The numbering of the statements below is from our paper:
For all $\alpha \in \mathbb{R}_+ \setminus \mathbb{N}$, we have \[ \operatorname{Range}\bigl(f_\alpha\bigr) \subseteq \{0,\ldots, \lceil\alpha\rceil\} . \]
The numbering of the statements below is from our paper:
Let $a\in\mathbb{N}$ and $b\in\mathbb{N}\mkern-2mu\setminus\mkern-2mu\{1\}$ be relatively prime, and set $\alpha=a/b$. Then \[ \operatorname{Range}(f_\alpha) = \{0\}\cup\bigl\{f_\alpha(1), \ldots, f_\alpha(b^2-1)\bigr\}. \]
The following proposition gives the ranges for the functions $f_\alpha$ for all $\alpha \in (0,1) \cap \mathbb{Q}$.
Let $a, b \in \mathbb{N}$ be relatively prime with $a \lt b$. Then \[ \operatorname{Range}\bigl(f_{1/b}\bigr) = \{0\} \quad \text{for all} \quad b \gt 1, \] and \[ \operatorname{Range}\bigl(f_{a/b}\bigr) = \{0,1\} \quad \text{for all} \quad b \gt a \gt 1. \]
The following proposition gives the ranges for the functions $f_\alpha$ for all rational numbers greater than $1$ whose denominator is $2$.
For all $s \in \mathbb{N}$ we have \[ \operatorname{Range}\bigl(f_{2 s + 1/2}\bigr) = \bigl\{0, s \bigr\} \] and \[ \operatorname{Range}\bigl(f_{2s - 1/2}\bigr) = \bigl\{0, s - 1, s \bigr\}. \]
The first equality in [Proposition 9] generalizes as follows.
Let $b\in\mathbb{N}\setminus\{1\},$ and $s \in\mathbb N \cup\{0\}$. Then \[ \operatorname{Range}\bigl(f_{s b+1/b}\bigr) = s \mkern 2mu \{0,\ldots, b-1\} . \]
The following corollary unifies Corollaries 14 and 15 in the paper.
Let $b \in\mathbb{N}\setminus\{1\}$ and $s \in\mathbb N \cup\{0\}$. Then \[ \operatorname{Range}\bigl(f_{sb+1+1/b}\bigr) = \bigl(s\{0, 1,\ldots, b-1\}\bigr)\cup \bigl(s\{1,\ldots, b-1\}+\{1\}\bigr). \]
Let $b \in\mathbb{N}\setminus\{1\}$ and $s \in\mathbb{N}$. Then \[ \operatorname{Range}\bigl(f_{sb-1+1/b}\bigr) = \bigl(s\{0, 1,\ldots, b-1\}\bigr)\cup \bigl(s\{1,\ldots, b-1\}+\{-1\}\bigr). \]
The above results were exact. The following is an inclusion result.
Let $b \in\mathbb{N}\setminus\{1\}$ and $s \in\mathbb N \cup\{0\}$ and $u \in \{0,\ldots,b-1\}$. Then \[ \operatorname{Range}\bigl(f_{sb+u+1/b}\bigr) \subseteq \bigl(s\{0, \ldots, b-1\}\bigr) + \{0,\ldots, u\}. \]
The numbering of the statements below is from our paper:
For all $t \in \mathbb{N}$ and \[ \alpha(t) = \frac{1+\sqrt{1+4t}}{2}, \] we have \[ \operatorname{Range}\bigl(f_{\alpha(t)}\bigr) = \bigl\{1,\ldots, \lfloor \alpha(t) \rfloor \}. \]
Based on Wolfram Mathematica experimentation presented below, we state the following conjecture.
For all positive irrational parameters $\alpha$ we have \[ \{1,\ldots, \lfloor\alpha\rfloor\} \subseteq \operatorname{Range}\bigl(f_\alpha\bigr). \]
If [Conjecture 25] is true, then for all positive irrational parameters \(\alpha\), one of the following equalities holds:
For all irrational numbers $\alpha \in (0,1)$ we have \[ \operatorname{Range}(f_\alpha) = \{0,1\}. \] That is, all irrational $\alpha \in (0,1)$ satisfy the equality in item D in [Proposition 26].
