Homogeneous van der Waerden
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The Erdős Discrepancy Problem is whether in any two-coloring of the naturals for any $C$ there is a sequence $d, 2d, ldots nd$ such that the difference of red and blue numbers in it is more than $C$.
This was recently shown to be true by Tao (building on Polymath5).
Now consider the following stronger conjecture, which also generalizes van der Waerden.
In any $k$-coloring of the naturals for any $n$ there is a monochromatic sequence $(i+1)d, (i+2)d, ldots (i+n)d$.
If true, this would of course be quite a strong result, so I more expect that someone might be able to show a simple counterexample to it.
What about the even stronger density version?
nt.number-theory co.combinatorics arithmetic-progression polymath5 colorings
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
$endgroup$
add a comment |
$begingroup$
The Erdős Discrepancy Problem is whether in any two-coloring of the naturals for any $C$ there is a sequence $d, 2d, ldots nd$ such that the difference of red and blue numbers in it is more than $C$.
This was recently shown to be true by Tao (building on Polymath5).
Now consider the following stronger conjecture, which also generalizes van der Waerden.
In any $k$-coloring of the naturals for any $n$ there is a monochromatic sequence $(i+1)d, (i+2)d, ldots (i+n)d$.
If true, this would of course be quite a strong result, so I more expect that someone might be able to show a simple counterexample to it.
What about the even stronger density version?
nt.number-theory co.combinatorics arithmetic-progression polymath5 colorings
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
$endgroup$
add a comment |
$begingroup$
The Erdős Discrepancy Problem is whether in any two-coloring of the naturals for any $C$ there is a sequence $d, 2d, ldots nd$ such that the difference of red and blue numbers in it is more than $C$.
This was recently shown to be true by Tao (building on Polymath5).
Now consider the following stronger conjecture, which also generalizes van der Waerden.
In any $k$-coloring of the naturals for any $n$ there is a monochromatic sequence $(i+1)d, (i+2)d, ldots (i+n)d$.
If true, this would of course be quite a strong result, so I more expect that someone might be able to show a simple counterexample to it.
What about the even stronger density version?
nt.number-theory co.combinatorics arithmetic-progression polymath5 colorings
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
$endgroup$
The Erdős Discrepancy Problem is whether in any two-coloring of the naturals for any $C$ there is a sequence $d, 2d, ldots nd$ such that the difference of red and blue numbers in it is more than $C$.
This was recently shown to be true by Tao (building on Polymath5).
Now consider the following stronger conjecture, which also generalizes van der Waerden.
In any $k$-coloring of the naturals for any $n$ there is a monochromatic sequence $(i+1)d, (i+2)d, ldots (i+n)d$.
If true, this would of course be quite a strong result, so I more expect that someone might be able to show a simple counterexample to it.
What about the even stronger density version?
nt.number-theory co.combinatorics arithmetic-progression polymath5 colorings
nt.number-theory co.combinatorics arithmetic-progression polymath5 colorings
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
edited Dec 27 '18 at 7:32
domotorp
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
asked Dec 24 '18 at 9:06
domotorpdomotorp
9,9773388
9,9773388
add a comment |
add a comment |
1 Answer
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This is false already for $k=2,n=4$. Color an integer $m$ according to the parity of the exponent of $2$ in the prime factorization. Among $i+1,i+2,i+3,i+4$ at least one number is odd, and at least one is divisible by $2$ and not by $4$. Those two numbers have the parity of the exponent different, hence so do the corresponding two among $(i+1)d,(i+2)d,(i+3)d,(i+4)d$. Hence those four numbers can't have the same color.
Edit: for completeness, it's false for $k=2,n=3$ as well. For integer $m$, write it as $3^icdot j,3nmid j$ and color $m$ according to $jmod 3$. Then among $i+1,i+2,i+3$ the two which are not divisible by $3$ will have different remainders, do multiplying them by $d$ gives numbers of different colors. Hence your conjecture only holds in trivial cases $k=1$ and $nleq 2$.
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
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1 Answer
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1 Answer
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$begingroup$
This is false already for $k=2,n=4$. Color an integer $m$ according to the parity of the exponent of $2$ in the prime factorization. Among $i+1,i+2,i+3,i+4$ at least one number is odd, and at least one is divisible by $2$ and not by $4$. Those two numbers have the parity of the exponent different, hence so do the corresponding two among $(i+1)d,(i+2)d,(i+3)d,(i+4)d$. Hence those four numbers can't have the same color.
Edit: for completeness, it's false for $k=2,n=3$ as well. For integer $m$, write it as $3^icdot j,3nmid j$ and color $m$ according to $jmod 3$. Then among $i+1,i+2,i+3$ the two which are not divisible by $3$ will have different remainders, do multiplying them by $d$ gives numbers of different colors. Hence your conjecture only holds in trivial cases $k=1$ and $nleq 2$.
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
$endgroup$
add a comment |
$begingroup$
This is false already for $k=2,n=4$. Color an integer $m$ according to the parity of the exponent of $2$ in the prime factorization. Among $i+1,i+2,i+3,i+4$ at least one number is odd, and at least one is divisible by $2$ and not by $4$. Those two numbers have the parity of the exponent different, hence so do the corresponding two among $(i+1)d,(i+2)d,(i+3)d,(i+4)d$. Hence those four numbers can't have the same color.
Edit: for completeness, it's false for $k=2,n=3$ as well. For integer $m$, write it as $3^icdot j,3nmid j$ and color $m$ according to $jmod 3$. Then among $i+1,i+2,i+3$ the two which are not divisible by $3$ will have different remainders, do multiplying them by $d$ gives numbers of different colors. Hence your conjecture only holds in trivial cases $k=1$ and $nleq 2$.
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
$endgroup$
add a comment |
$begingroup$
This is false already for $k=2,n=4$. Color an integer $m$ according to the parity of the exponent of $2$ in the prime factorization. Among $i+1,i+2,i+3,i+4$ at least one number is odd, and at least one is divisible by $2$ and not by $4$. Those two numbers have the parity of the exponent different, hence so do the corresponding two among $(i+1)d,(i+2)d,(i+3)d,(i+4)d$. Hence those four numbers can't have the same color.
Edit: for completeness, it's false for $k=2,n=3$ as well. For integer $m$, write it as $3^icdot j,3nmid j$ and color $m$ according to $jmod 3$. Then among $i+1,i+2,i+3$ the two which are not divisible by $3$ will have different remainders, do multiplying them by $d$ gives numbers of different colors. Hence your conjecture only holds in trivial cases $k=1$ and $nleq 2$.
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
$endgroup$
This is false already for $k=2,n=4$. Color an integer $m$ according to the parity of the exponent of $2$ in the prime factorization. Among $i+1,i+2,i+3,i+4$ at least one number is odd, and at least one is divisible by $2$ and not by $4$. Those two numbers have the parity of the exponent different, hence so do the corresponding two among $(i+1)d,(i+2)d,(i+3)d,(i+4)d$. Hence those four numbers can't have the same color.
Edit: for completeness, it's false for $k=2,n=3$ as well. For integer $m$, write it as $3^icdot j,3nmid j$ and color $m$ according to $jmod 3$. Then among $i+1,i+2,i+3$ the two which are not divisible by $3$ will have different remainders, do multiplying them by $d$ gives numbers of different colors. Hence your conjecture only holds in trivial cases $k=1$ and $nleq 2$.
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
edited Dec 24 '18 at 11:28
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
answered Dec 24 '18 at 9:58
WojowuWojowu
7,17313256
7,17313256
add a comment |
add a comment |
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