Mixing
Multiplying to make sorted sequence - limit on multiplicand
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Given a list of positive integers $ x_1, x_2, x_3, ldots, x_n $, we can find positive integer coefficients
$ c_1, c_2, c_3, ldots, c_n $ such that
$$ c_1x_1 lt c_2x_2 lt c_3x_3 lt ldots lt c_nx_n $$
(For eg., for $ 5, 4, 12, 1, 3 $, the coefficients $ 1, 2, 1, 13, 5 $ produce the increasing list $ 5, 8, 12, 13, 15 $.)
There is a smallest (i.e. least in value) set* of such coefficients for a given list. What is an upper bound, in terms of values in the given list, on the largest coefficient in such a minimum-value set of coefficients?
* By smallest, what I have in mind is the result of this iterative process:
$ c_1 = 1 $.
$ c_2 $ is the smallest positive integer such that $ c_2 x_2 gt x_1 $.
$ c_3 $ is the smallest positive integer such that $ c_3 x_3 gt c_2 x_2 $.
$ c_4 $ is the smallest positive integer such that $ c_4 x_4 gt c_3 x_3 $.
And so on.
I think this might be the same as saying the set of coefficients with the least sum, but in case it is not, the above is the actual expected answer.
linear-transformations order-theory transformation
asked Aug 5 at 23:59
sundar
1166
add a comment |Â
up vote
1
down vote
favorite
Given a list of positive integers $ x_1, x_2, x_3, ldots, x_n $, we can find positive integer coefficients
$ c_1, c_2, c_3, ldots, c_n $ such that
$$ c_1x_1 lt c_2x_2 lt c_3x_3 lt ldots lt c_nx_n $$
(For eg., for $ 5, 4, 12, 1, 3 $, the coefficients $ 1, 2, 1, 13, 5 $ produce the increasing list $ 5, 8, 12, 13, 15 $.)
There is a smallest (i.e. least in value) set* of such coefficients for a given list. What is an upper bound, in terms of values in the given list, on the largest coefficient in such a minimum-value set of coefficients?
* By smallest, what I have in mind is the result of this iterative process:
$ c_1 = 1 $.
$ c_2 $ is the smallest positive integer such that $ c_2 x_2 gt x_1 $.
$ c_3 $ is the smallest positive integer such that $ c_3 x_3 gt c_2 x_2 $.
$ c_4 $ is the smallest positive integer such that $ c_4 x_4 gt c_3 x_3 $.
And so on.
I think this might be the same as saying the set of coefficients with the least sum, but in case it is not, the above is the actual expected answer.
linear-transformations order-theory transformation
asked Aug 5 at 23:59
sundar
1166
Questions : 1) are $c_1,dots,c_n$ supposed to be integers ? 2) If this is the case, what good order do you define on $mathbb N^n$ to define your "smallest" set of integers ?
– Nicolas FRANCOIS
Aug 6 at 0:22
@NicolasFRANCOIS 1) Yes, updated the question on that. 2) I was afraid someone was going to ask that! I think what I need is the same as the set of such integer coefficients with the smallest sum, but I'll update the question with details.
– sundar
Aug 6 at 0:27
If any of the $c$s is $1$ you can ignore everything before it when considering the later part. You clearly get the largest $c$ by having $x_n=1$ and working the numbers before it to make $c_n-1x_n-1$ as large as possible.
– Ross Millikan
Aug 6 at 0:31
One "measure" of the largeness of your coefficients could be $max_k c_k$, in which case your "algorithm" may not be optimal...
– Nicolas FRANCOIS
Aug 6 at 0:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a list of positive integers $ x_1, x_2, x_3, ldots, x_n $, we can find positive integer coefficients
$ c_1, c_2, c_3, ldots, c_n $ such that
$$ c_1x_1 lt c_2x_2 lt c_3x_3 lt ldots lt c_nx_n $$
(For eg., for $ 5, 4, 12, 1, 3 $, the coefficients $ 1, 2, 1, 13, 5 $ produce the increasing list $ 5, 8, 12, 13, 15 $.)
There is a smallest (i.e. least in value) set* of such coefficients for a given list. What is an upper bound, in terms of values in the given list, on the largest coefficient in such a minimum-value set of coefficients?
* By smallest, what I have in mind is the result of this iterative process:
$ c_1 = 1 $.
$ c_2 $ is the smallest positive integer such that $ c_2 x_2 gt x_1 $.
