Counting monomials in skew-symmetric+diagonal matricesEquivalence class of permutations based on cycle...



Counting monomials in skew-symmetric+diagonal matrices


Equivalence class of permutations based on cycle decomposition and their inversesShowing a matrix is negative definite [formerly Showing a sum is always positive]What is the number of $2ntimes 2n$-matrices $g=begin{pmatrix} A & B \ C & -A^{t} end{pmatrix}$, $B$ and $C$ symmetric, over the finite field $mathbb{F}_{q}$ with $mathrm{rank}(g)=k$?Estimating a sum involving binomial coefficients [refined]Why does this antisymmetric product factor out a determinant?Set counting problem with a cap on the intersection between the set and a fixed partitionNumber Theory and p-Power-Partitioned NumbersHankel determinant evaluation of special lattice pathsArgmax of weighted sum of binomialsDeterminant of “skew-symmetric” matricesThe vanishing of sum of coefficients: symmetric polynomials













1












$begingroup$


This question is motivated by Richard Stanley's answer to this MO question.



Let $g(n)$ be the number of distinct monomials in the expansion of the determinant of an $ntimes n$ generic "skew-symmetric $+$ diagonal" matrix.



For example, $g(3)=4$ since
begin{align*}
detbegin{pmatrix} x_{1,1}&x_{1,2}&x_{1,3} \
-x_{1,2}&x_{2,2}&x_{2,3} \ -x_{1,3}&-x_{2,3}&x_{3,3}
end{pmatrix}
&=x_{1, 1}x_{2, 2}x_{3, 3}+x_{1, 1}x_{2, 3}^2+x_{1, 2}^2x_{3, 3}
+x_{1, 3}^2x_{2, 2}.
end{align*}
The sequence $g(n)$ seems to have found a match in OEIS with the generating function
$$ sum_{ngeq 0} g(n)frac{x^n}{n!} = frac{e^x}{1-frac12x^2}.$$




QUESTION. Is it true and can you furnish a proof for
$$g(n)=sum_{k=0}^{lfloor frac{n}2rfloor}frac{n!}{(n-2k)!,,2^k}?$$




POSTSCRIPT. I'm convinced by Stanley's reply below, so let's correct the above as follows:
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n/2-krfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






reference-request co.combinatorics linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If we fix all $n-2k$ diagonal elements, it remains a $2ktimes 2k$ squared Pfaffian. It should contain $(2k)!/2^k$ monomials. On purely combinatorial language, the squared Pfaffian monomials correspond to permutations with even cycles only, but we count these permutations up to the change of order of cycles. In other words, we count the spanning subgraphs of $K_{2n}$ in which every component is either an even cycle or an edge.
    $endgroup$
    – Fedor Petrov
    21 hours ago










  • $begingroup$
    Why do you still call it a guess? The exponential generating function from the answer of Richard Stanley proves it.
    $endgroup$
    – Fedor Petrov
    14 hours ago
















1












$begingroup$


This question is motivated by Richard Stanley's answer to this MO question.



Let $g(n)$ be the number of distinct monomials in the expansion of the determinant of an $ntimes n$ generic "skew-symmetric $+$ diagonal" matrix.



For example, $g(3)=4$ since
begin{align*}
detbegin{pmatrix} x_{1,1}&x_{1,2}&x_{1,3} \
-x_{1,2}&x_{2,2}&x_{2,3} \ -x_{1,3}&-x_{2,3}&x_{3,3}
end{pmatrix}
&=x_{1, 1}x_{2, 2}x_{3, 3}+x_{1, 1}x_{2, 3}^2+x_{1, 2}^2x_{3, 3}
+x_{1, 3}^2x_{2, 2}.
end{align*}
The sequence $g(n)$ seems to have found a match in OEIS with the generating function
$$ sum_{ngeq 0} g(n)frac{x^n}{n!} = frac{e^x}{1-frac12x^2}.$$




QUESTION. Is it true and can you furnish a proof for
$$g(n)=sum_{k=0}^{lfloor frac{n}2rfloor}frac{n!}{(n-2k)!,,2^k}?$$




POSTSCRIPT. I'm convinced by Stanley's reply below, so let's correct the above as follows:
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n/2-krfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






reference-request co.combinatorics linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If we fix all $n-2k$ diagonal elements, it remains a $2ktimes 2k$ squared Pfaffian. It should contain $(2k)!/2^k$ monomials. On purely combinatorial language, the squared Pfaffian monomials correspond to permutations with even cycles only, but we count these permutations up to the change of order of cycles. In other words, we count the spanning subgraphs of $K_{2n}$ in which every component is either an even cycle or an edge.
    $endgroup$
    – Fedor Petrov
    21 hours ago










  • $begingroup$
    Why do you still call it a guess? The exponential generating function from the answer of Richard Stanley proves it.
    $endgroup$
    – Fedor Petrov
    14 hours ago














1












1








1


1



$begingroup$


This question is motivated by Richard Stanley's answer to this MO question.



