Explanation of a regular pattern only occuring for prime numbersVisualized group tables for $mathbb{Z}$ and...
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Explanation of a regular pattern only occuring for prime numbers
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Explanation of a regular pattern only occuring for prime numbers
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$begingroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
Colors are assigned to numbers $0 leq k leq n$ from
$color{black}{textsf{black}}$ for $k=0$ over
$color{red}{textsf{red}}$ for $k=lfloor n/4rfloor$ and
$color{silver}{textsf{white}}$ for $k=lfloor n/2rfloor$ and
$color{blue}{textsf{blue}}$ for $k=lfloor 3n/4rfloor$ back to
$color{black}{textsf{black}}$ for $k = n$
Sizes are assigned to numbers $0 leq k leq n$ by
$textsf{1.5}$ if $k=lfloor n/4rfloor$ or $lfloor 3n/4rfloor$
$textsf{1.0}$ otherwise
Positions are shifted by $(n/2,n/2)$ modulo $n$ to bring $(0,0)$ to the center of the table.
Visualized this way, you will occasionally find (for some $n$) highly regular multiplication group tables like these (with $n=12,20,28,44,52,68$):
My question is:
Why do these patterns occur exactly when $n = 4p$ with a prime number $p$?
Find here some examples for $n neq 4p$, e.g. $n=61, 62, 63, 64$:
Here for some other prime numbers: $n = 4cdot 31 = 124$ and $n = 4cdot 37 = 148$:
One may observe that for $n = 4m$ and $x,y = m$ or $x,y = 3m$ the "size 1.5" dots are systematically separated by $0$ (= black) and $n/2$ (= white) dots:
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
$endgroup$
add a comment |
$begingroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
Colors are assigned to numbers $0 leq k leq n$ from
$color{black}{textsf{black}}$ for $k=0$ over
$color{red}{textsf{red}}$ for $k=lfloor n/4rfloor$ and
$color{silver}{textsf{white}}$ for $k=lfloor n/2rfloor$ and
$color{blue}{textsf{blue}}$ for $k=lfloor 3n/4rfloor$ back to
$color{black}{textsf{black}}$ for $k = n$
Sizes are assigned to numbers $0 leq k leq n$ by
$textsf{1.5}$ if $k=lfloor n/4rfloor$ or $lfloor 3n/4rfloor$
$textsf{1.0}$ otherwise
Positions are shifted by $(n/2,n/2)$ modulo $n$ to bring $(0,0)$ to the center of the table.
Visualized this way, you will occasionally find (for some $n$) highly regular multiplication group tables like these (with $n=12,20,28,44,52,68$):
My question is:
Why do these patterns occur exactly when $n = 4p$ with a prime number $p$?
Find here some examples for $n neq 4p$, e.g. $n=61, 62, 63, 64$:
Here for some other prime numbers: $n = 4cdot 31 = 124$ and $n = 4cdot 37 = 148$:
One may observe that for $n = 4m$ and $x,y = m$ or $x,y = 3m$ the "size 1.5" dots are systematically separated by $0$ (= black) and $n/2$ (= white) dots:
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
$endgroup$
3
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
2 hours ago
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
1 hour ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
$begingroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
Colors are assigned to numbers $0 leq k leq n$ from
$color{black}{textsf{black}}$ for $k=0$ over
$color{red}{textsf{red}}$ for $k=lfloor n/4rfloor$ and
$color{silver}{textsf{white}}$ for $k=lfloor n/2rfloor$ and
$color{blue}{textsf{blue}}$ for $k=lfloor 3n/4rfloor$ back to
$color{black}{textsf{black}}$ for $k = n$
Sizes are assigned to numbers $0 leq k leq n$ by
$textsf{1.5}$ if $k=lfloor n/4rfloor$ or $lfloor 3n/4rfloor$
$textsf{1.0}$ otherwise
Positions are shifted by $(n/2,n/2)$ modulo $n$ to bring $(0,0)$ to the center of the table.
Visualized this way, you will occasionally find (for some $n$) highly regular multiplication group tables like these (with $n=12,20,28,44,52,68$):
My question is:
Why do these patterns occur exactly when $n = 4p$ with a prime number $p$?
Find here some examples for $n neq 4p$, e.g. $n=61, 62, 63, 64$:
Here for some other prime numbers: $n = 4cdot 31 = 124$ and $n = 4cdot 37 = 148$:
One may observe that for $n = 4m$ and $x,y = m$ or $x,y = 3m$ the "size 1.5" dots are systematically separated by $0$ (= black) and $n/2$ (= white) dots:
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
$endgroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
Colors are assigned to numbers $0 leq k leq n$ from
$color{black}{textsf{black}}$ for $k=0$ over
$color{red}{textsf{red}}$ for $k=lfloor n/4rfloor$ and
$color{silver}{textsf{white}}$ for $k=lfloor n/2rfloor$ and
$color{blue}{textsf{blue}}$ for $k=lfloor 3n/4rfloor$ back to
$color{black}{textsf{black}}$ for $k = n$
Sizes are assigned to numbers $0 leq k leq n$ by
$textsf{1.5}$ if $k=lfloor n/4rfloor$ or $lfloor 3n/4rfloor$
$textsf{1.0}$ otherwise
Positions are shifted by $(n/2,n/2)$ modulo $n$ to bring $(0,0)$ to the center of the table.