$ c_3 $ is the smallest positive integer such that $ c_3 x_3 gt c_2 x_2 $.
$ c_4 $ is the smallest positive integer such that $ c_4 x_4 gt c_3 x_3 $.
And so on.
I think this might be the same as saying the set of coefficients with the least sum, but in case it is not, the above is the actual expected answer.
linear-transformations order-theory transformation
asked Aug 5 at 23:59
sundar
1166
Given a list of positive integers $ x_1, x_2, x_3, ldots, x_n $, we can find positive integer coefficients
$ c_1, c_2, c_3, ldots, c_n $ such that
$$ c_1x_1 lt c_2x_2 lt c_3x_3 lt ldots lt c_nx_n $$
(For eg., for $ 5, 4, 12, 1, 3 $, the coefficients $ 1, 2, 1, 13, 5 $ produce the increasing list $ 5, 8, 12, 13, 15 $.)
There is a smallest (i.e. least in value) set* of such coefficients for a given list. What is an upper bound, in terms of values in the given list, on the largest coefficient in such a minimum-value set of coefficients?
* By smallest, what I have in mind is the result of this iterative process:
$ c_1 = 1 $.
$ c_2 $ is the smallest positive integer such that $ c_2 x_2 gt x_1 $.
$ c_3 $ is the smallest positive integer such that $ c_3 x_3 gt c_2 x_2 $.
$ c_4 $ is the smallest positive integer such that $ c_4 x_4 gt c_3 x_3 $.
And so on.
I think this might be the same as saying the set of coefficients with the least sum, but in case it is not, the above is the actual expected answer.
linear-transformations order-theory transformation
asked Aug 5 at 23:59
sundar
1166
edited Aug 6 at 0:34
asked Aug 5 at 23:59
sundar
1166
asked Aug 5 at 23:59
sundar
1166
asked Aug 5 at 23:59
sundar
1166
1166
Questions : 1) are $c_1,dots,c_n$ supposed to be integers ? 2) If this is the case, what good order do you define on $mathbb N^n$ to define your "smallest" set of integers ?
– Nicolas FRANCOIS
Aug 6 at 0:22
@NicolasFRANCOIS 1) Yes, updated the question on that. 2) I was afraid someone was going to ask that! I think what I need is the same as the set of such integer coefficients with the smallest sum, but I'll update the question with details.
– sundar
Aug 6 at 0:27
If any of the $c$s is $1$ you can ignore everything before it when considering the later part. You clearly get the largest $c$ by having $x_n=1$ and working the numbers before it to make $c_n-1x_n-1$ as large as possible.
– Ross Millikan
Aug 6 at 0:31
One "measure" of the largeness of your coefficients could be $max_k c_k$, in which case your "algorithm" may not be optimal...
– Nicolas FRANCOIS
Aug 6 at 0:56
add a comment |Â
Questions : 1) are $c_1,dots,c_n$ supposed to be integers ? 2) If this is the case, what good order do you define on $mathbb N^n$ to define your "smallest" set of integers ?
– Nicolas FRANCOIS
Aug 6 at 0:22
@NicolasFRANCOIS 1) Yes, updated the question on that. 2) I was afraid someone was going to ask that! I think what I need is the same as the set of such integer coefficients with the smallest sum, but I'll update the question with details.
– sundar
Aug 6 at 0:27
If any of the $c$s is $1$ you can ignore everything before it when considering the later part. You clearly get the largest $c$ by having $x_n=1$ and working the numbers before it to make $c_n-1x_n-1$ as large as possible.
– Ross Millikan
Aug 6 at 0:31
One "measure" of the largeness of your coefficients could be $max_k c_k$, in which case your "algorithm" may not be optimal...
– Nicolas FRANCOIS
Aug 6 at 0:56
Questions : 1) are $c_1,dots,c_n$ supposed to be integers ? 2) If this is the case, what good order do you define on $mathbb N^n$ to define your "smallest" set of integers ?
– Nicolas FRANCOIS
Aug 6 at 0:22
Questions : 1) are $c_1,dots,c_n$ supposed to be integers ? 2) If this is the case, what good order do you define on $mathbb N^n$ to define your "smallest" set of integers ?
– Nicolas FRANCOIS
Aug 6 at 0:22
@NicolasFRANCOIS 1) Yes, updated the question on that. 2) I was afraid someone was going to ask that! I think what I need is the same as the set of such integer coefficients with the smallest sum, but I'll update the question with details.
