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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
Visualized group tables for $mathbb{Z}$ and $mathbb{Z}/nmathbb{Z}$Why do some Fibonacci numbers appear in an approximation for $e^{pisqrt{163}}$?Symmetry and trivial solutions to Pell equationsFinding Divisibility of Sequence of Numbers Generated RecursivelyPalindromic Numbers - Pattern “inside” Prime Numbers?Improving clarity and argumentation with hard-to-describe combinatorial proofVisualizing and understanding the roots of $f(z) = z^2 - e^{ivarphi}$Another color scheme for 3D visualizations of complex functionsVisualizing quadratic residues and their structureVisualized group tables for $mathbb{Z}$ and $mathbb{Z}/nmathbb{Z}$Strangely but closely related parametrized curves
$begingroup$
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, i.e. that there are only and exactly $4$ values along these lines.
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
6,34943991
$endgroup$
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, i.e. that there are only and exactly $4$ values along these lines.
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
6,34943991
$endgroup$
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
2 hours ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
2 hours 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, i.e. that there are only and exactly $4$ values along these lines.
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
6,34943991
$endgroup$
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, i.e. that there are only and exactly $4$ values along these lines.
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
6,34943991
asked 2 hours ago
Hans StrickerHans Stricker
6,34943991
edited 14 mins ago
Hans Stricker
asked 2 hours ago
Hans StrickerHans Stricker
6,34943991
asked 2 hours ago
Hans StrickerHans Stricker
6,34943991
asked 2 hours ago
Hans StrickerHans Stricker
6,34943991
6,34943991
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
2 hours ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
2 hours 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
2 hours ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
2 hours 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
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
2 hours ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
2 hours ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
2 hours 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 2 hours ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours 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. We have the lines $x,y=pm2,pm4,pm8,pm16$ with varying density of "size 1.5" points.
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 50 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
43 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
40 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
37 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
31 mins ago
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 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 2 hours ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours 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 2 hours ago
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours 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 2 hours 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 2 hours ago
Robert IsraelRobert Israel
325k23214468
answered 2 hours ago
Robert IsraelRobert Israel
325k23214468
answered 2 hours ago
Robert IsraelRobert Israel
325k23214468
answered 2 hours ago
Robert IsraelRobert Israel
325k23214468
325k23214468
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours ago
add a comment |
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours ago
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours ago
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans Stricker
2 hours 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. We have the lines $x,y=pm2,pm4,pm8,pm16$ with varying density of "size 1.5" points.
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 50 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
43 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
40 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
37 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
31 mins ago
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 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. We have the lines $x,y=pm2,pm4,pm8,pm16$ with varying density of "size 1.5" points.
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 50 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
43 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
40 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
37 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
31 mins ago
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 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. We have the lines $x,y=pm2,pm4,pm8,pm16$ with varying density of "size 1.5" points.
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 50 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. We have the lines $x,y=pm2,pm4,pm8,pm16$ with varying density of "size 1.5" points.
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 50 mins ago
StringString
13.8k32756
edited 21 mins ago
answered 50 mins ago
StringString
13.8k32756
answered 50 mins ago
StringString
13.8k32756
answered 50 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
43 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
40 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
37 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
31 mins ago
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 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
43 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
40 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
37 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
31 mins ago
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 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
43 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
43 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
40 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
40 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
37 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
37 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
31 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
31 mins ago
1
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 mins ago
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
26 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
2 hours ago
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans Stricker
2 hours ago