Visualized this way, you will occasionally find (for some $n$) highly regular multiplication group tables like these (with $n=12,20,28,44,52,68$):
My question is:
Why do these patterns occur exactly when $n = 4p$ with a prime number $p$?
Find here some examples for $n neq 4p$, e.g. $n=61, 62, 63, 64$:
Here for some other prime numbers: $n = 4cdot 31 = 124$ and $n = 4cdot 37 = 148$:
One may observe that for $n = 4m$ and $x,y = m$ or $x,y = 3m$ the "size 1.5" dots are systematically separated by $0$ (= black) and $n/2$ (= white) dots:
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
group-theory number-theory visualization
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
edited 19 mins ago
Hans Stricker
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
asked 2 hours ago
Hans StrickerHans Stricker
6,34443991
6,34443991
3
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
2 hours ago
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
1 hour ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
3
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
2 hours ago
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
1 hour ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
1 hour ago
3
3
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
2 hours ago
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
2 hours ago
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
1 hour ago
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
1 hour ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
1 hour ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $n=4p$, then for $xy equiv p$ or $3p$ mod $n$ you need $p$ to divide $x$ or $y$ but $2$ to divide neither: thus the "size $1.5$" dots are all on the lines $x = p$, $x = 3p$, $y = p$ and $y = 3p$.
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
$begingroup$
To elaborate a bit on Robert Israels fine answer, first note that:
$$
begin{align}
xy&equiv n/4\
xy&equiv 3n/4
end{align}
$$
implies that $n$ must be divisible by $4$. Hence we only have such situations when $n=4q$ for some $q$. This is true whether $q$ is prime or not. So let us consider $n=4q$ a bit further. Then we look at:
$$
begin{align}
xy&equiv q\
xy&equiv 3q
end{align}
$$
which can be summarized as:
$$
xy=q(2m+1)
$$
for some $m$. This actually reveals why $n=64=4cdot 16$ also appears to somewhat work. Here $q=16$ so any solutions to $xy=16(2m+1)$ will work.
When $q$ is prime, the picture becomes simpler since all solutions to $xy=q(2m+1)$ have to lie on one of the four lines $x=pm q,y=pm q$.
I am still a bit unsure of what happens for numbers that are not of the form $n=4q$. One thing is that $xyequivlfloor n/4rfloor$ implies $(n-x)y,x(n-y)equivlceil 3n/4rceil$ so this breaks the symmetry from the cases $n=4q$ to some extend. This breaks parts of the patterns constituting the horizontal and vertical lines.
answered 22 mins ago
StringString
13.8k32756
$endgroup$
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $n=4p$, then for $xy equiv p$ or $3p$ mod $n$ you need $p$ to divide $x$ or $y$ but $2$ to divide neither: thus the "size $1.5$" dots are all on the lines $x = p$, $x = 3p$, $y = p$ and $y = 3p$.
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
$begingroup$
If $n=4p$, then for $xy equiv p$ or $3p$ mod $n$ you need $p$ to divide $x$ or $y$ but $2$ to divide neither: thus the "size $1.5$" dots are all on the lines $x = p$, $x = 3p$, $y = p$ and $y = 3p$.
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
$begingroup$
If $n=4p$, then for $xy equiv p$ or $3p$ mod $n$ you need $p$ to divide $x$ or $y$ but $2$ to divide neither: thus the "size $1.5$" dots are all on the lines $x = p$, $x = 3p$, $y = p$ and $y = 3p$.
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
If $n=4p$, then for $xy equiv p$ or $3p$ mod $n$ you need $p$ to divide $x$ or $y$ but $2$ to divide neither: thus the "size $1.5$" dots are all on the lines $x = p$, $x = 3p$, $y = p$ and $y = 3p$.
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
answered 1 hour ago
Robert IsraelRobert Israel
325k23214468
325k23214468
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
1 hour ago
add a comment |
$begingroup$
To elaborate a bit on Robert Israels fine answer, first note that:
$$
begin{align}
xy&equiv n/4\
xy&equiv 3n/4
end{align}
$$
implies that $n$ must be divisible by $4$. Hence we only have such situations when $n=4q$ for some $q$. This is true whether $q$ is prime or not. So let us consider $n=4q$ a bit further. Then we look at:
$$
begin{align}
xy&equiv q\
xy&equiv 3q
end{align}
$$
which can be summarized as:
$$
xy=q(2m+1)
$$
for some $m$. This actually reveals why $n=64=4cdot 16$ also appears to somewhat work. Here $q=16$ so any solutions to $xy=16(2m+1)$ will work.
