The pattern of bitstrings of square numbers
Multi tool use
$begingroup$
Here is a picture of "square numbers" from mathworld website.
A plot of the first few square numbers represented as a sequence of binary bits is shown above. The top portion shows $1^2$ to $255^2$, and the bottom shows the next 510 values. (Black is for "one" bit, and while is for "zero" bit)
I want to learn more about this pattern. However, there is no further explanation about the pattern of this plot from the website.
Is there any known result about this pattern ?
number-theory
$endgroup$
|
show 1 more comment
$begingroup$
Here is a picture of "square numbers" from mathworld website.
A plot of the first few square numbers represented as a sequence of binary bits is shown above. The top portion shows $1^2$ to $255^2$, and the bottom shows the next 510 values. (Black is for "one" bit, and while is for "zero" bit)
I want to learn more about this pattern. However, there is no further explanation about the pattern of this plot from the website.
Is there any known result about this pattern ?
number-theory
$endgroup$
1
$begingroup$
Counting from the bottom from $0$, row $0$ has period two, and row $n$ when $n>0$ has period which is a divisor of $2^n$.
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:30
1
$begingroup$
It's also the case that row $n$ values are symmetric around the multiples of $2^n$ - that is, the squares of $a2^n-b$ and $a2^n+b$ have the same $n$th bit, for all $a,b$ with $b<a2^n$
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:35
$begingroup$
Vertical periodicity: $(2^mn)^2=2^{2m}n^2=n^2<<2m,$ where $a<<b$ is a shifted to the left by b binary places (i.e. the pattern is shifted upwards).
$endgroup$
– Loki Clock
Apr 3 '13 at 14:37
$begingroup$
I deleted my previous comment; the lengths of the rows of most significant digits are not powers of 2.
$endgroup$
– Loki Clock
Apr 3 '13 at 20:26
$begingroup$
Has anyone tried this for other radixes?
$endgroup$
– Jack M
Apr 3 '13 at 21:22
|
show 1 more comment
$begingroup$
Here is a picture of "square numbers" from mathworld website.
A plot of the first few square numbers represented as a sequence of binary bits is shown above. The top portion shows $1^2$ to $255^2$, and the bottom shows the next 510 values. (Black is for "one" bit, and while is for "zero" bit)
I want to learn more about this pattern. However, there is no further explanation about the pattern of this plot from the website.
Is there any known result about this pattern ?
number-theory
$endgroup$
Here is a picture of "square numbers" from mathworld website.
A plot of the first few square numbers represented as a sequence of binary bits is shown above. The top portion shows $1^2$ to $255^2$, and the bottom shows the next 510 values. (Black is for "one" bit, and while is for "zero" bit)
I want to learn more about this pattern. However, there is no further explanation about the pattern of this plot from the website.
Is there any known result about this pattern ?
number-theory
number-theory
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
3,41981830
asked Apr 3 '13 at 14:08
eigeig
32029
32029
1
$begingroup$
Counting from the bottom from $0$, row $0$ has period two, and row $n$ when $n>0$ has period which is a divisor of $2^n$.
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:30
1
$begingroup$
It's also the case that row $n$ values are symmetric around the multiples of $2^n$ - that is, the squares of $a2^n-b$ and $a2^n+b$ have the same $n$th bit, for all $a,b$ with $b<a2^n$
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:35
$begingroup$
Vertical periodicity: $(2^mn)^2=2^{2m}n^2=n^2<<2m,$ where $a<<b$ is a shifted to the left by b binary places (i.e. the pattern is shifted upwards).
$endgroup$
– Loki Clock
Apr 3 '13 at 14:37
$begingroup$
I deleted my previous comment; the lengths of the rows of most significant digits are not powers of 2.
$endgroup$
– Loki Clock
Apr 3 '13 at 20:26
$begingroup$
Has anyone tried this for other radixes?
$endgroup$
– Jack M
Apr 3 '13 at 21:22
|
show 1 more comment
1
$begingroup$
Counting from the bottom from $0$, row $0$ has period two, and row $n$ when $n>0$ has period which is a divisor of $2^n$.
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:30
1
$begingroup$
It's also the case that row $n$ values are symmetric around the multiples of $2^n$ - that is, the squares of $a2^n-b$ and $a2^n+b$ have the same $n$th bit, for all $a,b$ with $b<a2^n$
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:35
$begingroup$
Vertical periodicity: $(2^mn)^2=2^{2m}n^2=n^2<<2m,$ where $a<<b$ is a shifted to the left by b binary places (i.e. the pattern is shifted upwards).
$endgroup$
– Loki Clock
Apr 3 '13 at 14:37
$begingroup$
I deleted my previous comment; the lengths of the rows of most significant digits are not powers of 2.
$endgroup$
– Loki Clock
Apr 3 '13 at 20:26
$begingroup$
Has anyone tried this for other radixes?
$endgroup$
– Jack M
Apr 3 '13 at 21:22
1
1
$begingroup$
Counting from the bottom from $0$, row $0$ has period two, and row $n$ when $n>0$ has period which is a divisor of $2^n$.
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:30
$begingroup$
Counting from the bottom from $0$, row $0$ has period two, and row $n$ when $n>0$ has period which is a divisor of $2^n$.
