Finding determinant of matrix involving binomial coefficients [duplicate]
$begingroup$
This question already has an answer here:
determinant of a matrix with binomial coefficient entries
3 answers
I have been trying to solve a problem and have given my answer in the form of a determinant, namely:
begin{align*}
detbegin{pmatrix}
binom{q}{2} & q & 1 & 0 & dots & 0 \
binom{q}{3} & binom{q}{2} & q & 1 & dots & 0 \
binom{q}{4} & binom{q}{3} & binom{q}{2} & q & dots & 0 \
binom{q}{5} & binom{q}{4} & binom{q}{3} & binom{q}{2} & dots & 0 \
vdots & vdots & vdots & dots & ddots & q \
binom{q}{n+1} & binom{q}{n} & binom{q}{n-1} & binom{q}{n-2} & dots & binom{q}{2}
end{pmatrix}
end{align*}
This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...
Is there a good way to tackle this problem, or is this as good as it gets?
Any help would be much appreciated!
linear-algebra binomial-coefficients determinant
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
$endgroup$
marked as duplicate by darij grinberg, user1551 linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 19 at 0:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
determinant of a matrix with binomial coefficient entries
3 answers
I have been trying to solve a problem and have given my answer in the form of a determinant, namely:
begin{align*}
detbegin{pmatrix}
binom{q}{2} & q & 1 & 0 & dots & 0 \
binom{q}{3} & binom{q}{2} & q & 1 & dots & 0 \
binom{q}{4} & binom{q}{3} & binom{q}{2} & q & dots & 0 \
binom{q}{5} & binom{q}{4} & binom{q}{3} & binom{q}{2} & dots & 0 \
vdots & vdots & vdots & dots & ddots & q \
binom{q}{n+1} & binom{q}{n} & binom{q}{n-1} & binom{q}{n-2} & dots & binom{q}{2}
end{pmatrix}
end{align*}
This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...
Is there a good way to tackle this problem, or is this as good as it gets?
Any help would be much appreciated!
linear-algebra binomial-coefficients determinant
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
$endgroup$
marked as duplicate by darij grinberg, user1551 linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 19 at 0:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$.
$endgroup$
– darij grinberg
Jan 18 at 23:49
add a comment |
$begingroup$
This question already has an answer here:
determinant of a matrix with binomial coefficient entries
3 answers
I have been trying to solve a problem and have given my answer in the form of a determinant, namely:
begin{align*}
detbegin{pmatrix}
binom{q}{2} & q & 1 & 0 & dots & 0 \
binom{q}{3} & binom{q}{2} & q & 1 & dots & 0 \
binom{q}{4} & binom{q}{3} & binom{q}{2} & q & dots & 0 \
binom{q}{5} & binom{q}{4} & binom{q}{3} & binom{q}{2} & dots & 0 \
vdots & vdots & vdots & dots & ddots & q \
binom{q}{n+1} & binom{q}{n} & binom{q}{n-1} & binom{q}{n-2} & dots & binom{q}{2}
end{pmatrix}
end{align*}
This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...
Is there a good way to tackle this problem, or is this as good as it gets?
Any help would be much appreciated!
linear-algebra binomial-coefficients determinant
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
$endgroup$
This question already has an answer here:
determinant of a matrix with binomial coefficient entries
3 answers
I have been trying to solve a problem and have given my answer in the form of a determinant, namely:
begin{align*}
detbegin{pmatrix}
binom{q}{2} & q & 1 & 0 & dots & 0 \
binom{q}{3} & binom{q}{2} & q & 1 & dots & 0 \
binom{q}{4} & binom{q}{3} & binom{q}{2} & q & dots & 0 \
binom{q}{5} & binom{q}{4} & binom{q}{3} & binom{q}{2} & dots & 0 \
vdots & vdots & vdots & dots & ddots & q \
binom{q}{n+1} & binom{q}{n} & binom{q}{n-1} & binom{q}{n-2} & dots & binom{q}{2}
end{pmatrix}
end{align*}
This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...
Is there a good way to tackle this problem, or is this as good as it gets?
Any help would be much appreciated!
This question already has an answer here:
determinant of a matrix with binomial coefficient entries
3 answers
linear-algebra binomial-coefficients determinant
linear-algebra binomial-coefficients determinant
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
asked Jan 17 at 18:59
KurtKnödelKurtKnödel
474
474
marked as duplicate by darij grinberg, user1551 linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 19 at 0:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by darij grinberg, user1551 linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 19 at 0:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$.
$endgroup$
– darij grinberg
Jan 18 at 23:49
add a comment |
$begingroup$
This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$.
$endgroup$
– darij grinberg
Jan 18 at 23:49
$begingroup$
This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$.
$endgroup$
– darij grinberg
Jan 18 at 23:49
$begingroup$
This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$.
