Closed form expression for a recursion relation with binomial coefficients












6












$begingroup$


I am interested in the following sequence: $$ T_n = sumlimits^{n-1}_{k=0} begin{pmatrix} n \ k end{pmatrix} T_{k},     T_0 = C in mathbb{N} $$
I would like to express it as a function of n, but none of the method I have tried work.



Asymptotically, I can tell that $T_n = mathcal{O}(2^{frac{k^2}{2}})$.
One method that failed was to see $T_n$ as the $n$-th term in a series, but those terms grow to fast for it to work.



Do you know how to solve it, or have an intuition regarding how it might get solved?



Thank you.






co.combinatorics sequences-and-series binomial-coefficients integer-sequences







share|cite|improve this question









New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 5




    $begingroup$
    Here is a recipe: sage: T = lambda n: x if n == 0 else sum(binomial(n,k)*T(k) for k in range(n)) sage: T = cached_function(T) sage: [T(n) for n in range(5)] [x, x, 3*x, 13*x, 75*x] sage: oeis([T(n)/x for n in range(20)]) 0: A000670: Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
    $endgroup$
    – Martin Rubey
    14 hours ago


















6












$begingroup$


I am interested in the following sequence: $$ T_n = sumlimits^{n-1}_{k=0} begin{pmatrix} n \ k end{pmatrix} T_{k},     T_0 = C in mathbb{N} $$
I would like to express it as a function of n, but none of the method I have tried work.



Asymptotically, I can tell that $T_n = mathcal{O}(2^{frac{k^2}{2}})$.
One method that failed was to see $T_n$ as the $n$-th term in a series, but those terms grow to fast for it to work.



Do you know how to solve it, or have an intuition regarding how it might get solved?



Thank you.






co.combinatorics sequences-and-series binomial-coefficients integer-sequences







share|cite|improve this question









New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 5




    $begingroup$
    Here is a recipe: sage: T = lambda n: x if n == 0 else sum(binomial(n,k)*T(k) for k in range(n)) sage: T = cached_function(T) sage: [T(n) for n in range(5)] [x, x, 3*x, 13*x, 75*x] sage: oeis([T(n)/x for n in range(20)]) 0: A000670: Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
    $endgroup$
    – Martin Rubey
    14 hours ago
















6












6








6





$begingroup$


I am interested in the following sequence: $$ T_n = sumlimits^{n-1}_{k=0} begin{pmatrix} n \ k end{pmatrix} T_{k},     T_0 = C in mathbb{N} $$
I would like to express it as a function of n, but none of the method I have tried work.



Asymptotically, I can tell that $T_n = mathcal{O}(2^{frac{k^2}{2}})$.
One method that failed was to see $T_n$ as the $n$-th term in a series, but those terms grow to fast for it to work.



Do you know how to solve it, or have an intuition regarding how it might get solved?



Thank you.






co.combinatorics sequences-and-series binomial-coefficients integer-sequences







share|cite|improve this question









New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am interested in the following sequence: $$ T_n = sumlimits^{n-1}_{k=0} begin{pmatrix} n \ k end{pmatrix} T_{k},     T_0 = C in mathbb{N} $$
I would like to express it as a function of n, but none of the method I have tried work.



Asymptotically, I can tell that $T_n = mathcal{O}(2^{frac{k^2}{2}})$.
One method that failed was to see $T_n$ as the $n$-th term in a series, but those terms grow to fast for it to work.



Do you know how to solve it, or have an intuition regarding how it might get solved?



