Why does $fracu_5 operatornamemod u_4u_4 operatornamemod u_3 = u_3$ where $u_n$ is the number of Gray codes on $n$ bits

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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 $$fracu_5 operatornamemod u_4u_4 operatornamemod u_3 = u_3$$



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 $$beginalignedu_3 operatornamemod u_2 &= 0\ u_4 operatornamemod u_3 &= 96endaligned$$



$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatornamemod u_4 = [textnumber of Gray codes for n=3] times [textnumber of cyclic Gray codes for n=3]$$




combinatorics modular-arithmetic coding-theory coincidences gray-code




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  • 1




    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.
    – Gerry Myerson
    Jul 21 at 22:56














up vote
0
down vote

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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 $$fracu_5 operatornamemod u_4u_4 operatornamemod u_3 = u_3$$



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 $$beginalignedu_3 operatornamemod u_2 &= 0\ u_4 operatornamemod u_3 &= 96endaligned$$



$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatornamemod u_4 = [textnumber of Gray codes for n=3] times [textnumber of cyclic Gray codes for n=3]$$




combinatorics modular-arithmetic coding-theory coincidences gray-code




share|cite|improve this question

















  • 1




    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.
    – Gerry Myerson
    Jul 21 at 22:56












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 $$fracu_5 operatornamemod u_4u_4 operatornamemod u_3 = u_3$$



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 $$beginalignedu_3 operatornamemod u_2 &= 0\ u_4 operatornamemod u_3 &= 96endaligned$$



$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatornamemod u_4 = [textnumber of Gray codes for n=3] times [textnumber of cyclic Gray codes for n=3]$$




combinatorics modular-arithmetic coding-theory coincidences gray-code




share|cite|improve this question













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 $$fracu_5 operatornamemod u_4u_4 operatornamemod u_3 = u_3$$



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 $$beginalignedu_3 operatornamemod u_2 &= 0\ u_4 operatornamemod u_3 &= 96endaligned$$



$96$ is also the number of cyclic Gray codes for $n=3$. So the coincidence is also that $$u_5 operatornamemod u_4 = [textnumber of Gray codes for n=3] times [textnumber of cyclic Gray codes for n=3]$$





combinatorics modular-arithmetic coding-theory coincidences gray-code





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 15:01
























asked Jul 21 at 20:50









ogogmad

5,49211238




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  • 1




    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.
    – Gerry Myerson
    Jul 21 at 22:56












  • 1




    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.
    – Gerry Myerson
    Jul 21 at 22:56







1




1




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.
– Gerry Myerson
Jul 21 at 22:56




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.
– Gerry Myerson
Jul 21 at 22:56















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