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let $a_{n+1}=a_{n}+na_{n-1},a_{1}=1，a_{2}=2$. find the $a_{n}=?$

my ideas: $\dfrac{a_{n+1}}{(n+1)!}=\dfrac{1}{n+1}\dfrac{a_{n}}{n!}+\dfrac{1}{n+1}\dfrac{a_{n-1}}{(n-1)!},$

and let $b_{n}=\dfrac{a_{n}}{n!}$,then we have $(n+1)b_{n+1}=b_{n}+b_{n-1},b_{1}=1,b_{2}=1$,but I failure,

so let $f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n$,and assume that $b_{0}=0$

$f'(x)=\displaystyle\sum_{n=0}^{\infty}nb_{n}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(n+1)b_{n+1}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(b_{n}+b_{n-1})x^{n-1}=1+\dfrac{f(x)}{x}+f(x)$

then $f'(x)-(1+1/x)f(x)=1,f'(0)=1$

so $f(x)=xe^x(\displaystyle\int \dfrac{e^{-x}}{x}dx+c)$

so $c=0$

and $f(x)=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$

and my problem：How can prove that:

if:$f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$

then we have $b_{n}=\displaystyle\sum_{k=0}^{[n/2]}\dfrac{1}{（n-2k)!2^kk!}?$

math110
  • 93,304

2 Answers2

3

Let $f(x)$ be the exponential generating function of $(a_n)$:

$$ f(x) = \sum_{n=1}^{\infty} \frac{a_n}{n!} x^n = \sum_{n=1}^{\infty} b_n x^n, $$

where $(b_n)$ is as in OP's notation. Then

\begin{align*} f'(x) &= 1 + 2x + \sum_{n=1}^{\infty} (n+2) b_{n+2} x^{n+1} \\ &= 1 + 2x + \sum_{n=1}^{\infty} (b_{n+1} + b_n) x^{n+1} \\ &= 1 + 2x + (f(x) - x) + x f(x). \end{align*}

Thus we have

$$ f'(x) - (x+1) f(x) = 1 + x. $$

Multiplying the integration factor $e^{-(x+1)^2 / 2}$, we have

$$ \left\{ e^{-(x+1)^2 / 2} f(x) \right\}' = (1 + x) e^{-(x+1)^2 / 2}. $$

Integrating, we have

$$ e^{-(x+1)^2 / 2} f(x) = - e^{-(x+1)^2 / 2} + c, $$

or equivalently

$$ f(x) = c e^{(x+1)^2 / 2} - 1. $$

By noting that $f(0) = 0$, we must have $c = e^{-1/2}$ and therefore

\begin{align*} f(x) = e^{\frac{x^2}{2}+x} - 1 &= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{x^2 + 2x}{2} \right)^n \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} 2^{k-n} x^{2n-k} \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \sum_{l=n}^{2n} \binom{n}{2n-l} 2^{n-l} x^{l} \\ &= \sum_{l=1}^{\infty} \left( \sum_{\frac{l}{2}\leq n \leq l} \frac{1}{n!} \binom{n}{2n-l} 2^{n-l} \right) x^{l} \end{align*}

Therefore we have

$$ a_n = n! b_n = \sum_{\frac{n}{2}\leq k \leq n} \frac{n!}{(2k-n)!(n-k)!2^{n-k}}. $$

Finally, substituting $k \mapsto n-k$, we have

$$ a_n = \sum_{0\leq k \leq \frac{n}{2}} \frac{n!}{(n-2k)!k!2^k} = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{(n-2k)!k!2^k}. $$

The followings are computer calculations:

enter image description here

Sangchul Lee
  • 167,468
0

If $f(x)=\sum_{n\ge 1}a_nx^{n+1}$ then $$ x^3f'(x)+(x-1)f(x)+x^3+x^2=0 $$

Boris Novikov
  • 17,470