What are the prime divisors of $det(A_n)$, where $A_n$ is the $ntimes n$ matrix given by...
$begingroup$
For $ninmathbb{Z}_{ge 1}$, let $A_n$ be the $ntimes n$ matrix given by $(A_n)_{i,j}={nchoose |i-j|}$. From this post it is clear that
$$det(A_n)=prod_{k=0}^{n-1}left[left(expleft(frac{2pi k i}{n}right)+1right)^n-1right]=prod_{k=0}^{n-1}left(2^n(-1)^kcos^nleft(frac{pi k}nright)-1right).$$
Also, $det(A_n)$ is obviously integer for all $n$, and $det(A_n)=0$ iff $6mid n$.
Question: What is known about the (prime) divisors of $det(A_n)$?
If this is too broad, I am particularly interested in pairs $(p,d)$ where $p$ is prime with $dmid p-1$ and $pnmiddet(A_{(p-1)/d})$.
Edit: It seems like the product with cosines in it is integer independent of the exponent, see this question of mine.
Edit 2: The first few values of $det A_n$ are:
begin{align*}
det(A_1) &= 1\
det(A_2) &= -3=-(2^2-1)cdot 1^2\
det(A_3) &= 28 = (2^3-1)cdot2^2\
det(A_4) &= -375 = (2^4-1)cdot5^2\
det(A_5) &= 3751 = (2^5-1)cdot11^2\
det(A_6) &= 0\
det(A_7) &= 6835648 = (2^7-1)cdot232^2\
det(A_8) &=-1343091375 = -(2^8-1)cdot2295^2\
det(A_9) &= 364668913756 = (2^9-1)cdot26714^2
end{align*}
Edit 3: It looks like this is called 'Wendt's determinant'. My entire motivation for asking this question had to do with a possible proof for Fermat's last theorem and this link is apparently known as 'Wendt's theorem'
matrices number-theory binomial-coefficients determinant products
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
$endgroup$
add a comment |
$begingroup$
For $ninmathbb{Z}_{ge 1}$, let $A_n$ be the $ntimes n$ matrix given by $(A_n)_{i,j}={nchoose |i-j|}$. From this post it is clear that
$$det(A_n)=prod_{k=0}^{n-1}left[left(expleft(frac{2pi k i}{n}right)+1right)^n-1right]=prod_{k=0}^{n-1}left(2^n(-1)^kcos^nleft(frac{pi k}nright)-1right).$$
Also, $det(A_n)$ is obviously integer for all $n$, and $det(A_n)=0$ iff $6mid n$.
Question: What is known about the (prime) divisors of $det(A_n)$?
If this is too broad, I am particularly interested in pairs $(p,d)$ where $p$ is prime with $dmid p-1$ and $pnmiddet(A_{(p-1)/d})$.
Edit: It seems like the product with cosines in it is integer independent of the exponent, see this question of mine.
Edit 2: The first few values of $det A_n$ are:
begin{align*}
det(A_1) &= 1\
det(A_2) &= -3=-(2^2-1)cdot 1^2\
det(A_3) &= 28 = (2^3-1)cdot2^2\
det(A_4) &= -375 = (2^4-1)cdot5^2\
det(A_5) &= 3751 = (2^5-1)cdot11^2\
det(A_6) &= 0\
det(A_7) &= 6835648 = (2^7-1)cdot232^2\
det(A_8) &=-1343091375 = -(2^8-1)cdot2295^2\
det(A_9) &= 364668913756 = (2^9-1)cdot26714^2
end{align*}
Edit 3: It looks like this is called 'Wendt's determinant'. My entire motivation for asking this question had to do with a possible proof for Fermat's last theorem and this link is apparently known as 'Wendt's theorem'
matrices number-theory binomial-coefficients determinant products
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
$endgroup$
$begingroup$
It looks like $det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:mathbb{Z}tomathbb{Z}$.
$endgroup$
– Mastrem
Dec 20 '18 at 21:21
$begingroup$
$m(n)$ must be the product over $0<2k<n$.
