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Let $a_0 = 0$, and let $a_{n+1} = (n-1)a_n + n!$ for $n \ge 0$. Find an explicit formula for $a_n$.

I have gotten to the point where I have $\sum_{n \ge 0}a_{n+1}\frac{x^{n+1}}{(n+1)!}=\sum_{n \ge 0}a_{n}\frac{x^{n+1}}{n!}+\sum_{n \ge 0}\frac{x^{n+1}}{n+1}$.

I also set up the exponential generating function as $A(x) = \sum_{n \ge 0}a_n\frac{x^n}{n!}$, which is used to rewrite the above as $A(x)[1-x]=-ln(1-x)$ or $A(x)=-\frac{ln(1-x)}{1-x}$.

This is where I am stuck trying to find the explicit formula for $a_n$. Can you help me find the power series representation of the right-hand side? If so, I should be able to finish the question.

user106219
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  • A related problem. – Mhenni Benghorbal Nov 07 '13 at 19:54
  • @MhenniBenghorbal I looked at the related problem and all that I was able to get is that it can be written as $\sum_{n \ge 0}\frac{x^{n+1}}{n+1}\cdot\sum_{n \ge 0}x^n$, which expanded out gives $x+\frac{3}{2}x^2+\frac{11}{6}x^3+\frac{50}{24}x^4+\cdot\cdot\cdot$. This I can see has the form $\sum_{n \ge 0}C\cdot\frac{x^n}{n!}$ where C is some expression that I have not been able to come up with to generate the sequence {0,1,3,11,50,274,...}. – user106219 Nov 07 '13 at 21:26

1 Answers1

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I’m assuming that the recurrence is actually $a_{n+1}=(n+1)a_n+n!$, in order to justify your equation

$$\sum_{n\ge 0}a_{n+1}\frac{x^{n+1}}{(n+1)!}=\sum_{n\ge 0}a_n\frac{x^{n+1}}{n!}+\sum_{n\ge 0}\frac{x^{n+1}}{n+1}\;.$$

Taking $A(x)$ to be the generating function, I also get

$$A(x)=-\frac{\ln(1-x)}{1-x}=\frac1{1-x}\cdot\ln\frac1{1-x}=\sum_{n\ge 0}x^n\sum_{n\ge 1}\frac{x^n}n\;.$$

Now take the product:

$$\sum_{n\ge 0}x^n\sum_{n\ge 1}\frac{x^n}n=\sum_{n\ge 1}\sum_{k=1}^n\frac1kx^n=\sum_{n\ge 1}H_nx^n\;,$$

where $H_n=\sum_{k=1}^n\frac1k$ is the $n$-th harmonic number. Finally,

$$\sum_{n\ge 1}H_nx^n=\sum_{n\ge 1}n!H_n\frac{x^n}{n!}\;,$$

so $$a_n=n!H_n=n!\sum_{k=1}^n\frac1k\;.$$

In fact $\displaystyle a_n={{n+1}\brack 2}$, a Stirling number of the first kind, and the recurrence is a special case of the general recurrence

$${{n+1}\brack k}=n{n\brack k}+{n\brack{k-1}}\;,$$

since $\displaystyle{n\brack 1}=(n-1)!$.

Brian M. Scott
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