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:30
1
1
$begingroup$
It's also the case that row $n$ values are symmetric around the multiples of $2^n$ - that is, the squares of $a2^n-b$ and $a2^n+b$ have the same $n$th bit, for all $a,b$ with $b<a2^n$
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:35
$begingroup$
It's also the case that row $n$ values are symmetric around the multiples of $2^n$ - that is, the squares of $a2^n-b$ and $a2^n+b$ have the same $n$th bit, for all $a,b$ with $b<a2^n$
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:35
$begingroup$
Vertical periodicity: $(2^mn)^2=2^{2m}n^2=n^2<<2m,$ where $a<<b$ is a shifted to the left by b binary places (i.e. the pattern is shifted upwards).
$endgroup$
– Loki Clock
Apr 3 '13 at 14:37
$begingroup$
Vertical periodicity: $(2^mn)^2=2^{2m}n^2=n^2<<2m,$ where $a<<b$ is a shifted to the left by b binary places (i.e. the pattern is shifted upwards).
$endgroup$
– Loki Clock
Apr 3 '13 at 14:37
$begingroup$
I deleted my previous comment; the lengths of the rows of most significant digits are not powers of 2.
$endgroup$
– Loki Clock
Apr 3 '13 at 20:26
$begingroup$
I deleted my previous comment; the lengths of the rows of most significant digits are not powers of 2.
$endgroup$
– Loki Clock
Apr 3 '13 at 20:26
$begingroup$
Has anyone tried this for other radixes?
$endgroup$
– Jack M
Apr 3 '13 at 21:22
$begingroup$
Has anyone tried this for other radixes?
$endgroup$
– Jack M
Apr 3 '13 at 21:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
This gives the lengths of the highest black lines:
http://oeis.org/A126726
$endgroup$
add a comment |
$begingroup$
Note that each line is periodic. The bottom line alternates white and black, as even and odd numbers have even and odd squares. The next line is the leading bit of the numbers $pmod 4$. As the squares $pmod 4$ go $0,1,0,1$ it will always be white. The squares $pmod {16}$ go $0,1,4,9,0,1,4,9,0,1,4,9,0,1,4,9$ so the next line up will go white, white, black, white and repeat. The next will be white, white, white, black. You can continue .
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This gives the lengths of the highest black lines:
http://oeis.org/A126726
$endgroup$
add a comment |
$begingroup$
This gives the lengths of the highest black lines:
http://oeis.org/A126726
$endgroup$
add a comment |
$begingroup$
This gives the lengths of the highest black lines:
http://oeis.org/A126726
$endgroup$
This gives the lengths of the highest black lines:
http://oeis.org/A126726
answered Apr 3 '13 at 21:14
Loki ClockLoki Clock
1,914813
1,914813
add a comment |
add a comment |
$begingroup$
Note that each line is periodic. The bottom line alternates white and black, as even and odd numbers have even and odd squares. The next line is the leading bit of the numbers $pmod 4$. As the squares $pmod 4$ go $0,1,0,1$ it will always be white. The squares $pmod {16}$ go $0,1,4,9,0,1,4,9,0,1,4,9,0,1,4,9$ so the next line up will go white, white, black, white and repeat. The next will be white, white, white, black. You can continue .
$endgroup$
add a comment |
$begingroup$
Note that each line is periodic. The bottom line alternates white and black, as even and odd numbers have even and odd squares. The next line is the leading bit of the numbers $pmod 4$. As the squares $pmod 4$ go $0,1,0,1$ it will always be white. The squares $pmod {16}$ go $0,1,4,9,0,1,4,9,0,1,4,9,0,1,4,9$ so the next line up will go white, white, black, white and repeat. The next will be white, white, white, black. You can continue .
$endgroup$
add a comment |
$begingroup$
Note that each line is periodic. The bottom line alternates white and black, as even and odd numbers have even and odd squares. The next line is the leading bit of the numbers $pmod 4$. As the squares $pmod 4$ go $0,1,0,1$ it will always be white. The squares $pmod {16}$ go $0,1,4,9,0,1,4,9,0,1,4,9,0,1,4,9$ so the next line up will go white, white, black, white and repeat. The next will be white, white, white, black. You can continue .
$endgroup$
Note that each line is periodic. The bottom line alternates white and black, as even and odd numbers have even and odd squares. The next line is the leading bit of the numbers $pmod 4$. As the squares $pmod 4$ go $0,1,0,1$ it will always be white. The squares $pmod {16}$ go $0,1,4,9,0,1,4,9,0,1,4,9,0,1,4,9$ so the next line up will go white, white, black, white and repeat. The next will be white, white, white, black. You can continue .
answered Apr 3 '13 at 21:23
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
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TumSu,WLJprWjj1TNAETgPuF6cl55ykT2ldGMexN
1
$begingroup$
Counting from the bottom from $0$, row $0$ has period two, and row $n$ when $n>0$ has period which is a divisor of $2^n$.
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:30
1
$begingroup$
It's also the case that row $n$ values are symmetric around the multiples of $2^n$ - that is, the squares of $a2^n-b$ and $a2^n+b$ have the same $n$th bit, for all $a,b$ with $b<a2^n$
$endgroup$
– Thomas Andrews
Apr 3 '13 at 14:35
$begingroup$
Vertical periodicity: $(2^mn)^2=2^{2m}n^2=n^2<<2m,$ where $a<<b$ is a shifted to the left by b binary places (i.e. the pattern is shifted upwards).
$endgroup$
– Loki Clock
Apr 3 '13 at 14:37
$begingroup$
I deleted my previous comment; the lengths of the rows of most significant digits are not powers of 2.
$endgroup$
– Loki Clock
Apr 3 '13 at 20:26
$begingroup$
Has anyone tried this for other radixes?
$endgroup$
– Jack M
Apr 3 '13 at 21:22