– sundar
Aug 6 at 0:27
@NicolasFRANCOIS 1) Yes, updated the question on that. 2) I was afraid someone was going to ask that! I think what I need is the same as the set of such integer coefficients with the smallest sum, but I'll update the question with details.
– sundar
Aug 6 at 0:27
If any of the $c$s is $1$ you can ignore everything before it when considering the later part. You clearly get the largest $c$ by having $x_n=1$ and working the numbers before it to make $c_n-1x_n-1$ as large as possible.
– Ross Millikan
Aug 6 at 0:31
If any of the $c$s is $1$ you can ignore everything before it when considering the later part. You clearly get the largest $c$ by having $x_n=1$ and working the numbers before it to make $c_n-1x_n-1$ as large as possible.
– Ross Millikan
Aug 6 at 0:31
One "measure" of the largeness of your coefficients could be $max_k c_k$, in which case your "algorithm" may not be optimal...
– Nicolas FRANCOIS
Aug 6 at 0:56
One "measure" of the largeness of your coefficients could be $max_k c_k$, in which case your "algorithm" may not be optimal...
– Nicolas FRANCOIS
Aug 6 at 0:56
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
With your "algorithm", you have, for all $k$, by choosing the least possible $c_k+1$ ($lfloor xrfloor$ being the greatest integer lower than $x$):
$$c_k+1=leftlfloorfracc_kx_kx_k+1rightrfloor+1le fracc_kx_kx_k+1+1$$
So for example :
$$c_2le fracc_1x_1x_2+1 = fracx_1+x_2x_2$$
$$c_3le fracc_2x_2x_3+1=fracx_1+x_2+x_3x_3$$
And so on. More generally :
$$(forall k) c_kle fracx_1+x_2+dots+x_kx_k$$
So one simple majoration for your coefficients is :
$$(forall k) c_kle sum_i=1^n x_i$$
Is that the kind of majoration you are looking for ?
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
add a comment |Â
up vote
0
down vote
If any of the $x_i$ are large enough that we can get away with $c_i=1$ it essentially breaks the sequence. We can ignore all the $x$'s before and start the sequence there. If we start with some number for $x_1$ we can imagine choosing the rest of the $x$s to make the products as large as possible while not having any $c_i$ except the last be $1$. We want $x_2=x_1-1$, which gives $c_2=2$ and the second product is $2(x_1-1)$. The pattern continues with $x_3=2(x_1-1)-1, c_3=2,$ product $4x_1-6$. Then $x_4=4x_1-7,$ product $8x_1-14$. We get that the $k^th$ product is $2^k-1x_1-2^k+1+2$. Finally we choose $x_n=1$, which forces $c_n=2^n-2x_1-2^n+3$. I think that is as large as you can force for the given $x_1$. All the $c$s except the first and last are $2$. Starting with $x_1=10$ and letting $n=8$ we get the following
$$begin array r r r ri&x_i&c_i&x_ic_i\ hline1&10&1&10\
2&9&2&18\3&17&2&34\4&33&2&66\5&65&2&130\6&129&2&258\7&257&2&514\8&1&515&515 end array$$
answered Aug 6 at 0:45
Ross Millikan
276k21187352
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
With your "algorithm", you have, for all $k$, by choosing the least possible $c_k+1$ ($lfloor xrfloor$ being the greatest integer lower than $x$):
$$c_k+1=leftlfloorfracc_kx_kx_k+1rightrfloor+1le fracc_kx_kx_k+1+1$$
So for example :
$$c_2le fracc_1x_1x_2+1 = fracx_1+x_2x_2$$
$$c_3le fracc_2x_2x_3+1=fracx_1+x_2+x_3x_3$$
And so on. More generally :
$$(forall k) c_kle fracx_1+x_2+dots+x_kx_k$$
So one simple majoration for your coefficients is :
$$(forall k) c_kle sum_i=1^n x_i$$
Is that the kind of majoration you are looking for ?
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
add a comment |Â
up vote
1
down vote
accepted
With your "algorithm", you have, for all $k$, by choosing the least possible $c_k+1$ ($lfloor xrfloor$ being the greatest integer lower than $x$):
$$c_k+1=leftlfloorfracc_kx_kx_k+1rightrfloor+1le fracc_kx_kx_k+1+1$$
So for example :
$$c_2le fracc_1x_1x_2+1 = fracx_1+x_2x_2$$
$$c_3le fracc_2x_2x_3+1=fracx_1+x_2+x_3x_3$$
And so on. More generally :
$$(forall k) c_kle fracx_1+x_2+dots+x_kx_k$$
So one simple majoration for your coefficients is :
$$(forall k) c_kle sum_i=1^n x_i$$
Is that the kind of majoration you are looking for ?