$endgroup$
– metamorphy
Dec 20 '18 at 22:03
$begingroup$
@metamorhpy. Oh, nevermind, I got it. Also, if $dmid k$, then $det A_dmid det A_k$
$endgroup$
– Mastrem
Dec 20 '18 at 22:16
$begingroup$
I found oeis.org/search?q=232%2C2295&language=english&go=Search
$endgroup$
– Mastrem
Dec 21 '18 at 10:27
$begingroup$
See oeis.org/A048954 and oeis.org/A215615
$endgroup$
– lhf
Dec 21 '18 at 10:28
add a comment |
$begingroup$
For $ninmathbb{Z}_{ge 1}$, let $A_n$ be the $ntimes n$ matrix given by $(A_n)_{i,j}={nchoose |i-j|}$. From this post it is clear that
$$det(A_n)=prod_{k=0}^{n-1}left[left(expleft(frac{2pi k i}{n}right)+1right)^n-1right]=prod_{k=0}^{n-1}left(2^n(-1)^kcos^nleft(frac{pi k}nright)-1right).$$
Also, $det(A_n)$ is obviously integer for all $n$, and $det(A_n)=0$ iff $6mid n$.
Question: What is known about the (prime) divisors of $det(A_n)$?
If this is too broad, I am particularly interested in pairs $(p,d)$ where $p$ is prime with $dmid p-1$ and $pnmiddet(A_{(p-1)/d})$.
Edit: It seems like the product with cosines in it is integer independent of the exponent, see this question of mine.
Edit 2: The first few values of $det A_n$ are:
begin{align*}
det(A_1) &= 1\
det(A_2) &= -3=-(2^2-1)cdot 1^2\
det(A_3) &= 28 = (2^3-1)cdot2^2\
det(A_4) &= -375 = (2^4-1)cdot5^2\
det(A_5) &= 3751 = (2^5-1)cdot11^2\
det(A_6) &= 0\
det(A_7) &= 6835648 = (2^7-1)cdot232^2\
det(A_8) &=-1343091375 = -(2^8-1)cdot2295^2\
det(A_9) &= 364668913756 = (2^9-1)cdot26714^2
end{align*}
Edit 3: It looks like this is called 'Wendt's determinant'. My entire motivation for asking this question had to do with a possible proof for Fermat's last theorem and this link is apparently known as 'Wendt's theorem'
matrices number-theory binomial-coefficients determinant products
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
$endgroup$
For $ninmathbb{Z}_{ge 1}$, let $A_n$ be the $ntimes n$ matrix given by $(A_n)_{i,j}={nchoose |i-j|}$. From this post it is clear that
$$det(A_n)=prod_{k=0}^{n-1}left[left(expleft(frac{2pi k i}{n}right)+1right)^n-1right]=prod_{k=0}^{n-1}left(2^n(-1)^kcos^nleft(frac{pi k}nright)-1right).$$
Also, $det(A_n)$ is obviously integer for all $n$, and $det(A_n)=0$ iff $6mid n$.
Question: What is known about the (prime) divisors of $det(A_n)$?
If this is too broad, I am particularly interested in pairs $(p,d)$ where $p$ is prime with $dmid p-1$ and $pnmiddet(A_{(p-1)/d})$.
Edit: It seems like the product with cosines in it is integer independent of the exponent, see this question of mine.
Edit 2: The first few values of $det A_n$ are:
begin{align*}
det(A_1) &= 1\
det(A_2) &= -3=-(2^2-1)cdot 1^2\
det(A_3) &= 28 = (2^3-1)cdot2^2\
det(A_4) &= -375 = (2^4-1)cdot5^2\
det(A_5) &= 3751 = (2^5-1)cdot11^2\
det(A_6) &= 0\
det(A_7) &= 6835648 = (2^7-1)cdot232^2\
det(A_8) &=-1343091375 = -(2^8-1)cdot2295^2\
det(A_9) &= 364668913756 = (2^9-1)cdot26714^2
end{align*}
Edit 3: It looks like this is called 'Wendt's determinant'. My entire motivation for asking this question had to do with a possible proof for Fermat's last theorem and this link is apparently known as 'Wendt's theorem'
matrices number-theory binomial-coefficients determinant products
matrices number-theory binomial-coefficients determinant products
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
edited Dec 21 '18 at 10:30
Mastrem
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
asked Dec 20 '18 at 20:36
MastremMastrem
3,79211230
3,79211230
$begingroup$
It looks like $det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:mathbb{Z}tomathbb{Z}$.
