Partition of twelve integers with a special property.
$begingroup$
Twelve numbers are selected from positive integers $1le nle27$ and partitioned into the following two sets $A$ and $B$ such that the power sums of $A $and $B$ are the same up to the fifth power.
$$A={1,5,10,18,23,27}$$
$$B={2,3,13,15,25,26}$$
We notice that we can also select twelve positive integers $1le nle23 $ and form sets $C$ and $D$ with the same property.
$$C={1,6,7,17,18,23}$$
$$D={2,3,11,13,21,22}$$
My question is what is the minimum $k$ for which this selection of $12$ integers $1le nle k$ is possible.
number-theory
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
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add a comment |
$begingroup$
Twelve numbers are selected from positive integers $1le nle27$ and partitioned into the following two sets $A$ and $B$ such that the power sums of $A $and $B$ are the same up to the fifth power.
$$A={1,5,10,18,23,27}$$
$$B={2,3,13,15,25,26}$$
We notice that we can also select twelve positive integers $1le nle23 $ and form sets $C$ and $D$ with the same property.
$$C={1,6,7,17,18,23}$$
$$D={2,3,11,13,21,22}$$
My question is what is the minimum $k$ for which this selection of $12$ integers $1le nle k$ is possible.
number-theory
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
$endgroup$
add a comment |
$begingroup$
Twelve numbers are selected from positive integers $1le nle27$ and partitioned into the following two sets $A$ and $B$ such that the power sums of $A $and $B$ are the same up to the fifth power.
$$A={1,5,10,18,23,27}$$
$$B={2,3,13,15,25,26}$$
We notice that we can also select twelve positive integers $1le nle23 $ and form sets $C$ and $D$ with the same property.
$$C={1,6,7,17,18,23}$$
$$D={2,3,11,13,21,22}$$
My question is what is the minimum $k$ for which this selection of $12$ integers $1le nle k$ is possible.
number-theory
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
$endgroup$
Twelve numbers are selected from positive integers $1le nle27$ and partitioned into the following two sets $A$ and $B$ such that the power sums of $A $and $B$ are the same up to the fifth power.
$$A={1,5,10,18,23,27}$$
$$B={2,3,13,15,25,26}$$
We notice that we can also select twelve positive integers $1le nle23 $ and form sets $C$ and $D$ with the same property.
$$C={1,6,7,17,18,23}$$
$$D={2,3,11,13,21,22}$$
My question is what is the minimum $k$ for which this selection of $12$ integers $1le nle k$ is possible.
number-theory
number-theory
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
asked Dec 25 '18 at 2:15
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.3k42061
42.3k42061
add a comment |
add a comment |
1 Answer
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$begingroup$
for the very symmetric way you constructed those, you have already found the best possible. For each, you can subtract the midpoint to get sextuples symmetric around zero:
? g=(x^2 - 25) * ( x^2 -36) * ( x^2 - 121)
%9 = x^6 - 182*x^4 + 8281*x^2 - 108900
? h=(x^2 - 1) * ( x^2 -81) * ( x^2 - 100)
%10 = x^6 - 182*x^4 + 8281*x^2 - 8100
?
? i=(x^2 - 1) * ( x^2 -121) * ( x^2 - 144)
%11 = x^6 - 266*x^4 + 17689*x^2 - 17424
? j=(x^2 - 16) * ( x^2 -81) * ( x^2 - 169)
%12 = x^6 - 266*x^4 + 17689*x^2 - 219024
? j-i
%13 = -201600
?
Here are your two and some worse examples:
jagy@phobeusjunior:~$ ./mse
11 6 5 10 9 1
13 9 4 12 11 1
17 9 8 16 13 3
18 11 7 17 14 3
18 13 5 17 15 2
19 11 8 17 16 1
20 11 9 19 15 4
It seems possible that some less symmetric examples might do slightly better, but that is a much more elaborate and slow program
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
for the very symmetric way you constructed those, you have already found the best possible. For each, you can subtract the midpoint to get sextuples symmetric around zero:
? g=(x^2 - 25) * ( x^2 -36) * ( x^2 - 121)
%9 = x^6 - 182*x^4 + 8281*x^2 - 108900
? h=(x^2 - 1) * ( x^2 -81) * ( x^2 - 100)
%10 = x^6 - 182*x^4 + 8281*x^2 - 8100
?