For all positive irrational $\alpha$ we have \[ \operatorname{Range}(f_\alpha) = \{1\} \qquad \Leftrightarrow \qquad \alpha = \frac{1+\sqrt{5}}{2}. \]
If you are not familiar with Wolfram Mathematica, for a quick start see my Mathematica webpage. There are several movies linked at Mathematica webpage which might make the start easier. Also, one can search for newer, better movies. The web resources for Wolfram Mathematica are huge.
Below is the Mathematica code for the function $f_\alpha (n)$ defined in the paper. You can simply copy and paste the code below to a Mathematica notebook. It should work.
Clear[ff, n, \[Alpha]]; ff[n_, \[Alpha]_] := Floor[\[Alpha]^2 n] - Floor[\[Alpha] Floor[\[Alpha] n]]
Whenever one defines a new function in Mathematica, it is a good idea to test it on a simplest possible example.
ff[7, Pi]
To get the range of the function $f_\alpha$ for a specific $\alpha$, for example $\alpha = 3 \pi$, we utilize the following code:
Sort[DeleteDuplicates[Table[ff[k, 3 Pi], {k, 1, 1000}]]]
The output of the preceding code is \[ \{0,1,2,3,4,5,6,7,8,9,10\}. \] That this is the range of $f_{3 \pi}$ follows from [Proposition 1]. In the case of irrational \(\alpha\), determining how far we need to extend the variable \(k\) in the code to capture the exact range of \(f_{\alpha}\) is not straightforward. [Proposition 3] in our paper provides an upper bound for the case when \(\alpha\) is rational.
To illustrate [Proposition 9] we execute the following code
TableForm[({(If[EvenQ[# - 1/2], TableForm[{2, "*", (# - 1/2)/2, "+", 1/2}, TableDirections -> Row, TableSpacing -> 0.1], TableForm[{2, "*", (# + 1/2)/2, "-", 1/2}, TableDirections -> Row, TableSpacing -> 0.1]]), Union[Table[ff[k, #], {k, 1, 2^2 - 1}]]}) & /@ Table[(2 j + 1)/2, {j, 1, 12}], TableDepth -> 2]
which outputs a Wolfram Mathematica table, here presented converted to an HTML table by ChatGPT:
The Rational Number | The Range |
---|---|
\(2\cdot 1 - \tfrac{1}{2}\) | \(\{0,1\}\) |
\(2\cdot 1 + \tfrac{1}{2}\) | \(\{0,1\}\) |
\(2\cdot 2 - \tfrac{1}{2}\) | \(\{0,1,2\}\) |
\(2\cdot 2 + \tfrac{1}{2}\) | \(\{0,2\}\) |
\(2\cdot 3 - \tfrac{1}{2}\) | \(\{0,2,3\}\) |
\(2\cdot 3 + \tfrac{1}{2}\) | \(\{0,3\}\) |
\(2\cdot 4 - \tfrac{1}{2}\) | \(\{0,3,4\}\) |
\(2\cdot 4 + \tfrac{1}{2}\) | \(\{0,4\}\) |
\(2\cdot 5 - \tfrac{1}{2}\) | \(\{0,4,5\}\) |
\(2\cdot 5 + \tfrac{1}{2}\) | \(\{0,5\}\) |
\(2\cdot 6 - \tfrac{1}{2}\) | \(\{0,5,6\}\) |
\(2\cdot 6 + \tfrac{1}{2}\) | \(\{0,6\}\) |
To illustrate [Corollary 12] we execute the following code
bb = 3; TableForm[({TableForm[{#, "*", bb, "+", 1/bb}, TableDirections -> Row, TableSpacing -> 0], Union[Table[ff[k, #*bb + 1/bb], {k, 1, bb^2}]]} & /@ Range[1, 12]), TableDepth -> 2]