$endgroup$
– darij grinberg
Jan 18 at 23:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as
$$ 144 = 1 cdot 2^2 cdot 3^2 cdot 4, $$
$$ 2880 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5, $$
$$ 86400 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6, $$
$$ 3628800 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6^2 cdot 7, $$
That leads to a solid recursion, let me call my number $w,$
$$ d_{w+1} = left( frac{(q+w-2)(q+w-3)}{w(w-1)} right) ; d_w $$
See if you can write that securely in your notation.
One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out
$$
frac{1}{n+1}
left(
begin{array}{c}
q+n-1 \
n
end{array}
right)
left(
begin{array}{c}
q+n-2 \
n
end{array}
right)
$$
? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2
===================================================================
? factor(det4 * 144)
%16 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 1]
? factor(det5 * 2880)
%15 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 1]
? factor(det6 * 86400)
%20 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 1]
?
? factor(det7 * 3628800)
%27 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 2]
[q + 5 1]
=========================================================================
? factor(144)
%21 =
[2 4]
[3 2]
? factor(2880)
%22 =
[2 6]
[3 2]
[5 1]
? factor(86400)
%23 =
[2 7]
[3 3]
[5 2]
?
? factor( 3628800)
%28 =
[2 8]
[3 4]
[5 2]
[7 1]
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
$endgroup$
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
1
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
add a comment |
$begingroup$
I cannot give an analytic proof, but this seems to give the correct results:
$$D=frac{1}{n!(n+1)!}frac{(q+n-1)!}{(q-1)!}frac{(q+n-2)!}{(q-2)!}$$
answered Jan 18 at 23:16
xidgelxidgel
24815
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as
$$ 144 = 1 cdot 2^2 cdot 3^2 cdot 4, $$
$$ 2880 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5, $$
$$ 86400 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6, $$
$$ 3628800 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6^2 cdot 7, $$
That leads to a solid recursion, let me call my number $w,$
$$ d_{w+1} = left( frac{(q+w-2)(q+w-3)}{w(w-1)} right) ; d_w $$
See if you can write that securely in your notation.
One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out
$$
frac{1}{n+1}
left(
begin{array}{c}
q+n-1 \
n
end{array}
right)
left(
begin{array}{c}
q+n-2 \
n
end{array}
right)
$$
? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2
===================================================================
? factor(det4 * 144)
%16 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 1]
? factor(det5 * 2880)
%15 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 1]
? factor(det6 * 86400)
%20 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 1]
?
? factor(det7 * 3628800)
%27 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 2]
[q + 5 1]
=========================================================================
? factor(144)
%21 =
[2 4]
[3 2]
? factor(2880)
%22 =
[2 6]
[3 2]
[5 1]
? factor(86400)
%23 =
[2 7]
[3 3]
[5 2]
?
? factor( 3628800)
%28 =
[2 8]
[3 4]
[5 2]
[7 1]
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
$endgroup$
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
1
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
add a comment |
$begingroup$
seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as
$$ 144 = 1 cdot 2^2 cdot 3^2 cdot 4, $$
$$ 2880 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5, $$
$$ 86400 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6, $$
$$ 3628800 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6^2 cdot 7, $$
That leads to a solid recursion, let me call my number $w,$
$$ d_{w+1} = left( frac{(q+w-2)(q+w-3)}{w(w-1)} right) ; d_w $$
See if you can write that securely in your notation.
One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out
$$
frac{1}{n+1}
left(
begin{array}{c}
q+n-1 \
n
end{array}
right)
left(
begin{array}{c}
q+n-2 \
n
end{array}
right)
$$
? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2
===================================================================
? factor(det4 * 144)
%16 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 1]
? factor(det5 * 2880)
%15 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 1]
? factor(det6 * 86400)
%20 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 1]
?
? factor(det7 * 3628800)
%27 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 2]
[q + 5 1]
=========================================================================
? factor(144)
%21 =
[2 4]
[3 2]
? factor(2880)
%22 =
[2 6]
[3 2]
[5 1]
? factor(86400)
%23 =
[2 7]
[3 3]
[5 2]
?
? factor( 3628800)
%28 =
[2 8]
[3 4]
[5 2]
[7 1]
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
$endgroup$
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
1
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
add a comment |
$begingroup$
seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as
$$ 144 = 1 cdot 2^2 cdot 3^2 cdot 4, $$
$$ 2880 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5, $$
$$ 86400 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6, $$
$$ 3628800 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6^2 cdot 7, $$
That leads to a solid recursion, let me call my number $w,$
$$ d_{w+1} = left( frac{(q+w-2)(q+w-3)}{w(w-1)} right) ; d_w $$
See if you can write that securely in your notation.