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
With your "algorithm", you have, for all $k$, by choosing the least possible $c_k+1$ ($lfloor xrfloor$ being the greatest integer lower than $x$):
$$c_k+1=leftlfloorfracc_kx_kx_k+1rightrfloor+1le fracc_kx_kx_k+1+1$$
So for example :
$$c_2le fracc_1x_1x_2+1 = fracx_1+x_2x_2$$
$$c_3le fracc_2x_2x_3+1=fracx_1+x_2+x_3x_3$$
And so on. More generally :
$$(forall k) c_kle fracx_1+x_2+dots+x_kx_k$$
So one simple majoration for your coefficients is :
$$(forall k) c_kle sum_i=1^n x_i$$
Is that the kind of majoration you are looking for ?
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
With your "algorithm", you have, for all $k$, by choosing the least possible $c_k+1$ ($lfloor xrfloor$ being the greatest integer lower than $x$):
$$c_k+1=leftlfloorfracc_kx_kx_k+1rightrfloor+1le fracc_kx_kx_k+1+1$$
So for example :
$$c_2le fracc_1x_1x_2+1 = fracx_1+x_2x_2$$
$$c_3le fracc_2x_2x_3+1=fracx_1+x_2+x_3x_3$$
And so on. More generally :
$$(forall k) c_kle fracx_1+x_2+dots+x_kx_k$$
So one simple majoration for your coefficients is :
$$(forall k) c_kle sum_i=1^n x_i$$
Is that the kind of majoration you are looking for ?
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
answered Aug 6 at 0:54
Nicolas FRANCOIS
3,2501415
3,2501415
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
add a comment |Â
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
Yes, that's exactly the sort of bound I was looking for. Good that it comes down to a nicely simple expression. Thank you!
– sundar
Aug 6 at 1:05
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
This wasn't part of the question so feel free to ignore it, but: what would be an upper bound on the largest (i.e. final) value in the list resulting from the multiplication? One obvious one is $ x_n $ multiplied by the above summation, but is there a tighter bound than that?
– sundar
Aug 6 at 1:16
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
As Ross answer showed, the sum of all values seems to be reachable for a particular set of values $(x_1,dots,x_n)$, so my upper bound looks accurate :-)
– Nicolas FRANCOIS
Aug 6 at 2:46
add a comment |Â
up vote
0
down vote
If any of the $x_i$ are large enough that we can get away with $c_i=1$ it essentially breaks the sequence. We can ignore all the $x$'s before and start the sequence there. If we start with some number for $x_1$ we can imagine choosing the rest of the $x$s to make the products as large as possible while not having any $c_i$ except the last be $1$. We want $x_2=x_1-1$, which gives $c_2=2$ and the second product is $2(x_1-1)$. The pattern continues with $x_3=2(x_1-1)-1, c_3=2,$ product $4x_1-6$. Then $x_4=4x_1-7,$ product $8x_1-14$. We get that the $k^th$ product is $2^k-1x_1-2^k+1+2$. Finally we choose $x_n=1$, which forces $c_n=2^n-2x_1-2^n+3$. I think that is as large as you can force for the given $x_1$. All the $c$s except the first and last are $2$. Starting with $x_1=10$ and letting $n=8$ we get the following
$$begin array r r r ri&x_i&c_i&x_ic_i\ hline1&10&1&10\
2&9&2&18\3&17&2&34\4&33&2&66\5&65&2&130\6&129&2&258\7&257&2&514\8&1&515&515 end array$$
answered Aug 6 at 0:45
Ross Millikan
276k21187352
add a comment |Â
up vote
0
down vote
If any of the $x_i$ are large enough that we can get away with $c_i=1$ it essentially breaks the sequence. We can ignore all the $x$'s before and start the sequence there. If we start with some number for $x_1$ we can imagine choosing the rest of the $x$s to make the products as large as possible while not having any $c_i$ except the last be $1$. We want $x_2=x_1-1$, which gives $c_2=2$ and the second product is $2(x_1-1)$. The pattern continues with $x_3=2(x_1-1)-1, c_3=2,$ product $4x_1-6$. Then $x_4=4x_1-7,$ product $8x_1-14$. We get that the $k^th$ product is $2^k-1x_1-2^k+1+2$. Finally we choose $x_n=1$, which forces $c_n=2^n-2x_1-2^n+3$. I think that is as large as you can force for the given $x_1$. All the $c$s except the first and last are $2$. Starting with $x_1=10$ and letting $n=8$ we get the following
