most recent 30 from http://mathoverflow.net 2013-05-19T17:01:55Z http://mathoverflow.net/feeds/question/94440 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94440/find-generating-function Melania 2012-04-18T18:10:21Z 2012-04-20T20:43:59Z

<p>Define a transformation $T_s$ of integer sequence ${ a_n }$ by $$ b_n=T_s(a_n)={n \choose s} \sum_{i=s}^{n-1} \frac{a_i}{{i \choose s}}, $$ for a fixed $s \in \mathbb{N}.$</p> <p>For instance, if we aplly the transformation $T_2$ to the sequence $a_n=1$ then we get the sequence $b_n=n(n-2).$ Maple code for the sample</p> <pre><code> T:=(a,s)-&gt;factor(simplify(product(n-i,i=0..s-1)*sum(a(k)/product(k-i,i=0..s-1),k=s..n-1))); a:=n-&gt;1:T(a,2); n(n-2) </code></pre> <p><strong>Question.</strong> Suppose that $a_n$ has a generating function (ordinary or exponential or another one) $a(x).$ What is the generating function of transformed sequence? </p>

http://mathoverflow.net/questions/94440/find-generating-function/94685#94685 i707107 2012-04-20T20:43:59Z 2012-04-20T20:43:59Z

<p>This can be done step by step. First note that $\binom{n}{s}/\binom{i}{s}$ can be written as $n(n-1)\cdots(n-s+1)/i(i-1)\cdots(i-s+1)$ Since we have the generating function (with assuming $a_i=0$ for $i&lt; s$) $$a(x)x^{-s}=\sum_{i=s}^{\infty} a_ix^{i-s}$$ We obtain the following by integrating $s$ times. Let $A_0(x)=a(x)x^{-s}$, and $A_{k+1}(x)= \int_0^x A_k(t)dt$. Then $$A_s(x)=\sum_{i=s}^{\infty} a_i\frac{x^i}{i(i-1)\cdots (i-s+1)}$$ The generating function for $b_n/ n(n-1)\cdots (n-s+1)$ can be obtained from the product $$\left(\sum_{i=s}^{\infty} a_i\frac{x^i}{i(i-1)\cdots (i-s+1)}\right)\left(\sum_{j=1}^{\infty} x^j\right)$$</p> <p>Now the generating function for $b_n$ follows from differentiating s times Again assuming $b_i =0$ for $i&lt; s$, we have $$B_0(x)= A_s(x) \frac{x}{1-x}$$ $$B_{k+1}(x)=\frac{d}{dx} B_k(x)$$ $$\sum_{n=s}^{\infty} b_n x^n = B_s(x) x^s$$</p>