bb = 5; TableForm[({TableForm[{#, "*", bb, "+", 1/bb}, TableDirections -> Row, TableSpacing -> 0], Union[Table[ff[k, #*bb + 1/bb], {k, 1, bb^2}]]} & /@ Range[1, 12]), TableDepth -> 2]
which outputs a Wolfram Mathematica table, here presented converted to an HTML table by ChatGPT:
The Rational Number | The range |
---|---|
\(1\cdot5 + \tfrac{1}{5}\) | \(\{0,1,2,3,4\}\) |
\(2\cdot5 + \tfrac{1}{5}\) | \(\{0,2,4,6,8\}\) |
\(3\cdot5 + \tfrac{1}{5}\) | \(\{0,3,6,9,12\}\) |
\(4\cdot5 + \tfrac{1}{5}\) | \(\{0,4,8,12,16\}\) |
\(5\cdot5 + \tfrac{1}{5}\) | \(\{0,5,10,15,20\}\) |
\(6\cdot5 + \tfrac{1}{5}\) | \(\{0,6,12,18,24\}\) |
\(7\cdot5 + \tfrac{1}{5}\) | \(\{0,7,14,21,28\}\) |
\(8\cdot5 + \tfrac{1}{5}\) | \(\{0,8,16,24,32\}\) |
\(9\cdot5 + \tfrac{1}{5}\) | \(\{0,9,18,27,36\}\) |
\(10\cdot5 + \tfrac{1}{5}\) | \(\{0,10,20,30,40\}\) |
\(11\cdot5 + \tfrac{1}{5}\) | \(\{0,11,22,33,44\}\) |
\(12\cdot5 + \tfrac{1}{5}\) | \(\{0,12,24,36,48\}\) |
bb = 11; TableForm[({TableForm[{#, "*", bb, "+", 1/bb}, TableDirections -> Row, TableSpacing -> 0], Union[Table[ff[k, #*bb + 1/bb], {k, 1, bb^2}]]} & /@ Range[1, 12]), TableDepth -> 2]
which outputs a Wolfram Mathematica table, here presented converted to an HTML table by ChatGPT:
The Rational Number | The Range |
---|---|
\(1\cdot 11 + \tfrac{1}{11}\) | \(\{0,1,2,3,4,5,6,7,8,9,10\}\) |
\(2\cdot 11 + \tfrac{1}{11}\) | \(\{0,2,4,6,8,10,12,14,16,18,20\}\) |
\(3\cdot 11 + \tfrac{1}{11}\) | \(\{0,3,6,9,12,15,18,21,24,27,30\}\) |
\(4\cdot 11 + \tfrac{1}{11}\) | \(\{0,4,8,12,16,20,24,28,32,36,40\}\) |
\(5\cdot 11 + \tfrac{1}{11}\) | \(\{0,5,10,15,20,25,30,35,40,45,50\}\) |
\(6\cdot 11 + \tfrac{1}{11}\) | \(\{0,6,12,18,24,30,36,42,48,54,60\}\) |
\(7\cdot 11 + \tfrac{1}{11}\) | \(\{0,7,14,21,28,35,42,49,56,63,70\}\) |
\(8\cdot 11 + \tfrac{1}{11}\) | \(\{0,8,16,24,32,40,48,56,64,72,80\}\) |
\(9\cdot 11 + \tfrac{1}{11}\) | \(\{0,9,18,27,36,45,54,63,72,81,90\}\) |
\(10\cdot 11 + \tfrac{1}{11}\) | \(\{0,10,20,30,40,50,60,70,80,90,100\}\) |
\(11\cdot 11 + \tfrac{1}{11}\) | \(\{0,11,22,33,44,55,66,77,88,99,110\}\) |
\(12\cdot 11 + \tfrac{1}{11}\) | \(\{0,12,24,36,48,60,72,84,96,108,120\}\) |
To illustrate [Corollary 18] we execute the following code
bb = 5; TableForm[({TableForm[{#, "*", bb, "+", 1, "+", 1/bb}, TableDirections -> Row, TableSpacing -> 0], Union[Table[ff[k, #*bb + 1 + 1/bb], {k, 1, bb^2}]], Union[#*Range[0, bb - 1], #*Range[1, bb - 1] + 1]} & /@ Range[1, 12]), TableDepth -> 2]
which outputs a Wolfram Mathematica table, here presented converted to an HTML table by ChatGPT:
The Rational Number | The Range | The Union from Corollary 16 |
---|---|---|
\(0\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,1\}\) | \(\{0,1\}\) |
\(1\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,1,2,3,4,5\}\) | \(\{0,1,2,3,4,5\}\) |
\(2\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,2,3,4,5,6,7,8,9\}\) | \(\{0,2,3,4,5,6,7,8,9\}\) |
