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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.

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  • $\begingroup$ If $n=1$ then the sum in $(2)$ is empty, hence equal to $0$. Also for $A_2=[2,2]$ the expression in $(2)$ equals $0$. In fact, identity $(1)$ immediately implies that the sum in identity $(2)$ must be an integer between $0$ and $n-1$, inclusive. $\endgroup$
    – Servaes
    Commented Apr 24, 2019 at 9:11
  • $\begingroup$ ah right the first bit was a typo I've fixed $\endgroup$ Commented Apr 24, 2019 at 9:15
  • $\begingroup$ and the second was also a typo I'm sorry about that $\endgroup$ Commented Apr 24, 2019 at 9:21
  • $\begingroup$ That doesn't change the fact that for $A_2=[2,2]$ you get $0$. $\endgroup$
    – Servaes
    Commented Apr 24, 2019 at 9:23
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    $\begingroup$ I'm very sorry about that it was a very lazy post $\endgroup$ Commented Apr 24, 2019 at 9:39

2 Answers 2

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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 $$\frac1n\sum_{j=1}^{n-1}\mathcal{R}\left(f_j\left(\frac{1}{\mathcal{P}(n)}\right)\right)=0.$$

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  • $\begingroup$ Sorry is the $f_1$ a typo? $\endgroup$ Commented Apr 24, 2019 at 9:45
  • $\begingroup$ Which $f_1$ where? $\endgroup$
    – Servaes
    Commented Apr 24, 2019 at 9:45
  • $\begingroup$ $$\frac1n\sum_{j=1}^{n-1}\mathcal{R}\left(f_1\left(\frac{1}{\mathcal{P}(n)}\right)\right) =\begin{cases} \tfrac{n-1}{n}&\text{ if }\ \forall i:\,a_i\in\{1,2\}\\ \tfrac{n-2}{n}&\text{ otherwise }. \end{cases}$$ $\endgroup$ Commented Apr 24, 2019 at 9:46
  • $\begingroup$ @Adam Yes, fixed. $\endgroup$
    – Servaes
    Commented Apr 24, 2019 at 9:47
  • $\begingroup$ Ok so yes I agree with the counter example without the exclusion of $a_j=1$, but I'm not sure how you were able to make the assertion regarding elementary calculus can you please explain this part for me thankyou very much for taking the time to look at my problem by the way $\endgroup$ Commented Apr 24, 2019 at 9:49
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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:

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

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