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Why does $frac{u_5 operatorname{mod} u_4}{u_4 operatorname{mod} u_3} = u_3$ where $u_n$ is the number of Gray codes on $n$ bits

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$begingroup$

Let $u_n$ be the number of Gray codes on $n$ bits. See the entry on OEIS - only five values are listed.

I noticed that $$frac{u_5 operatorname{mod} u_4}{u_4 operatorname{mod} u_3} = u_3$$

Or in IPython:

In [2]: import fractions



In [3]: fractions.Fraction(187499658240 % 91392, 91392 % 144)

Out[3]: Fraction(144, 1)

Is there a reason for that, or is it just a coincidence?

If the numbers 5, 4 and 3 are replaced with 4, 3 and 2 respectively, then the analagous claim doesn't hold because $$begin{aligned}u_3 operatorname{mod} u_2 &= 0\ u_4 operatorname{mod} u_3 &= 96end{aligned}$$

$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatorname{mod} u_4 = [text{number of Gray codes for }n=3] times [text{number of cyclic Gray codes for }n=3]$$

combinatorics modular-arithmetic coding-theory gray-code

share|cite|improve this question

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm betting on coincidence. Given that it works once, and fails once, I see no reason to believe it's anything but coincidence. Given three positive integers, it's not hard to find an equation they satisfy.
  $endgroup$
  – Gerry Myerson
  Jul 21 '18 at 22:56
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $u_n$ be the number of Gray codes on $n$ bits. See the entry on OEIS - only five values are listed.

I noticed that $$frac{u_5 operatorname{mod} u_4}{u_4 operatorname{mod} u_3} = u_3$$

Or in IPython:

In [2]: import fractions



In [3]: fractions.Fraction(187499658240 % 91392, 91392 % 144)

Out[3]: Fraction(144, 1)

Is there a reason for that, or is it just a coincidence?

If the numbers 5, 4 and 3 are replaced with 4, 3 and 2 respectively, then the analagous claim doesn't hold because $$begin{aligned}u_3 operatorname{mod} u_2 &= 0\ u_4 operatorname{mod} u_3 &= 96end{aligned}$$

$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatorname{mod} u_4 = [text{number of Gray codes for }n=3] times [text{number of cyclic Gray codes for }n=3]$$

combinatorics modular-arithmetic coding-theory gray-code

share|cite|improve this question

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm betting on coincidence. Given that it works once, and fails once, I see no reason to believe it's anything but coincidence. Given three positive integers, it's not hard to find an equation they satisfy.
  $endgroup$
  – Gerry Myerson
  Jul 21 '18 at 22:56
  

  
  
  
  

add a comment | 

$begingroup$

Let $u_n$ be the number of Gray codes on $n$ bits. See the entry on OEIS - only five values are listed.

I noticed that $$frac{u_5 operatorname{mod} u_4}{u_4 operatorname{mod} u_3} = u_3$$

Or in IPython:

In [2]: import fractions



In [3]: fractions.Fraction(187499658240 % 91392, 91392 % 144)

Out[3]: Fraction(144, 1)

Is there a reason for that, or is it just a coincidence?

If the numbers 5, 4 and 3 are replaced with 4, 3 and 2 respectively, then the analagous claim doesn't hold because $$begin{aligned}u_3 operatorname{mod} u_2 &= 0\ u_4 operatorname{mod} u_3 &= 96end{aligned}$$

$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatorname{mod} u_4 = [text{number of Gray codes for }n=3] times [text{number of cyclic Gray codes for }n=3]$$

combinatorics modular-arithmetic coding-theory gray-code

share|cite|improve this question

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438

$endgroup$

In [2]: import fractions



In [3]: fractions.Fraction(187499658240 % 91392, 91392 % 144)

Out[3]: Fraction(144, 1)

share|cite|improve this question

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438

share|cite|improve this question

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

edited Mar 15 at 9:41

YuiTo Cheng

2,1362837

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438

asked Jul 21 '18 at 20:50

man on laptopman on laptop

5,78611438