\(3\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,3,4,6,7,9,10,12,13\}\) | \(\{0,3,4,6,7,9,10,12,13\}\) |
\(4\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,4,5,8,9,12,13,16,17\}\) | \(\{0,4,5,8,9,12,13,16,17\}\) |
\(5\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,5,6,10,11,15,16,20,21\}\) | \(\{0,5,6,10,11,15,16,20,21\}\) |
\(6\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,6,7,12,13,18,19,24,25\}\) | \(\{0,6,7,12,13,18,19,24,25\}\) |
\(7\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,7,8,14,15,21,22,28,29\}\) | \(\{0,7,8,14,15,21,22,28,29\}\) |
\(8\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,8,9,16,17,24,25,32,33\}\) | \(\{0,8,9,16,17,24,25,32,33\}\) |
\(9\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,9,10,18,19,27,28,36,37\}\) | \(\{0,9,10,18,19,27,28,36,37\}\) |
\(10\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,10,11,20,21,30,31,40,41\}\) | \(\{0,10,11,20,21,30,31,40,41\}\) |
\(11\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,11,12,22,23,33,34,44,45\}\) | \(\{0,11,12,22,23,33,34,44,45\}\) |
\(12\cdot5 + 1 + \tfrac{1}{5}\) | \(\{0,12,13,24,25,36,37,48,49\}\) | \(\{0,12,13,24,25,36,37,48,49\}\) |
To illustrate [Corollary 19] we execute the following code
bb = 5; TableForm[({TableForm[{#, "*", bb, "-", 1, "+", 1/bb}, TableDirections -> Row, TableSpacing -> 0], Union[Table[ff[k, #*bb - 1 + 1/bb], {k, 1, bb^2}]], Union[#*Range[0, bb - 1], #*Range[1, bb - 1] - 1]} & /@ Range[1, 12]), TableDepth -> 2]
which outputs a Wolfram Mathematica table, here presented converted to an HTML table by ChatGPT:
The Rational Number | The Range | The Union from Corollary 17 |
---|---|---|
\(1\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,1,2,3,4\}\) | \(\{0,1,2,3,4\}\) |
\(2\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,1,2,3,4,5,6,7,8\}\) | \(\{0,1,2,3,4,5,6,7,8\}\) |
\(3\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,2,3,5,6,8,9,11,12\}\) | \(\{0,2,3,5,6,8,9,11,12\}\) |
\(4\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,3,4,7,8,11,12,15,16\}\) | \(\{0,3,4,7,8,11,12,15,16\}\) |
\(5\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,4,5,9,10,14,15,19,20\}\) | \(\{0,4,5,9,10,14,15,19,20\}\) |
\(6\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,5,6,11,12,17,18,23,24\}\) | \(\{0,5,6,11,12,17,18,23,24\}\) |
\(7\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,6,7,13,14,20,21,27,28\}\) | \(\{0,6,7,13,14,20,21,27,28\}\) |
\(8\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,7,8,15,16,23,24,31,32\}\) | \(\{0,7,8,15,16,23,24,31,32\}\) |
\(9\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,8,9,17,18,26,27,35,36\}\) | \(\{0,8,9,17,18,26,27,35,36\}\) |