Let $g(n)$ be the number of distinct monomials in the expansion of the determinant of an $ntimes n$ generic "skew-symmetric $+$ diagonal" matrix.



For example, $g(3)=4$ since
begin{align*}
detbegin{pmatrix} x_{1,1}&x_{1,2}&x_{1,3} \
-x_{1,2}&x_{2,2}&x_{2,3} \ -x_{1,3}&-x_{2,3}&x_{3,3}
end{pmatrix}
&=x_{1, 1}x_{2, 2}x_{3, 3}+x_{1, 1}x_{2, 3}^2+x_{1, 2}^2x_{3, 3}
+x_{1, 3}^2x_{2, 2}.
end{align*}
The sequence $g(n)$ seems to have found a match in OEIS with the generating function
$$ sum_{ngeq 0} g(n)frac{x^n}{n!} = frac{e^x}{1-frac12x^2}.$$




QUESTION. Is it true and can you furnish a proof for
$$g(n)=sum_{k=0}^{lfloor frac{n}2rfloor}frac{n!}{(n-2k)!,,2^k}?$$




POSTSCRIPT. I'm convinced by Stanley's reply below, so let's correct the above as follows:
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n/2-krfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






reference-request co.combinatorics linear-algebra







share|cite|improve this question











$endgroup$




This question is motivated by Richard Stanley's answer to this MO question.



Let $g(n)$ be the number of distinct monomials in the expansion of the determinant of an $ntimes n$ generic "skew-symmetric $+$ diagonal" matrix.



For example, $g(3)=4$ since
begin{align*}
detbegin{pmatrix} x_{1,1}&x_{1,2}&x_{1,3} \
-x_{1,2}&x_{2,2}&x_{2,3} \ -x_{1,3}&-x_{2,3}&x_{3,3}
end{pmatrix}
&=x_{1, 1}x_{2, 2}x_{3, 3}+x_{1, 1}x_{2, 3}^2+x_{1, 2}^2x_{3, 3}
+x_{1, 3}^2x_{2, 2}.
end{align*}
The sequence $g(n)$ seems to have found a match in OEIS with the generating function
$$ sum_{ngeq 0} g(n)frac{x^n}{n!} = frac{e^x}{1-frac12x^2}.$$




QUESTION. Is it true and can you furnish a proof for
$$g(n)=sum_{k=0}^{lfloor frac{n}2rfloor}frac{n!}{(n-2k)!,,2^k}?$$




POSTSCRIPT. I'm convinced by Stanley's reply below, so let's correct the above as follows:
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n/2-krfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






reference-request co.combinatorics linear-algebra




reference-request co.combinatorics linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







T. Amdeberhan

















asked 22 hours ago









T. AmdeberhanT. Amdeberhan

17.7k229131




17.7k229131








  • 1




    $begingroup$
    If we fix all $n-2k$ diagonal elements, it remains a $2ktimes 2k$ squared Pfaffian. It should contain $(2k)!/2^k$ monomials. On purely combinatorial language, the squared Pfaffian monomials correspond to permutations with even cycles only, but we count these permutations up to the change of order of cycles. In other words, we count the spanning subgraphs of $K_{2n}$ in which every component is either an even cycle or an edge.
    $endgroup$
    – Fedor Petrov
    21 hours ago










  • $begingroup$
    Why do you still call it a guess? The exponential generating function from the answer of Richard Stanley proves it.
    $endgroup$
    – Fedor Petrov
    14 hours ago














  • 1




    $begingroup$
    If we fix all $n-2k$ diagonal elements, it remains a $2ktimes 2k$ squared Pfaffian. It should contain $(2k)!/2^k$ monomials. On purely combinatorial language, the squared Pfaffian monomials correspond to permutations with even cycles only, but we count these permutations up to the change of order of cycles. In other words, we count the spanning subgraphs of $K_{2n}$ in which every component is either an even cycle or an edge.
    $endgroup$
    – Fedor Petrov
    21 hours ago










  • $begingroup$
    Why do you still call it a guess? The exponential generating function from the answer of Richard Stanley proves it.
    $endgroup$
    – Fedor Petrov
    14 hours ago