? i=(x^2 - 1) * ( x^2 -121) * ( x^2 - 144)
%11 = x^6 - 266*x^4 + 17689*x^2 - 17424
? j=(x^2 - 16) * ( x^2 -81) * ( x^2 - 169)
%12 = x^6 - 266*x^4 + 17689*x^2 - 219024
? j-i
%13 = -201600
?
Here are your two and some worse examples:
jagy@phobeusjunior:~$ ./mse
11 6 5 10 9 1
13 9 4 12 11 1
17 9 8 16 13 3
18 11 7 17 14 3
18 13 5 17 15 2
19 11 8 17 16 1
20 11 9 19 15 4
It seems possible that some less symmetric examples might do slightly better, but that is a much more elaborate and slow program
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
$endgroup$
add a comment |
$begingroup$
for the very symmetric way you constructed those, you have already found the best possible. For each, you can subtract the midpoint to get sextuples symmetric around zero:
? g=(x^2 - 25) * ( x^2 -36) * ( x^2 - 121)
%9 = x^6 - 182*x^4 + 8281*x^2 - 108900
? h=(x^2 - 1) * ( x^2 -81) * ( x^2 - 100)
%10 = x^6 - 182*x^4 + 8281*x^2 - 8100
?
? i=(x^2 - 1) * ( x^2 -121) * ( x^2 - 144)
%11 = x^6 - 266*x^4 + 17689*x^2 - 17424
? j=(x^2 - 16) * ( x^2 -81) * ( x^2 - 169)
%12 = x^6 - 266*x^4 + 17689*x^2 - 219024
? j-i
%13 = -201600
?
Here are your two and some worse examples:
jagy@phobeusjunior:~$ ./mse
11 6 5 10 9 1
13 9 4 12 11 1
17 9 8 16 13 3
18 11 7 17 14 3
18 13 5 17 15 2
19 11 8 17 16 1
20 11 9 19 15 4
It seems possible that some less symmetric examples might do slightly better, but that is a much more elaborate and slow program
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
$endgroup$
add a comment |
$begingroup$
for the very symmetric way you constructed those, you have already found the best possible. For each, you can subtract the midpoint to get sextuples symmetric around zero:
? g=(x^2 - 25) * ( x^2 -36) * ( x^2 - 121)
%9 = x^6 - 182*x^4 + 8281*x^2 - 108900
? h=(x^2 - 1) * ( x^2 -81) * ( x^2 - 100)
%10 = x^6 - 182*x^4 + 8281*x^2 - 8100
?
? i=(x^2 - 1) * ( x^2 -121) * ( x^2 - 144)
%11 = x^6 - 266*x^4 + 17689*x^2 - 17424
? j=(x^2 - 16) * ( x^2 -81) * ( x^2 - 169)
%12 = x^6 - 266*x^4 + 17689*x^2 - 219024
? j-i
%13 = -201600
?
Here are your two and some worse examples:
jagy@phobeusjunior:~$ ./mse
11 6 5 10 9 1
13 9 4 12 11 1
17 9 8 16 13 3
18 11 7 17 14 3
18 13 5 17 15 2
19 11 8 17 16 1
20 11 9 19 15 4
It seems possible that some less symmetric examples might do slightly better, but that is a much more elaborate and slow program
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
$endgroup$
for the very symmetric way you constructed those, you have already found the best possible. For each, you can subtract the midpoint to get sextuples symmetric around zero:
? g=(x^2 - 25) * ( x^2 -36) * ( x^2 - 121)
%9 = x^6 - 182*x^4 + 8281*x^2 - 108900
? h=(x^2 - 1) * ( x^2 -81) * ( x^2 - 100)
%10 = x^6 - 182*x^4 + 8281*x^2 - 8100
?
? i=(x^2 - 1) * ( x^2 -121) * ( x^2 - 144)
%11 = x^6 - 266*x^4 + 17689*x^2 - 17424
? j=(x^2 - 16) * ( x^2 -81) * ( x^2 - 169)
%12 = x^6 - 266*x^4 + 17689*x^2 - 219024
? j-i
%13 = -201600
?
Here are your two and some worse examples:
jagy@phobeusjunior:~$ ./mse
11 6 5 10 9 1
13 9 4 12 11 1
17 9 8 16 13 3
18 11 7 17 14 3
18 13 5 17 15 2
19 11 8 17 16 1
20 11 9 19 15 4
It seems possible that some less symmetric examples might do slightly better, but that is a much more elaborate and slow program
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
answered Dec 25 '18 at 4:19
Will JagyWill Jagy
105k5103202
105k5103202
add a comment |
add a comment |
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