\(10\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,9,10,19,20,29,30,39,40\}\) | \(\{0,9,10,19,20,29,30,39,40\}\) |
\(11\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,10,11,21,22,32,33,43,44\}\) | \(\{0,10,11,21,22,32,33,43,44\}\) |
\(12\cdot5 - 1 + \tfrac{1}{5}\) | \(\{0,11,12,23,24,35,36,47,48\}\) | \(\{0,11,12,23,24,35,36,47,48\}\) |
To illustrate [Proposition 14] we execute the following code
bb = 4; uu = 3; TableForm[({TableForm[{#, "*", bb, "+", uu, "+", 1/ bb}, TableDirections -> Row, TableSpacing -> 0], Union[Table[ff[k, #*bb + uu + 1/bb], {k, 1, bb^2}]], Union[Flatten[Outer[Plus, #*Range[0, bb - 1], Range[0, uu]], 1]]} & /@ Range[0, 12]), TableDepth -> 2]
which outputs a Wolfram Mathematica table, here presented converted to an HTML table by ChatGPT:
The Rational Number | The Range | The Superset from Proposition 12 |
---|---|---|
\(0\cdot4 + 3 + \tfrac14\) | \(\{0,1,2,3\}\) | \(\{0,1,2,3\}\) |
\(1\cdot4 + 3 + \tfrac14\) | \(\{0,1,2,3,4,5,6\}\) | \(\{0,1,2,3,4,5,6\}\) |
\(2\cdot4 + 3 + \tfrac14\) | \(\{0,2,3,5,6,8,9\}\) | \(\{0,1,2,3,4,5,6,7,8,9\}\) |
\(3\cdot4 + 3 + \tfrac14\) | \(\{0,3,4,7,8,11,12\}\) | \(\{0,1,2,3,4,5,6,7,8,9,10,11,12\}\) |
\(4\cdot4 + 3 + \tfrac14\) | \(\{0,4,5,9,10,14,15\}\) | \(\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}\) |
\(5\cdot4 + 3 + \tfrac14\) | \(\{0,5,6,11,12,17,18\}\) | \(\{0, 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18\}\) |
\(6\cdot4 + 3 + \tfrac14\) | \(\{0,6,7,13,14,20,21\}\) | \(\{0, 1, 2, 3, 6, 7, 8, 9, 12, 13, 14, 15, 18, 19, 20, 21\}\) |
\(7\cdot4 + 3 + \tfrac14\) | \(\{0,7,8,15,16,23,24\}\) | \(\{0, 1, 2, 3, 7, 8, 9, 10, 14, 15, 16, 17, 21, 22, 23, 24\}\) |
\(8\cdot4 + 3 + \tfrac14\) | \(\{0,8,9,17,18,26,27\}\) | \(\{0, 1, 2, 3, 8, 9, 10, 11, 16, 17, 18, 19, 24, 25, 26, 27\}\) |
\(9\cdot4 + 3 + \tfrac14\) | \(\{0,9,10,18,19,27,28\}\) | \(\{0, 1, 2, 3, 9, 10, 11, 12, 18, 19, 20, 21, 27, 28, 29, 30\}\) |
\(10\cdot4 + 3 + \tfrac14\) | \(\{0,10,11,20,21,30,31\}\) | \(\{0, 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33\}\) |
\(11\cdot4 + 3 + \tfrac14\) | \(\{0,11,12,22,23,33,34\}\) | \(\{0, 1, 2, 3, 11, 12, 13, 14, 22, 23, 24, 25, 33, 34, 35, 36\}\) |
\(12\cdot4 + 3 + \tfrac14\) | \(\{0,12,13,25,26,38,39\}\) | \(\{0, 1, 2, 3, 12, 13, 14, 15, 24, 25, 26, 27, 36, 37, 38, 39\}\) |
By [Proposition 23], the equality in item A in [Proposition 26] holds for all irrational numbers of the form \[ \frac{1+\sqrt{1+4t}}{2} \quad \text{with} \quad t \in \mathbb{N}. \] In particular, with \[ \alpha = \frac{1+\sqrt{2017}}{2}, \] we have \[ \operatorname{Range}(f_{\alpha}) = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22\}, \] which is confirmed by the code
Sort[DeleteDuplicates[Table[ff[k, (1 + Sqrt[2017])/2], {k, 1, 100}]]]