1




1




$begingroup$
If we fix all $n-2k$ diagonal elements, it remains a $2ktimes 2k$ squared Pfaffian. It should contain $(2k)!/2^k$ monomials. On purely combinatorial language, the squared Pfaffian monomials correspond to permutations with even cycles only, but we count these permutations up to the change of order of cycles. In other words, we count the spanning subgraphs of $K_{2n}$ in which every component is either an even cycle or an edge.
$endgroup$
– Fedor Petrov
21 hours ago




$begingroup$
If we fix all $n-2k$ diagonal elements, it remains a $2ktimes 2k$ squared Pfaffian. It should contain $(2k)!/2^k$ monomials. On purely combinatorial language, the squared Pfaffian monomials correspond to permutations with even cycles only, but we count these permutations up to the change of order of cycles. In other words, we count the spanning subgraphs of $K_{2n}$ in which every component is either an even cycle or an edge.
$endgroup$
– Fedor Petrov
21 hours ago












$begingroup$
Why do you still call it a guess? The exponential generating function from the answer of Richard Stanley proves it.
$endgroup$
– Fedor Petrov
14 hours ago




$begingroup$
Why do you still call it a guess? The exponential generating function from the answer of Richard Stanley proves it.
$endgroup$
– Fedor Petrov
14 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

The correct generating function is
$$ expleft( x +frac{x^2}{2}+frac 12sum_{ngeq 2}frac{x^{2n}}{2n}right)
=frac{expleft(x+frac{x^2}{4}right)}{(1-x^2)^{1/4}}. $$
This appears in http://oeis.org/A243107, but without a combinatorial or algebraic interpretation. For skew symmetric matrices just multiply by $e^{-x}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, you are right, I missed out on my counting. Thank you!
    $endgroup$
    – T. Amdeberhan
    18 hours ago











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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









4












$begingroup$

The correct generating function is
$$ expleft( x +frac{x^2}{2}+frac 12sum_{ngeq 2}frac{x^{2n}}{2n}right)
=frac{expleft(x+frac{x^2}{4}right)}{(1-x^2)^{1/4}}. $$
This appears in http://oeis.org/A243107, but without a combinatorial or algebraic interpretation. For skew symmetric matrices just multiply by $e^{-x}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, you are right, I missed out on my counting. Thank you!
    $endgroup$
    – T. Amdeberhan
    18 hours ago
















4












$begingroup$

The correct generating function is
$$ expleft( x +frac{x^2}{2}+frac 12sum_{ngeq 2}frac{x^{2n}}{2n}right)
=frac{expleft(x+frac{x^2}{4}right)}{(1-x^2)^{1/4}}. $$
This appears in http://oeis.org/A243107, but without a combinatorial or algebraic interpretation. For skew symmetric matrices just multiply by $e^{-x}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, you are right, I missed out on my counting. Thank you!
    $endgroup$
    – T. Amdeberhan
    18 hours ago














4












4








4





$begingroup$

The correct generating function is
$$ expleft( x +frac{x^2}{2}+frac 12sum_{ngeq 2}frac{x^{2n}}{2n}right)
=frac{expleft(x+frac{x^2}{4}right)}{(1-x^2)^{1/4}}. $$
This appears in http://oeis.org/A243107, but without a combinatorial or algebraic interpretation. For skew symmetric matrices just multiply by $e^{-x}$.






share|cite|improve this answer









$endgroup$



The correct generating function is
$$ expleft( x +frac{x^2}{2}+frac 12sum_{ngeq 2}frac{x^{2n}}{2n}right)
=frac{expleft(x+frac{x^2}{4}right)}{(1-x^2)^{1/4}}. $$
This appears in http://oeis.org/A243107, but without a combinatorial or algebraic interpretation. For skew symmetric matrices just multiply by $e^{-x}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 20 hours ago









Richard StanleyRichard Stanley

28.8k9115189




28.8k9115189












  • $begingroup$
    Oh, you are right, I missed out on my counting. Thank you!
    $endgroup$
    – T. Amdeberhan
    18 hours ago


















  • $begingroup$
    Oh, you are right, I missed out on my counting. Thank you!
    $endgroup$
    – T. Amdeberhan
    18 hours ago
















$begingroup$
Oh, you are right, I missed out on my counting. Thank you!
$endgroup$
– T. Amdeberhan
18 hours ago




$begingroup$
Oh, you are right, I missed out on my counting. Thank you!
$endgroup$
– T. Amdeberhan
18 hours ago


















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