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 guess to
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n-k/2rfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






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
    5 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 guess to
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n-k/2rfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






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
    5 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 guess to
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n-k/2rfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






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 guess to
$$g(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n}{2k}frac{(2k)!}{k!}cdotprod_{i=1}^kfrac{4i-3}4cdotsum_{j=0}^{lfloor n-k/2rfloor}
binom{n-2k}{2j}frac{(2j)!}{4^j,j!}.$$






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 54 mins ago







T. Amdeberhan

















asked 6 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
    5 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
    5 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
5 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
5 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

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 hours ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324564%2fcounting-monomials-in-skew-symmetricdiagonal-matrices%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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 hours ago
















3












$begingroup$

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 hours ago














3












3








3





$begingroup$

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 4 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
    2 hours ago


















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
















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




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


















draft saved

draft discarded




















































Thanks for contributing an answer to MathOverflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324564%2fcounting-monomials-in-skew-symmetricdiagonal-matrices%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Щит и меч (фильм) Содержание Названия серий | Сюжет |...

Image
Фильмы по алфавитуФильмы 1968 годаФильмы СССР 1968 годаЧёрно-белые фильмы СССРФильмы студии «Мосфильм»Лидеры советского кинопрокатаФильмы-драмы СССРФильмы о Второй мировой войнеФильмы о шпионахЭкранизации произведений Вадима КожевниковаФильмы Владимира БасоваФильмы о фашизме 1968 годуодноимённому романуВадима КожевниковаВладимиром БасовымСтанислав ЛюбшинГерманиюАбверегауптштурмфюрераССБерлинслужбу СДSchwarzbraun ist die HaselnussXVIII векаЭрикиХермса НиляГитлерюгендаАбвераАх, мой милый АвгустинXVII векаHorst WesselDeutschland über alles У этого термина существуют и другие значения, см. Щит и меч. Щит и меч Жанр драма военный фильм Режиссёр Владимир Басов Автор сценария Владимир Басов Вадим Кожевников В главных ролях Станислав Любшин Валентина Титова Олег Янковский Георгий Мартынюк Владимир Басов Оператор Тимофей Лебешев Композитор Вениамин Баснер Кинокомпания Киностудия «Мосфильм». Творческое объединение писателей и киноработников ...
Read more

Венесуэла на летних Олимпийских играх 2000 Содержание Состав...

Image
БангладешБахрейнБутанБрунейВьетнамГонконгИндияИндонезияИорданияИракИранЙеменКазахстанКамбоджаКатарКитайКиргизияКНДРКувейтЛаосЛиванМалайзияМальдивыМонголияМьянмаНепалОАЭОман ПакистанГосударство ПалестинаСаудовская АравияСингапурСирияТаджикистанТаиландТайваньТуркменияУзбекистанФилиппиныШри-ЛанкаЮжная КореяЯпонияИндивидуальные олимпийские спортсменыАлжирАнголаБенинБотсванаБуркина-ФасоБурундиГабонГамбияГанаГвинеяГвинея-БисауДемократическая Республика КонгоДжибутиЕгипетЗамбияЗимбабвеКабо-ВердеКамерунКенияКоморские островаКот-д’ИвуарЛесотоЛиберияЛивияМаврикийМавританияМадагаскарМалавиМалиМароккоМозамбикНамибияНигерНигерияРеспублика КонгоРуандаСан-Томе и ПринсипиСвазилендСейшелыСенегалСомалиСуданСьерра-ЛеонеТанзанияТогоТунисУгандаЦентральноафриканская РеспубликаЧадЭкваториальная ГвинеяЭритреяЭфиопияЮАРАвстралияАмериканское СамоаВануатуГуамМикронезияНауруНовая ЗеландияОстрова КукаПалауПапуа — Новая ГвинеяСамоаСоломоновы ОстроваТонгаФиджи Венесуэла на Олимпийских играх2000 год в ВенесуэлеСтран...
Read more

Meter-Bus Содержание Параметры шины | Стандартизация |...

Image
RS-232IrDAHARTмодемтоковая петляИРПСMIDIIO-LinkRS-422RS-423RS-485LINM-BusKansas City StandardКомпакт-кассетагрампластинкаPPPSLIPIrDAHARTISO 7816MIDIIO-LinkModbusLINDMX-512P-NETProfibusIEC 608708250 UART16550 UART Сетевые протоколыПоследовательные шины физического уровняполевой шиныасинхронного интерфейсаOpen Metering SystemAESздесь У этого термина существуют и другие значения, см. M-bus (значения). M-Bus (Meter-Bus) — стандарт физического уровня для полевой шины на основе асинхронного интерфейса. Также под этим названием понимают коммуникационный протокол, используемый для связи устройств по этой шине. M-bus преимущественно применяется для приборов учета электрической энергии (электросчётчики), тепловой энергии (теплосчётчики), расходомеров воды и газа. Данные передаются на компьютерную станцию (сервер) напрямую или через концентраторы шины M-Bus, а также усилители-повторители сигнала. Содержание 1 Параметры шины 2 Стандартизация 2.1 Open Metering System ...
Read more