When $q$ is prime, the picture becomes simpler since all solutions to $xy=q(2m+1)$ have to lie on one of the four lines $x=pm q,y=pm q$.
I am still a bit unsure of what happens for numbers that are not of the form $n=4q$. One thing is that $xyequivlfloor n/4rfloor$ implies $(n-x)y,x(n-y)equivlceil 3n/4rceil$ so this breaks the symmetry from the cases $n=4q$ to some extend. This breaks parts of the patterns constituting the horizontal and vertical lines.
answered 22 mins ago
StringString
13.8k32756
$endgroup$
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
add a comment |
$begingroup$
To elaborate a bit on Robert Israels fine answer, first note that:
$$
begin{align}
xy&equiv n/4\
xy&equiv 3n/4
end{align}
$$
implies that $n$ must be divisible by $4$. Hence we only have such situations when $n=4q$ for some $q$. This is true whether $q$ is prime or not. So let us consider $n=4q$ a bit further. Then we look at:
$$
begin{align}
xy&equiv q\
xy&equiv 3q
end{align}
$$
which can be summarized as:
$$
xy=q(2m+1)
$$
for some $m$. This actually reveals why $n=64=4cdot 16$ also appears to somewhat work. Here $q=16$ so any solutions to $xy=16(2m+1)$ will work.
When $q$ is prime, the picture becomes simpler since all solutions to $xy=q(2m+1)$ have to lie on one of the four lines $x=pm q,y=pm q$.
I am still a bit unsure of what happens for numbers that are not of the form $n=4q$. One thing is that $xyequivlfloor n/4rfloor$ implies $(n-x)y,x(n-y)equivlceil 3n/4rceil$ so this breaks the symmetry from the cases $n=4q$ to some extend. This breaks parts of the patterns constituting the horizontal and vertical lines.
answered 22 mins ago
StringString
13.8k32756
$endgroup$
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
add a comment |
$begingroup$
To elaborate a bit on Robert Israels fine answer, first note that:
$$
begin{align}
xy&equiv n/4\
xy&equiv 3n/4
end{align}
$$
implies that $n$ must be divisible by $4$. Hence we only have such situations when $n=4q$ for some $q$. This is true whether $q$ is prime or not. So let us consider $n=4q$ a bit further. Then we look at:
$$
begin{align}
xy&equiv q\
xy&equiv 3q
end{align}
$$
which can be summarized as:
$$
xy=q(2m+1)
$$
for some $m$. This actually reveals why $n=64=4cdot 16$ also appears to somewhat work. Here $q=16$ so any solutions to $xy=16(2m+1)$ will work.
When $q$ is prime, the picture becomes simpler since all solutions to $xy=q(2m+1)$ have to lie on one of the four lines $x=pm q,y=pm q$.
I am still a bit unsure of what happens for numbers that are not of the form $n=4q$. One thing is that $xyequivlfloor n/4rfloor$ implies $(n-x)y,x(n-y)equivlceil 3n/4rceil$ so this breaks the symmetry from the cases $n=4q$ to some extend. This breaks parts of the patterns constituting the horizontal and vertical lines.
answered 22 mins ago
StringString
13.8k32756
$endgroup$
To elaborate a bit on Robert Israels fine answer, first note that:
$$
begin{align}
xy&equiv n/4\
xy&equiv 3n/4
end{align}
$$
implies that $n$ must be divisible by $4$. Hence we only have such situations when $n=4q$ for some $q$. This is true whether $q$ is prime or not. So let us consider $n=4q$ a bit further. Then we look at:
$$
begin{align}
xy&equiv q\
xy&equiv 3q
end{align}
$$
which can be summarized as:
$$
xy=q(2m+1)
$$
for some $m$. This actually reveals why $n=64=4cdot 16$ also appears to somewhat work. Here $q=16$ so any solutions to $xy=16(2m+1)$ will work.
When $q$ is prime, the picture becomes simpler since all solutions to $xy=q(2m+1)$ have to lie on one of the four lines $x=pm q,y=pm q$.
I am still a bit unsure of what happens for numbers that are not of the form $n=4q$. One thing is that $xyequivlfloor n/4rfloor$ implies $(n-x)y,x(n-y)equivlceil 3n/4rceil$ so this breaks the symmetry from the cases $n=4q$ to some extend. This breaks parts of the patterns constituting the horizontal and vertical lines.
answered 22 mins ago
StringString
13.8k32756
edited 6 mins ago
answered 22 mins ago
StringString
13.8k32756
answered 22 mins ago
StringString
13.8k32756
answered 22 mins ago
StringString
13.8k32756
13.8k32756
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
add a comment |
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans Stricker
15 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
12 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
9 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
$begingroup$
Ah! I didn't suspect the floor function. How would you suggest to get rid of it? Any other suggestions to "draw more clear conclusions" in order to "see different principles at play"? (Thanks anyway for your suggestions.)
$endgroup$
– Hans Stricker
3 mins ago
add a comment |
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$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
2 hours ago
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
1 hour ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
1 hour ago