One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out
$$
frac{1}{n+1}
left(
begin{array}{c}
q+n-1 \
n
end{array}
right)
left(
begin{array}{c}
q+n-2 \
n
end{array}
right)
$$
? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2
===================================================================
? factor(det4 * 144)
%16 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 1]
? factor(det5 * 2880)
%15 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 1]
? factor(det6 * 86400)
%20 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 1]
?
? factor(det7 * 3628800)
%27 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 2]
[q + 5 1]
=========================================================================
? factor(144)
%21 =
[2 4]
[3 2]
? factor(2880)
%22 =
[2 6]
[3 2]
[5 1]
? factor(86400)
%23 =
[2 7]
[3 3]
[5 2]
?
? factor( 3628800)
%28 =
[2 8]
[3 4]
[5 2]
[7 1]
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
$endgroup$
seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as
$$ 144 = 1 cdot 2^2 cdot 3^2 cdot 4, $$
$$ 2880 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5, $$
$$ 86400 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6, $$
$$ 3628800 = 1 cdot 2^2 cdot 3^2 cdot 4^2 cdot 5^2 cdot 6^2 cdot 7, $$
That leads to a solid recursion, let me call my number $w,$
$$ d_{w+1} = left( frac{(q+w-2)(q+w-3)}{w(w-1)} right) ; d_w $$
See if you can write that securely in your notation.
One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out
$$
frac{1}{n+1}
left(
begin{array}{c}
q+n-1 \
n
end{array}
right)
left(
begin{array}{c}
q+n-2 \
n
end{array}
right)
$$
? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2
===================================================================
? factor(det4 * 144)
%16 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 1]
? factor(det5 * 2880)
%15 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 1]
? factor(det6 * 86400)
%20 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 1]
?
? factor(det7 * 3628800)
%27 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 2]
[q + 5 1]
=========================================================================
? factor(144)
%21 =
[2 4]
[3 2]
? factor(2880)
%22 =
[2 6]
[3 2]
[5 1]
? factor(86400)
%23 =
[2 7]
[3 3]
[5 2]
?
? factor( 3628800)
%28 =
[2 8]
[3 4]
[5 2]
[7 1]
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
edited Jan 17 at 23:31
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
answered Jan 17 at 22:01
Will JagyWill Jagy
103k5101200
103k5101200
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
1
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
add a comment |
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
1
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
Thanks a lot, that is indeed a very nice recursion. I have to admit I am still having some difficulties proving the recursion. Is there any intuition behind it?
$endgroup$
– KurtKnödel
Jan 17 at 23:02
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
@KurtKnödel Time will tell. It does become a closed form, I typed that in now. Same message, see if you can match that up with the way you are writing things.
$endgroup$
– Will Jagy
Jan 17 at 23:33
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
$begingroup$
One more cheeky question. My current approach is expanding the determinant by its first row. Do you think that this is the right approach, or am I way off?
$endgroup$
– KurtKnödel
Jan 18 at 10:48
1
1
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
$begingroup$
Update: I managed to prove it! Thanks a lot!:)
$endgroup$
– KurtKnödel
Jan 18 at 14:18
add a comment |
$begingroup$
I cannot give an analytic proof, but this seems to give the correct results:
$$D=frac{1}{n!(n+1)!}frac{(q+n-1)!}{(q-1)!}frac{(q+n-2)!}{(q-2)!}$$
answered Jan 18 at 23:16
xidgelxidgel
24815
$endgroup$
add a comment |
$begingroup$
I cannot give an analytic proof, but this seems to give the correct results:
$$D=frac{1}{n!(n+1)!}frac{(q+n-1)!}{(q-1)!}frac{(q+n-2)!}{(q-2)!}$$
answered Jan 18 at 23:16
xidgelxidgel
24815
$endgroup$
add a comment |
$begingroup$
I cannot give an analytic proof, but this seems to give the correct results:
$$D=frac{1}{n!(n+1)!}frac{(q+n-1)!}{(q-1)!}frac{(q+n-2)!}{(q-2)!}$$
answered Jan 18 at 23:16
xidgelxidgel
24815
$endgroup$
I cannot give an analytic proof, but this seems to give the correct results:
$$D=frac{1}{n!(n+1)!}frac{(q+n-1)!}{(q-1)!}frac{(q+n-2)!}{(q-2)!}$$
answered Jan 18 at 23:16
xidgelxidgel
24815
answered Jan 18 at 23:16
xidgelxidgel
24815
answered Jan 18 at 23:16
xidgelxidgel
24815
answered Jan 18 at 23:16
xidgelxidgel
24815
24815
add a comment |
add a comment |
$begingroup$
This is the particular case of math.stackexchange.com/questions/2681139 for $l=q$ and $k=2$.
$endgroup$
– darij grinberg
Jan 18 at 23:49