Based on computations in Mathematica, we conjecture that for all irrational parameters of the form \[ p \mkern 2mu \frac{1+\sqrt{1+4t}}{2} \quad \text{with} \quad p, t \in \mathbb{N}, \] the equality in item A in [Proposition 26] holds as well. For example, with \[ 5 \varphi = 5 \mkern 2mu \frac{1+\sqrt{5}}{2} , \] we have \[ \operatorname{Range}(f_{5 \varphi }) = \{1,2,3,4,5,6,7,8\}, \] which is confirmed by the code
Sort[DeleteDuplicates[Table[ff[k, 5 GoldenRatio], {k, 1, 100}]]]
We calculated $f_{2+\sqrt{2}}(n)$ for $n \in \mathbb{N}$ with $n \leq 10^6$, and based on that we conjecture that the equality in item B in [Proposition 26] holds for $\alpha = 2+\sqrt{2}$. Here is the corresponding code
Sort[DeleteDuplicates[Table[ff[k, 2 + Sqrt[2]], {k, 1, 10^6}]]]
This command is quite slow, and it outputs \[ \{0, 1, 2, 3\}. \]
We calculated $f_{\sqrt{2}}(n)$ for $n \in \mathbb{N}$ with $n \leq 10^6$, and based on that we conjecture that the equality in item C in [Proposition 26] holds for $\alpha = \sqrt{2}$. Here is the corresponding code
Sort[DeleteDuplicates[Table[ff[k, Sqrt[2]], {k, 1, 10^6}]]]
This command is quite slow, and it outputs \[ \{1, 2\}. \]
The equality in item D in [Proposition 26] holds for $\alpha = e$ and $\alpha = \pi$. This is confirmed by the Mathematica code
Sort[DeleteDuplicates[Table[ff[k, E], {k, 1, 20}]]]
Sort[DeleteDuplicates[Table[ff[k, Pi], {k, 1, 20}]]]
In this item we present Wolfram Mathematica code that calculates exact intervals for the parameter \(\alpha \in (0,1)\) such that \[ f_\alpha(m) = 1 \] for a given \(m \in \mathbb{N}\setminus\{1\}\).
Clear[nn, AllIntervals]; AllIntervals[nn_Integer?Positive] := Module[{tt1, tt2}, tt1 = Table[{k, {Sqrt[k/nn], Sqrt[(k + 1)/nn]}}, {k, 1, nn - 1}]; tt2 = SplitBy[ SortBy[Prepend[ Flatten[Table[ Table[{s, {Max[{s/r, r/nn}], Min[{(s + 1)/r, (r + 1)/nn}]}}, {s, 0, r - 1}], {r, 1, nn - 1}], 1], {0, {0, 1/nn}}], First], First]; Flatten[Table[ Select[{Max[tt1[[j, 2, 1]], #[[2, 1]]], Min[tt1[[j, 2, 2]], #[[2, 2]]]} & /@ tt2[[j]], #[[1]] < #[[2]] &], {j, 1, nn - 1}], 1]]
To test this command execute
FullSimplify[AllIntervals[7]]
which outputs
{{1/Sqrt[7], 3/7}, {Sqrt[2/7], 4/7}, {Sqrt[3/7], 5/7}, {2/Sqrt[7], 4/ 5}, {Sqrt[5/7], 6/7}, {Sqrt[6/7], 1}}
To support the validity of [Conjecture 27] denote the sets calculated in the preceding item as \[ A_m = \bigl\{ \alpha \in (0,1) : f_\alpha(m) = 1 \bigr\}, \] where \(m \in \mathbb{N}\setminus\{1\}\). With this notation, and using [Proposition 6], [Conjecture 27] can be restated as \[ \bigcup \Bigl\{A_m : m\in\mathbb{N}\setminus\{1\} \Bigr\}= (0,1) \setminus \Bigl\{ \frac{1}{b} : b \in \mathbb{N} \Bigr\}. \]
The union \[ \bigcup \Bigl\{A_m : m\in\{2,\ldots,100\} \Bigr\} \] consists of 4845 intervals of which many overlap. Below is Wolfram Mathematica code that will calculate the exact union of a given list of intervals.