$$begin array r r r ri&x_i&c_i&x_ic_i\ hline1&10&1&10\
2&9&2&18\3&17&2&34\4&33&2&66\5&65&2&130\6&129&2&258\7&257&2&514\8&1&515&515 end array$$
answered Aug 6 at 0:45
Ross Millikan
276k21187352
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If any of the $x_i$ are large enough that we can get away with $c_i=1$ it essentially breaks the sequence. We can ignore all the $x$'s before and start the sequence there. If we start with some number for $x_1$ we can imagine choosing the rest of the $x$s to make the products as large as possible while not having any $c_i$ except the last be $1$. We want $x_2=x_1-1$, which gives $c_2=2$ and the second product is $2(x_1-1)$. The pattern continues with $x_3=2(x_1-1)-1, c_3=2,$ product $4x_1-6$. Then $x_4=4x_1-7,$ product $8x_1-14$. We get that the $k^th$ product is $2^k-1x_1-2^k+1+2$. Finally we choose $x_n=1$, which forces $c_n=2^n-2x_1-2^n+3$. I think that is as large as you can force for the given $x_1$. All the $c$s except the first and last are $2$. Starting with $x_1=10$ and letting $n=8$ we get the following
$$begin array r r r ri&x_i&c_i&x_ic_i\ hline1&10&1&10\
2&9&2&18\3&17&2&34\4&33&2&66\5&65&2&130\6&129&2&258\7&257&2&514\8&1&515&515 end array$$
answered Aug 6 at 0:45
Ross Millikan
276k21187352
If any of the $x_i$ are large enough that we can get away with $c_i=1$ it essentially breaks the sequence. We can ignore all the $x$'s before and start the sequence there. If we start with some number for $x_1$ we can imagine choosing the rest of the $x$s to make the products as large as possible while not having any $c_i$ except the last be $1$. We want $x_2=x_1-1$, which gives $c_2=2$ and the second product is $2(x_1-1)$. The pattern continues with $x_3=2(x_1-1)-1, c_3=2,$ product $4x_1-6$. Then $x_4=4x_1-7,$ product $8x_1-14$. We get that the $k^th$ product is $2^k-1x_1-2^k+1+2$. Finally we choose $x_n=1$, which forces $c_n=2^n-2x_1-2^n+3$. I think that is as large as you can force for the given $x_1$. All the $c$s except the first and last are $2$. Starting with $x_1=10$ and letting $n=8$ we get the following
$$begin array r r r ri&x_i&c_i&x_ic_i\ hline1&10&1&10\
2&9&2&18\3&17&2&34\4&33&2&66\5&65&2&130\6&129&2&258\7&257&2&514\8&1&515&515 end array$$
answered Aug 6 at 0:45
Ross Millikan
276k21187352
edited Aug 6 at 0:53
answered Aug 6 at 0:45
Ross Millikan
276k21187352
answered Aug 6 at 0:45
Ross Millikan
276k21187352
answered Aug 6 at 0:45
Ross Millikan
276k21187352
276k21187352
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Questions : 1) are $c_1,dots,c_n$ supposed to be integers ? 2) If this is the case, what good order do you define on $mathbb N^n$ to define your "smallest" set of integers ?
– Nicolas FRANCOIS
Aug 6 at 0:22
@NicolasFRANCOIS 1) Yes, updated the question on that. 2) I was afraid someone was going to ask that! I think what I need is the same as the set of such integer coefficients with the smallest sum, but I'll update the question with details.
– sundar
Aug 6 at 0:27
If any of the $c$s is $1$ you can ignore everything before it when considering the later part. You clearly get the largest $c$ by having $x_n=1$ and working the numbers before it to make $c_n-1x_n-1$ as large as possible.
– Ross Millikan
Aug 6 at 0:31
One "measure" of the largeness of your coefficients could be $max_k c_k$, in which case your "algorithm" may not be optimal...
– Nicolas FRANCOIS
Aug 6 at 0:56