$endgroup$
– Mastrem
Dec 20 '18 at 21:21
$begingroup$
$m(n)$ must be the product over $0<2k<n$.
$endgroup$
– metamorphy
Dec 20 '18 at 22:03
$begingroup$
@metamorhpy. Oh, nevermind, I got it. Also, if $dmid k$, then $det A_dmid det A_k$
$endgroup$
– Mastrem
Dec 20 '18 at 22:16
$begingroup$
I found oeis.org/search?q=232%2C2295&language=english&go=Search
$endgroup$
– Mastrem
Dec 21 '18 at 10:27
$begingroup$
See oeis.org/A048954 and oeis.org/A215615
$endgroup$
– lhf
Dec 21 '18 at 10:28
add a comment |
$begingroup$
It looks like $det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:mathbb{Z}tomathbb{Z}$.
$endgroup$
– Mastrem
Dec 20 '18 at 21:21
$begingroup$
$m(n)$ must be the product over $0<2k<n$.
$endgroup$
– metamorphy
Dec 20 '18 at 22:03
$begingroup$
@metamorhpy. Oh, nevermind, I got it. Also, if $dmid k$, then $det A_dmid det A_k$
$endgroup$
– Mastrem
Dec 20 '18 at 22:16
$begingroup$
I found oeis.org/search?q=232%2C2295&language=english&go=Search
$endgroup$
– Mastrem
Dec 21 '18 at 10:27
$begingroup$
See oeis.org/A048954 and oeis.org/A215615
$endgroup$
– lhf
Dec 21 '18 at 10:28
$begingroup$
It looks like $det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:mathbb{Z}tomathbb{Z}$.
$endgroup$
– Mastrem
Dec 20 '18 at 21:21
$begingroup$
It looks like $det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:mathbb{Z}tomathbb{Z}$.
$endgroup$
– Mastrem
Dec 20 '18 at 21:21
$begingroup$
$m(n)$ must be the product over $0<2k<n$.
$endgroup$
– metamorphy
Dec 20 '18 at 22:03
$begingroup$
$m(n)$ must be the product over $0<2k<n$.
$endgroup$
– metamorphy
Dec 20 '18 at 22:03
$begingroup$
@metamorhpy. Oh, nevermind, I got it. Also, if $dmid k$, then $det A_dmid det A_k$
$endgroup$
– Mastrem
Dec 20 '18 at 22:16
$begingroup$
@metamorhpy. Oh, nevermind, I got it. Also, if $dmid k$, then $det A_dmid det A_k$
$endgroup$
– Mastrem
Dec 20 '18 at 22:16
$begingroup$
I found oeis.org/search?q=232%2C2295&language=english&go=Search
$endgroup$
– Mastrem
Dec 21 '18 at 10:27
$begingroup$
I found oeis.org/search?q=232%2C2295&language=english&go=Search
$endgroup$
– Mastrem
Dec 21 '18 at 10:27
$begingroup$
See oeis.org/A048954 and oeis.org/A215615
$endgroup$
– lhf
Dec 21 '18 at 10:28
$begingroup$
See oeis.org/A048954 and oeis.org/A215615
$endgroup$
– lhf
Dec 21 '18 at 10:28
add a comment |
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$begingroup$
It looks like $det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:mathbb{Z}tomathbb{Z}$.
$endgroup$
– Mastrem
Dec 20 '18 at 21:21
$begingroup$
$m(n)$ must be the product over $0<2k<n$.
$endgroup$
– metamorphy
Dec 20 '18 at 22:03
$begingroup$
@metamorhpy. Oh, nevermind, I got it. Also, if $dmid k$, then $det A_dmid det A_k$
$endgroup$
– Mastrem
Dec 20 '18 at 22:16
$begingroup$
I found oeis.org/search?q=232%2C2295&language=english&go=Search
$endgroup$
– Mastrem
Dec 21 '18 at 10:27
$begingroup$
See oeis.org/A048954 and oeis.org/A215615
$endgroup$
– lhf
Dec 21 '18 at 10:28