Mixing
Divisibility of kissing numbers
$begingroup$
Denote by $ K(d) $ the kissing number in dimension $ d $.
I have two questions :
1) does $ dmid K(d) $ for all $ d $?
2) does $ dmid D $ imply $K(d)mid K(D) $?
number-theory symmetry integer-lattices
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
$endgroup$
add a comment |
$begingroup$
Denote by $ K(d) $ the kissing number in dimension $ d $.
I have two questions :
1) does $ dmid K(d) $ for all $ d $?
2) does $ dmid D $ imply $K(d)mid K(D) $?
number-theory symmetry integer-lattices
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
$endgroup$
4
$begingroup$
I think this question will be very difficult to answer. We do not currently have enough "data" on $K(n)$ for $nneq 1,2,3,4,8,24$, so we can't numerically ascertain either.
$endgroup$
– YiFan
Jan 31 at 21:15
add a comment |
$begingroup$
Denote by $ K(d) $ the kissing number in dimension $ d $.
I have two questions :
1) does $ dmid K(d) $ for all $ d $?
2) does $ dmid D $ imply $K(d)mid K(D) $?
number-theory symmetry integer-lattices
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
$endgroup$
Denote by $ K(d) $ the kissing number in dimension $ d $.
I have two questions :
1) does $ dmid K(d) $ for all $ d $?
2) does $ dmid D $ imply $K(d)mid K(D) $?
number-theory symmetry integer-lattices
number-theory symmetry integer-lattices
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
asked Jan 31 at 21:09
Sylvain JulienSylvain Julien
1,144918
1,144918
4
$begingroup$
I think this question will be very difficult to answer. We do not currently have enough "data" on $K(n)$ for $nneq 1,2,3,4,8,24$, so we can't numerically ascertain either.
$endgroup$
– YiFan
Jan 31 at 21:15
add a comment |
4
$begingroup$
I think this question will be very difficult to answer. We do not currently have enough "data" on $K(n)$ for $nneq 1,2,3,4,8,24$, so we can't numerically ascertain either.
$endgroup$
– YiFan
Jan 31 at 21:15
4
4
$begingroup$
I think this question will be very difficult to answer. We do not currently have enough "data" on $K(n)$ for $nneq 1,2,3,4,8,24$, so we can't numerically ascertain either.
$endgroup$
– YiFan
Jan 31 at 21:15
$begingroup$
I think this question will be very difficult to answer. We do not currently have enough "data" on $K(n)$ for $nneq 1,2,3,4,8,24$, so we can't numerically ascertain either.
$endgroup$
– YiFan
Jan 31 at 21:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As already noted in the comments, these are open questions. However, the answer cannot be "yes" to both.
If the answer to (1) is "yes", then since it is known $40leq K(5)leq 44$, we can conclude $K(5)=40$. It is also known that $500leq K(10)leq 554$, so we'd conclude $K(10)in{500,510,520,530,540,550}$.
If the answer to (2) is "yes", we have $5mid 10$, so we'd need $K(10)=520$. Yet also $2mid 10$, and $K(2)=6nmid 520$.
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As already noted in the comments, these are open questions. However, the answer cannot be "yes" to both.
If the answer to (1) is "yes", then since it is known $40leq K(5)leq 44$, we can conclude $K(5)=40$. It is also known that $500leq K(10)leq 554$, so we'd conclude $K(10)in{500,510,520,530,540,550}$.
If the answer to (2) is "yes", we have $5mid 10$, so we'd need $K(10)=520$. Yet also $2mid 10$, and $K(2)=6nmid 520$.
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
$begingroup$
As already noted in the comments, these are open questions. However, the answer cannot be "yes" to both.
If the answer to (1) is "yes", then since it is known $40leq K(5)leq 44$, we can conclude $K(5)=40$. It is also known that $500leq K(10)leq 554$, so we'd conclude $K(10)in{500,510,520,530,540,550}$.
If the answer to (2) is "yes", we have $5mid 10$, so we'd need $K(10)=520$. Yet also $2mid 10$, and $K(2)=6nmid 520$.
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
$begingroup$
As already noted in the comments, these are open questions. However, the answer cannot be "yes" to both.
If the answer to (1) is "yes", then since it is known $40leq K(5)leq 44$, we can conclude $K(5)=40$. It is also known that $500leq K(10)leq 554$, so we'd conclude $K(10)in{500,510,520,530,540,550}$.
If the answer to (2) is "yes", we have $5mid 10$, so we'd need $K(10)=520$. Yet also $2mid 10$, and $K(2)=6nmid 520$.
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
As already noted in the comments, these are open questions. However, the answer cannot be "yes" to both.
If the answer to (1) is "yes", then since it is known $40leq K(5)leq 44$, we can conclude $K(5)=40$. It is also known that $500leq K(10)leq 554$, so we'd conclude $K(10)in{500,510,520,530,540,550}$.
If the answer to (2) is "yes", we have $5mid 10$, so we'd need $K(10)=520$. Yet also $2mid 10$, and $K(2)=6nmid 520$.
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 31 at 21:57
nathan.j.mcdougallnathan.j.mcdougall
1,519818
1,519818
add a comment |
add a comment |
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4
$begingroup$
I think this question will be very difficult to answer. We do not currently have enough "data" on $K(n)$ for $nneq 1,2,3,4,8,24$, so we can't numerically ascertain either.
$endgroup$
– YiFan
Jan 31 at 21:15