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I am working on some independent research and I came across a recursive sequence of the form

$$ b_{n+2} = \sum_{k=1}^{n} \binom{n}{k-1} d_{n+3-k}b_{k}, $$

with $b_{1} = 1$, $b_{2} = 0$ and $d_{j}$ has an explicit formula... Is there a way to find $b_{n}$ explicitly? Or even in general, something like

$$ a_{n+1} = \sum_{k=0}^{n-m} c_{n,k} a_{k} $$

where $a_{0}, a_{1}, \dots, a_{m} $ are given and $c_{n,k}$ is explicit.

I know generating functions but I don't believe they would work here. Thanks for the help in advance!

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    "I know generating functions but I don't believe they would work here." Please show what you did in this direction, which would justify your belief. – 2017-02-25
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    Generation functions would involve $\sum_{n=1}^{\infty} \sum_{k=1}^{n} \binom{n}{k-1} d_{n+3-k} b_{k} x^n$ and I would expand the inner sum out, but I would get sums of $b_{2}x^n$, $b_{3}x^n$, ... and where would I stop the index of b from being a finite number to being $n-1$ or $n-2$, etc? Because it would go out to $b_{n}x^n$. – 2017-02-25
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    Ah, so it seems your problem is how to manipulate double sums, not that generating functions would fail to help solving your question. A promising start in this direction is to note that the sequences $$a_n=\frac{b_n}{(n-1)!}\qquad c_n=\frac{d_{n+3}}{(n+1)!}$$ with the proper initializations, solve $$(n+1)a_{n+2}=\sum_{k=1}^nc_{n-k}a_k$$ Can you push this further, using the generating functions $$A(x)=\sum_na_nx^n\qquad C(x)=\sum_nc_nx^n\ ?$$ – 2017-02-25
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    Yes my main dilemma is with the double sum. Hmm, I will have to think about it, it could work...too late for tonight. Thanks! – 2017-02-25
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    @Did I don't believe it works...I am running into the same problem. – 2017-02-27
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    What did you try? Please be specific and include the details in your question. – 2017-02-27
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    I just wrote it out. Tried summing $\sum_{k=1}^{n} c_{n-k}a_{k}$ but you run into the same problem where do you break the k values? Because you'll have $c_{n-1}a_{1}$, $c_{n-2}a_{2}$, 3 etc to start (which are good) and $c_{1}a_{n-1}$, $c_{0}a_{n}$ (which are also good) but what about the middle? Where does it transition from a subscript of $n$ with the sequence $c$ and the sequence $a$. All I could think of is to break it at $n/2$ if $n$ is even or $(n-1)/2$ if $n$ is odd. Or something like that. – 2017-02-28
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    Maybe you do not realize this but it is *impossible* to know exactly what you did as long as you merely paraphrase it in comments. Please write down completely and in details what you did (no comment on what you did, only what you did), append it to the question, and then we will be able to make progress. Otherwise, I fail to see the point. – 2017-02-28
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    @Did I've kept it as a recursive sequence and yesterday, I discovered I was able to connect this sequence quite easily to Bell polynomials. But I am still interested about the general case, if you have any input. – 2017-03-08
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    Right, obviously you are not taking into account my comments, perhaps are you not even *reading* them? – 2017-03-08
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    @Did I am reading them, they weren't too helpful after I wrote things out so I decided to keep the recursion as is. When you say "right", did you know it was related to Bell polynomials? That would have helped me greatly. – 2017-03-09

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