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Question:

let sequence $\{a_{n}\}$,such $a_{-1}=1,a_{0}=1$

and $$a_{n}=2a_{n-1}+(2n-1)^2a_{n-2},n\ge 1$$

Find the $a_{n}$

I find $$a_{0}=1,a_{1}=3,a_{2}=15,a_{3}=105$$ and I found $$a_{0}=1$$ $$a_{1}=1\cdot 3$$ $$a_{2}=1\cdot 3\cdot 5$$ $$a_{3}=1\cdot 3\cdot 5\cdot 7$$ so I guess $$a_{n}=1\cdot 3\cdot 5\cdots (2n+1)$$ and It is easy use Mathematical induction to prove it.

Now My question: can you someone have other methods?

why I want to see other methods,because I think this problem reslut is interesting.and this sequence form is seem not ugly,so I think this problem have without mathematical indution.

Community
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math110
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  • 1
    Should that say $a_n = 2a_{n-1}+(2n-1)^2a_{\color{red}{n-2}}$? – JimmyK4542 Sep 04 '14 at 07:59
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    I cannot see one. But if you have a nice easy way of solving a problem, why do you want a different way? If the method you have is horribly complicated, of course you want a simpler method, but that is not the case here. – almagest Sep 04 '14 at 08:42
  • Actually we can get an exponential generating function for the recurrence. I think $e^{\frac{x^2}2}$ works because of http://math.stackexchange.com/questions/917632/evaluate-the-sum/917653#917653 – Isomorphism Sep 04 '14 at 09:01
  • From your results, it is clear that $a_n=(2n+1)!!$ (sequence A001147 at OEIS) – Claude Leibovici Sep 04 '14 at 09:11

4 Answers4

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Taking $$ g_n = \frac{a_{n+1}}{a_n} $$ This gives us $$ g_{n-1}(g_n-2)=(2n+1)^2$$ This gives, $g_n = 2n+3$ Now , $$\frac{a_{n+1}}{a_{-1}} = g_{-1}g_1g_2\cdots g_n $$ $$ a_n = g_{n-1} \cdot g_{n-2} \cdots \cdot g_{-1} $$ $$ a_n = (2n+1)!! \Box $$

Shivang jindal
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3

Starting with $$ a_n=2a_{n-1}+(2n-1)^2a_{n-2}\tag{1} $$ Let $a_n=2^nb_n$, then we have $$ \begin{align} b_n&=b_{n-1}+(n-\tfrac12)^2b_{n-2}\tag{2a}\\ b_{n-1}&=b_{n-2}+(n-\tfrac32)^2b_{n-3}\tag{2b}\\ b_n-b_{n-1}&=(n-\tfrac12)^2b_{n-2}\tag{2c}\\ (n-\tfrac12)b_{n-1}&=(n-\tfrac12)b_{n-2}+(n-\tfrac12)(n-\tfrac32)^2b_{n-3}\tag{2d}\\ b_n-(n+\tfrac12)b_{n-1}&=(n-\tfrac12)(n-\tfrac32)b_{n-2}-(n-\tfrac12)(n-\tfrac32)^2b_{n-3}\tag{2e}\\ &=(n-\tfrac12)(n-\tfrac32)\left[b_{n-2}-(n-\tfrac32)b_{n-3}\right]\tag{2f} \end{align} $$ Explanation:
$\mathrm{(2a)}$: $a_n=2^nb_n$ applied to $(1)$
$\mathrm{(2b)}$: substitute $n\mapsto n-1$ in $\mathrm{(2a)}$
$\mathrm{(2c)}$: subtract $b_{n-1}$ from $\mathrm{(2b)}$
$\mathrm{(2d)}$: multiply $\mathrm{(2b)}$ by $(n-\frac12)$
$\mathrm{(2e)}$: subtract $\mathrm{(2d)}$ from $\mathrm{(2c)}$
$\mathrm{(2f)}$: factor the right hand side of $\mathrm{(2e)}$

Noting that $b_0-\frac12b_{-1}=1-\frac12\cdot2=0$ and $b_1-\frac32b_0=\frac32-\frac32\cdot1=0$, equation $\mathrm{(2f)}$ ensures that $b_n=(n+\frac12)b_{n-1}$ for $n\ge0$, which is equivalent to $$ a_n=(2n+1)a_{n-1}\tag{3} $$ Since $a_{-1}=1$, $(3)$ implies $$ a_n=(2n+1)!!\tag{4} $$

robjohn
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From your results, it is clear that $a_n=(2n+1)!!$ (which is sequence A001147 at OEIS) and which verifies the recurrence relation.

Another way to write it is $$a_n= \frac{2^{n+1} \left(n+\frac{1}{2}\right)!}{\sqrt{\pi }}$$

For the time being, I did not find any generating function except

$$\sum_{i=0}^\infty \frac{x^n}{a_n}={\sqrt{\frac{\pi }{2x}} e^{x/2} \text{erf}\left({\sqrt{\frac{x}{2}}}\right)}$$

In OEIS, they give $$a_n=\int_0^\infty \frac{e^{-x/2} x^{n}}{\sqrt{2 \pi x}}dx$$

Claude Leibovici
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$$ a_{n}=2a_{n-1}+(2n-1)^2a_{n-2}. $$

Let $b_n=a_n/a_{n-1}$. Then $\{b_n\}$ satisfies

$$ b_{n}=2+(2n-1)^2/b_{n-1}, \quad b_1=3. $$

Let $c_n=b_n/(2n+1)$, then $\{c_n\}$ satisfies

$$ (2n+1)c_{n}=2+\frac{2n-1}{c_{n-1}},\quad c_1=1, $$ which is satisfied only by the sequence $c_n=1$, $n\in\mathbb N$.

Yiorgos S. Smyrlis
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