/283)(3+1/425)(3+1/319)(3+1/479)(3+1/719)(3+1/1079)(3+1/1619)(3+1/2429)(3+1/911)(3+1/1367) – Collag3n Feb 22 '25 at 21:06
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    (3+1/5)(3+1/7)(3+1/11)(3+1/13)(3+1/17)(3+1/347)(3+1/521)(3+1/391)(3+1/587)(3+1/881)(3+1/661)(3+1/31)(3+1/47)(3+1/71)(3+1/107)(3+1/161)(3+1/121)(3+1/91)(3+1/137)(3+1/103)(3+1/155)(3+1/233)(3+1/175)(3+1/263)(3+1/395)(3+1/593)(3+1/445)(3+1/167)(3+1/251)(3+1/377)(3+1/283)(3+1/425)(3+1/319)(3+1/479)(3+1/719)(3+1/1079)(3+1/1619) – Collag3n Feb 22 '25 at 21:07
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    I didn't use PARI/GP to find them, I just took them from a previous post I made: https://math.stackexchange.com/questions/3586737/how-can-i-prove-that-b-prod-i-1n-3-frac1x-i-results-not-in-an-e?noredirect=1&lq=1 and multiplied them by (3+1/5)(3+1/7)(3+1/11)(3+1/13)(3+1/17) – Collag3n Feb 22 '25 at 21:07
  • This is lovely.... :-)) – Gottfried Helms Feb 23 '25 at 00:35
  • I'm still struggling to understand what the two identities that you (@Collag3n) have proposed mean. Because that means that the product equation is not exclusive for cycles, but does it tell something else about the problem? Is is because all the y_i end up eventually reaching 1, right? – Carlos Toscano-Ochoa Feb 24 '25 at 20:27
  • The set of $y_k$ (in the first of the two given solutions) cannot be in a cylic orbit because in the ordered list of them between $17$ and $47$ there is a hole: $47$ is larger than $ (3 \cdot 17+1)/2$. A similar hole is in the (ordered) second example between $17$ and $31$. Perhaps such a property can be found in the constructable products generally ... – Gottfried Helms Feb 24 '25 at 21:50
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    @Carlos Toscano-Ochoa, yes, it means that it is not exclusive to collatz cycles and at least in that case, you have not uniqueness. As Gottfried said, my examples are a concatenation of 2 different Collatz sequences, so there is a hole "in the Collatz sense" (2 holes in fact since the last element does not connect to the first either), and if the Collatz conjecture is true, you can't have any cases without "holes". – Collag3n Feb 25 '25 at 15:41
2

If you write it as $2^k=\frac{(3y_{1}+1)\cdot(3y_{2}+1)\cdot...\cdot(3y_{N}+1)}{y_{1}\cdot y_{2}\cdot...\cdot y_{N}}$ and rewrite every $(3y_{i}+1)$ as $2^{k_i}x_i$ where $x_i$ is odd, you can see that in order to cancel out the odd part in the numerator, you must have the same odd part in the denominator: $\prod x_i=\prod y_i$. The most likely answer would be that the set of $x_i$ is the same as the set of $y_i$, in which case, not only all those $y_i$ are consecutive numbers in a Collatz sequence, but they are in a cycle (the last one must be the first one to cancel out). In this case, uniqueness would means that there are no two cycles of the same length ($k$), which might be harder to prove than the Collatz conjecture itself.

Now, a proof that both sets are the same would require to show that we cannot have a situation like this: $\prod x_i=55\cdot 7=\prod y_i=5\cdot 77$ where some factors (e.g. 11) are not part of the same numbers (perhaps by using the fact that $x_i$ has no common factor with $y_i$ for any $i$)

EDIT:

Here is an example of factors "not part of the same numbers" , so my "most likely" was a bit hasty

$\prod_{i=1}^{n} (3 +\frac{1}{a_i})$ cannot be a power of 2 if $a_i \equiv \pm 1 \pmod 6$

Collag3n
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  • Hmm, I still don't have an idea what we actually must prove, to show, that duplicity is not possible. In my other comment I refer to the generalized Collatz-problem $y_2=(181 \cdot y_1 +1) / 2^A \text{ where } A=\nu_2(181 \cdot y_1 +1)$, and we have a) at least one cycle at all and moreover b) duplicity. – Gottfried Helms Feb 22 '25 at 18:32
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    @Gottfried Helms, having two collatz cycles of the same length would prove duplicity, but as you said, and as I said above, this would be very hard to prove unicity. Now, I said that in the case these numbers were part of a Collatz sequence, but it seems it is not necessarily the case: https://math.stackexchange.com/questions/4178090/prod-i-1n-3-frac1a-i-cannot-be-a-power-of-2-if-a-i-equiv-pm-1 – Collag3n Feb 22 '25 at 20:26
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    Yes, the multiplication-formula removes the condition, that the elements $y_k$ must follow the Collatz-transformation (one orbit). For example, the condition of being elements of a Collatz-orbit, exposes one specific example (of the "1-cycle" name), that all must follow that $y_{k+1} = 3/2y_k+1/2 $ forming nearly a geometric sequence with (a nearly) quotient of $q=3/2$. This will give a maximal stretch of the values of the productformula - such a restriction is not inherent to that productformula: the expansion between the minimal and maximal element is arbitrary. – Gottfried Helms Feb 23 '25 at 10:07