Clear[UnionInts, lstInts]; UnionInts[lstInts_List] := Apply[List, Interval@@lstInts]
To test this command execute
UnionInts[ SortBy[Flatten[Table[AllIntervals[k], {k, 2, 100}], 1], First]]
to get the output of fourteen intervals
{{1/(3 Sqrt[11]), 10/99}, {1/(7 Sqrt[2]), 1/9}, {1/(4 Sqrt[5]), 9/ 80}, {1/Sqrt[79], 1/8}, {1/(3 Sqrt[7]), 8/63}, {1/Sqrt[62], 1/ 7}, {Sqrt[2/97], 14/97}, {1/(4 Sqrt[3]), 1/6}, {Sqrt[2/71], 12/ 71}, {1/Sqrt[35], 1/5}, {2/(3 Sqrt[11]), 1/4}, {Sqrt[6/95], 1/ 3}, {Sqrt[11/2]/7, 1/2}, {5/(3 Sqrt[11]), 1}}
The union \[ \bigcup \Bigl\{A_m : m\in\{2,\ldots,500\} \Bigr\} \] consists of 124021 intervals of which many overlap. This union is in fact a disjoint union of 32 intervals.
The Exact Intervals | Approximations |
---|---|
\( \left[\dfrac{1}{10\sqrt5},\dfrac{1}{22}\right) \) | \( \bigl[0.0447214,\,0.0454545\bigr) \) |
\( \left[\dfrac{1}{\sqrt{483}},\,\dfrac{22}{483}\right)\bigcup \left[\dfrac{1}{\sqrt{482}},\,\dfrac{1}{21}\right) \) | \( \bigl[0.0455016,\,0.0455487\bigr)\cup\bigl[0.0455488,\,0.0476190\bigr) \) |
\( \left[\dfrac{1}{2\sqrt{110}},\,\dfrac{21}{440}\right)\bigcup \left[\dfrac{1}{\sqrt{439}},\,\dfrac{1}{20}\right) \) | \( \bigl[0.0476731,\,0.0477273\bigr)\cup\bigl[0.0477274,\,0.0500000\bigr) \) |
\( \left[\dfrac{1}{\sqrt{399}},\,\dfrac{20}{399}\right)\bigcup \left[\dfrac{1}{\sqrt{398}},\,\dfrac{1}{19}\right) \) | \( \bigl[0.0500626,\,0.0501253\bigr)\cup\bigl[0.0501255,\,0.0526316\bigr) \) |
\( \left[\dfrac{1}{6\sqrt{10}},\,\dfrac{19}{360}\right)\bigcup \left[\dfrac{1}{\sqrt{359}},\,\dfrac{1}{18}\right) \) | \( \bigl[0.0527046,\,0.0527778\bigr)\cup\bigl[0.0527780,\,0.0555556\bigr) \) |
\( \left[\dfrac{1}{\sqrt{323}},\,\dfrac{18}{323}\right)\bigcup \left[\dfrac{1}{\sqrt{322}},\,\dfrac{1}{17}\right) \) | \( \bigl[0.0556415,\,0.0557276\bigr)\cup\bigl[0.0557278,\,0.0588235\bigr) \) |
\( \left[\dfrac{1}{12\sqrt2},\,\dfrac{17}{288}\right)\bigcup \left[\dfrac{1}{\sqrt{287}},\,\dfrac{1}{16}\right) \) | \( \bigl[0.0589256,\,0.0590278\bigr)\cup\bigl[0.0590281,\,0.0625000\bigr) \) |
