Natural Number Multiset conjecture counter example sought

Question

Consider a multiset of natural numbers:

$$A_n=[a_j]_{j=1..n}$$

in the cases:

$a_j \not=a_k$ for all $j,k \in {\{1,2,3,...,n}\} \land n \gt 1\quad\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\quad\quad\quad\quad(\operatorname{i})$

$a_j \not=a_k$ for some $j,k \in {\{1,2,3,...,n}\} \land n \gt 1$ and $a_j \not= 1$ $\quad\quad\quad\quad\quad\,\,\,\,\,(\operatorname{ii})$

We define the product of all elements of $A_n$ as a function of $n$ as follows:

$$\mathcal P(n)=\prod_{j=1}^{n}a_j$$

If we define the function $f_k$ as a function of some fixed $x$ to be the raising to a power of $x$ $k-1$ times so for example :

$$f_1(x)=1$$ $$f_2(x)=x$$ $$f_3(x)=x^x$$ $$f_4(x)=(x^x)^x$$

Then we define a rounding function $\mathcal R$ as:

$$\mathcal R(x)=\cases{\lfloor x \rfloor&${\{x}\} <\frac{1}{2}$\cr \lceil x \rceil&${\{x}\}\geq\frac{1}{2} $\cr}$$

where ${\{x}\}$ is the fractional part of $x$

We have that for any $A_n$ as previously defined,

$$\mathcal R\Biggl(f_j\Bigl(\frac{1}{\mathcal P(n)}\Bigr)\Biggr) \in {\{0,1}\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,(1)$$

$$\sum_{j=1}^{n-1}\frac{\mathcal R\Biggl(f_j\Bigl(\frac{1}{\mathcal P(n)}\Bigr)\Biggr)}{n} \in {\Biggl\{1,\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9}}\Biggr\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$

$(1)$ is clearly true seeing that the sequence in $j$ is always a binary alternating sequence, But this does not explain the result of $(2)$, however it is supported by the results like:

$$\lim _{N\rightarrow \infty}\Biggr(\sum^{N}_{n=1}\frac{(1+(-1)^n)}{2N}\Biggl)=\frac{1}{2}$$

$$\lim _{N\rightarrow \infty}\Biggr(\sum^{N}_{n=0}\frac{(1+(-1)^n)}{2N}\Biggl)=\frac{1}{2}$$

So the proof will of course require the infinite, because we need to establish that it is true for any such $A_n$, which I think I will probably not be capable enough to do, so a counter example is actually a happy thing here I think

EDIT:

The counterexamples should occur as an example for $n=12$, as we expect sums like $\frac{6}{11},\frac{8}{13},\frac{9}{15}$ and so on to occur.

Answer 1

By elementary calculus we have $f_j(x)\geq\tfrac12$ for all $x\in[0,1]$ and all $j\geq2$. It follows that for $j\geq2$ and for all $n\geq0$ we have $$\mathcal{R}\left(f_j\left(\frac{1}{\mathcal{P}(n)}\right)\right)=1.$$ Of course it is clear that $$\mathcal{R}\left(f_1\left(\frac{1}{\mathcal{P}(n)}\right)\right) =\begin{cases} 1&\text{ if }\mathcal{P}(n)\leq2,\\ 0&\text{ if }\mathcal{P}(n)>2, \end{cases}$$ and that $\mathcal{P}(n)\leq2$ if and only if $a_i\in\{1,2\}$ for all $i$. It follows that $$\frac1n\sum_{j=1}^{n-1}\mathcal{R}\left(f_j\left(\frac{1}{\mathcal{P}(n)}\right)\right) =\begin{cases} \tfrac{n-1}{n}&\text{ if }\ \forall i:\,a_i\in\{1,2\}\\ \tfrac{n-2}{n}&\text{ otherwise }. \end{cases}$$ In particular, counterexamples to your claim $(2)$ are easily constructed. Take for example $A_3=\{1,2,3\}$. Then $\mathcal{P}(3)=6$ and $$\frac1n\sum_{j=1}^{n-1}\mathcal{R}\left(f_j\left(\frac{1}{\mathcal{P}(n)}\right)\right) =\frac13\left(\lfloor f_1(\tfrac16)\rfloor+\lfloor f_2(\tfrac16)\rfloor\right) =\frac13(0+1)=\frac13.$$ There are of course also the obvious counterexamples with $n=1$; in this case the sum is empty and so $$\frac1n\sum_{j=1}^{n-1}\mathcal{R}\left(f_j\left(\frac{1}{\mathcal{P}(n)}\right)\right)=0.$$

Answer 2

counterexample $A_n=[6554,5521,1862,3070,1200,5494,312,3301,9623,2333,9813,9338]$

$$\sum_{j=1}^{n-1}\frac{\mathcal R\Biggl(f_j\Bigl(\frac{1}{\mathcal P(n)}\Bigr)\Biggr)}{n} =\frac{6}{11}$$

The actual expression for the sum above for multisets that adhere to the condition $(\operatorname{ii})$ is:

$$\sum _{j=1}^{n-1}\frac{1}{2}{\frac { \left( -1 \right) ^{j+1}+1}{n}}= \cases{\frac{1}{2}&$n \in {\{2k-1:k\in \mathbb N}\} $\cr \frac{1}{2}{\frac {n}{n-1}}&$n \in {\{2k:k\in \mathbb N}\} $\cr}$$