Counting monomials in skew-symmetric+diagonal matrices












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
    2 days 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
    2 days 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
    2 days 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
    2 days 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 yesterday







T. Amdeberhan

















asked 2 days ago









T. AmdeberhanT. Amdeberhan

17.8k229131




17.8k229131








  • 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
    2 days 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
    2 days 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
    2 days 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
    2 days 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
2 days 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
2 days 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
2 days 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
2 days ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

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
    2 days 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









5












$begingroup$

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
    2 days ago
















5












$begingroup$

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
    2 days ago














5












5








5





$begingroup$

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 2 days ago









Richard StanleyRichard Stanley

28.8k9115189




28.8k9115189












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


















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
















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




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


















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