\( \left[\dfrac{1}{\sqrt{255}},\,\dfrac{16}{255}\right)\bigcup \left[\dfrac{1}{\sqrt{254}},\,\dfrac{1}{15}\right) \) | \( \bigl[0.0626224,\,0.0627451\bigr)\cup\bigl[0.0627456,\,0.0666667\bigr) \) |
\( \left[\sqrt{\dfrac{2}{449}},\,\dfrac{30}{449}\right)\bigcup \left[\dfrac{1}{4\sqrt{14}},\,\dfrac{1}{14}\right) \) | \( \bigl[0.0667409,\,0.0668151\bigr)\cup\bigl[0.0668153,\,0.0714286\bigr) \) |
\( \left[\sqrt{\dfrac{2}{391}},\,\dfrac{28}{391}\right)\bigcup \left[\dfrac{1}{\sqrt{195}},\,\dfrac{1}{13}\right) \) | \( \bigl[0.0715199,\,0.0716113\bigr)\cup\bigl[0.0716115,\,0.0769231\bigr) \) |
\( \left[\sqrt{\dfrac{2}{337}},\,\dfrac{26}{337}\right)\bigcup \left[\dfrac{1}{2\sqrt{42}},\,\dfrac{1}{12}\right) \) | \( \bigl[0.0770371,\,0.0771513\bigr)\cup\bigl[0.0771517,\,0.0833333\bigr) \) |
\( \left[\sqrt{\dfrac{3}{431}},\,\dfrac{1}{11}\right) \) | \( \bigl[0.0834300,\,0.0909091\bigr) \) |
\( \left[\dfrac{2}{\sqrt{483}},\,\dfrac{1}{10}\right) \) | \( \bigl[0.0910032,\,0.1000000\bigr) \) |
\( \left[\sqrt{\dfrac{5}{499}},\,\dfrac{1}{9}\right) \) | \( \bigl[0.1001002,\,0.1111111\bigr) \) |
\( \left[\sqrt{\dfrac{6}{485}},\,\dfrac{1}{8}\right) \) | \( \bigl[0.1112256,\,0.1250000\bigr) \) |
\( \left[\sqrt{\dfrac{7}{447}},\,\dfrac{1}{7}\right) \) | \( \bigl[0.1251397,\,0.1428571\bigr) \) |
\( \left[\sqrt{\dfrac{10}{489}},\,\dfrac{1}{6}\right) \) | \( \bigl[0.1430031,\,0.1666667\bigr) \) |
\( \left[\sqrt{\dfrac{13}{467}},\,\dfrac{1}{5}\right) \) | \( \bigl[0.1668450,\,0.2000000\bigr) \) |
\( \left[2 \sqrt{\dfrac{5}{499}},\,\dfrac{1}{4}\right) \) | \( \bigl[0.2002003,\,0.2500000\bigr) \) |
\( \left[\dfrac{1}{3}\sqrt{\dfrac{31}{55}},\,\dfrac{1}{3}\right) \) | \( \bigl[0.2502524,\,0.3333333\bigr) \) |
\( \left[\sqrt{\dfrac{55}{494}},\,\dfrac{1}{2}\right) \) | \( \bigl[0.3336705,\,0.5000000\bigr) \) |
\( \left[5\sqrt{\dfrac{5}{499}},\,1\right) \) | \( \bigl[0.5005008,\,1.0000000\bigr) \) |
The figure below displays the individual sets \(A_m\) for \(m \in \{2,\ldots,100\}\), rendered in rainbow colors and arranged vertically, with \(A_2\) at the bottom and \(A_{100}\) at the top. The union of these sets is shown in black at the very bottom of the image. The union consists of \(14\) intervals.