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







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  • 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






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asked 14 hours ago









SharkySharky

311




311




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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
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$begingroup$

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









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    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






















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