Thank you.






co.combinatorics sequences-and-series binomial-coefficients integer-sequences




co.combinatorics sequences-and-series binomial-coefficients integer-sequences






share|cite|improve this question









New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 13 hours ago









Carlo Beenakker

76.8k9180284




76.8k9180284






New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 14 hours ago









SharkySharky

311




311




New contributor




Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sharky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 5




    $begingroup$
    Here is a recipe: sage: T = lambda n: x if n == 0 else sum(binomial(n,k)*T(k) for k in range(n)) sage: T = cached_function(T) sage: [T(n) for n in range(5)] [x, x, 3*x, 13*x, 75*x] sage: oeis([T(n)/x for n in range(20)]) 0: A000670: Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
    $endgroup$
    – Martin Rubey
    14 hours ago
















  • 5




    $begingroup$
    Here is a recipe: sage: T = lambda n: x if n == 0 else sum(binomial(n,k)*T(k) for k in range(n)) sage: T = cached_function(T) sage: [T(n) for n in range(5)] [x, x, 3*x, 13*x, 75*x] sage: oeis([T(n)/x for n in range(20)]) 0: A000670: Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
    $endgroup$
    – Martin Rubey
    14 hours ago










5




5




$begingroup$
Here is a recipe: sage: T = lambda n: x if n == 0 else sum(binomial(n,k)*T(k) for k in range(n)) sage: T = cached_function(T) sage: [T(n) for n in range(5)] [x, x, 3*x, 13*x, 75*x] sage: oeis([T(n)/x for n in range(20)]) 0: A000670: Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
$endgroup$
– Martin Rubey
14 hours ago






$begingroup$
Here is a recipe: sage: T = lambda n: x if n == 0 else sum(binomial(n,k)*T(k) for k in range(n)) sage: T = cached_function(T) sage: [T(n) for n in range(5)] [x, x, 3*x, 13*x, 75*x] sage: oeis([T(n)/x for n in range(20)]) 0: A000670: Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
$endgroup$
– Martin Rubey
14 hours ago












1 Answer
1






active

oldest

votes


















9












$begingroup$

The $T_n$'s are equal to the product of $C$ and the Fubini numbers: number of ordered partitions of $n$, also known as ordered Bell numbers. The generating function is $(2-e^x)^{-1}$ and the large-$n$ asymptotics is
$$T_nsim C frac{n!}{2(ln 2)^{n+1}}.$$






share|cite|improve this answer









$endgroup$













    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
    });


    }
    });






    Sharky is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f323499%2fclosed-form-expression-for-a-recursion-relation-with-binomial-coefficients%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









    9












    $begingroup$

    The $T_n$'s are equal to the product of $C$ and the Fubini numbers: number of ordered partitions of $n$, also known as ordered Bell numbers. The generating function is $(2-e^x)^{-1}$ and the large-$n$ asymptotics is
    $$T_nsim C frac{n!}{2(ln 2)^{n+1}}.$$






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      The $T_n$'s are equal to the product of $C$ and the Fubini numbers: number of ordered partitions of $n$, also known as ordered Bell numbers. The generating function is $(2-e^x)^{-1}$ and the large-$n$ asymptotics is
      $$T_nsim C frac{n!}{2(ln 2)^{n+1}}.$$






      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        The $T_n$'s are equal to the product of $C$ and the Fubini numbers: number of ordered partitions of $n$, also known as ordered Bell numbers. The generating function is $(2-e^x)^{-1}$ and the large-$n$ asymptotics is
        $$T_nsim C frac{n!}{2(ln 2)^{n+1}}.$$






        share|cite|improve this answer









        $endgroup$



        The $T_n$'s are equal to the product of $C$ and the Fubini numbers: number of ordered partitions of $n$, also known as ordered Bell numbers. The generating function is $(2-e^x)^{-1}$ and the large-$n$ asymptotics is
        $$T_nsim C frac{n!}{2(ln 2)^{n+1}}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 14 hours ago









        Carlo BeenakkerCarlo Beenakker

        76.8k9180284




        76.8k9180284






















            Sharky is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            Sharky is a new contributor. Be nice, and check out our Code of Conduct.













            Sharky is a new contributor. Be nice, and check out our Code of Conduct.












            Sharky is a new contributor. Be nice, and check out our Code of Conduct.
















            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%2f323499%2fclosed-form-expression-for-a-recursion-relation-with-binomial-coefficients%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

            Fluorita

            Image
            Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
            Read more

            Hulsita

            Image
            Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
            Read more

            Península de Txukotka

            Image
            Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
            Read more