For any positive integer $n$ define the sequence $(x_k)_{k\ge0}$ by $x_0=0,x_1=1$ and $$x_{k+2}=\frac{cx_{k+1}-(n-1)x_k}{k+1},k\ge0.$$ Fix $n$ and let $c$ be the largest real number such that $x_{n+1}=0.$ Find $x_k$ in terms of $n$ and $k,1\le k\le n.$
I can see that $x_{n+1}$ is a polynomial of degree $n$ in $c.$ So I just need to find those $n$ values of $c$ and choose the largest to find $c$. Some help please!
On
In one method we put $x_k=x^k $. It gives a second order polynomial if $x_k\neq 0$. Since the equation is linear in x a solution is obtained by a linear combination of the two roots power k.
On
Let $b = \sqrt{\frac{n-1}{2}}$, $\alpha = \frac{c}{2b}$ and $y_k = x_{k+1}$ for $k \ge -1$. We can rewrite the recurrence relation as
$$(k+1)y_{k+1} = \alpha b y_k - 2 b^2 y_{k-1} = 0 \quad\text{ for } k \ge 0\tag{*1}$$
Let $y(t)$ be the OGF of the sequence $y_k$. i.e.
$\displaystyle\;y(t) \stackrel{def}{=} \sum_{k=0}^\infty y_k t^k\;$.
Multiply both sides of $(*1)$ by $t^k$ and start to sum from $0$, we find $y(t)$ satisfies
$$\frac{d}{dt}y(t) = (2\alpha b - 2b^2 t)y(t) \quad\text{ with initial condition }\quad y(0) = y_0 = x_1 = 1$$ This leads to $$\sum_{k=0}^\infty y_k t^k = y(t) = e^{2\alpha b t - b^2t^2} = \sum_{k=0}^\infty \frac{1}{k!} H_k(\alpha)(bt)^tag{*2} $$ where $H_k(x)$ is the physicist's version of Hermite polynomials defined by $$H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} = \left(2x- \frac{d}{dx}\right)^n 1$$
Compare coefficients of $t^k$ from both sides of $(*2)$, we get
$$x_{k+1} = y_k = \frac{b^k}{k!}H_k(\alpha)$$
The condition $x_{n+1} = 0$ becomes $H_n(\alpha) = 0$. Let $\alpha_n$ is the largest root of $H_n(x) = 0$. We obtain an expression of $x_k$ in terms of the Hermite polynomials and its root:
$$ c = 2b\alpha_n = \sqrt{2(n-1)}\alpha_n \quad\text{ and }\quad x_{k+1} = \frac{1}{k!}H_k(\alpha_n)\left(\frac{n-1}{2}\right)^{k/2}, \forall k \ge 0 $$ Following is the result for first few $n$ (numbers rounded to 4 decimal places): $$ \begin{array}{rrr:rrr} n & \alpha_n & c_n & x_1 & x_2 & x_3 & x_4 & x_5 & x_6\\ \hline 2 & \frac{1}{\sqrt{2}} & 1 & 1 & 0\\ 3 & \sqrt{\frac32} & 2.4495 & 1 & 2.4495 & 2 & 0\\ 4 & \sqrt{\frac{\sqrt{6}+3}{2}} & 4.0433 & 1 & 4.0433 & 6.6742 & 4.9520 & 0\\ 5 & \sqrt{\frac{\sqrt{10}+5}{2}} & 5.7139 & 1 & 5.7139 & 14.3246 & 19.6646 & 13.7661 & 0\\ 6 & 2.3506 & 7.4333 & 1 & 7.4333 & 25.1267 & 49.8691 & 61.2641 & 41.2094 \end{array} $$
In my opinion, the explicit expression of $x_k$ in terms of Hermite polynomials is not that useful. If one have access to an accurate value of $\alpha_n$, it is computational more efficient just compute $x_k$ by the recurrence relations.
If one only need $\alpha_n$ for a few fixed $n$, one can use the FindRoot command on WA (wolfram alpha) to find them numerically. As an example, following is a short table of $\alpha_n$ for $3 \le n \le 20$. The first column is generated by throwing following command to WA.
Table[FindRoot[HermiteH[n,z]{z,Sqrt[2*n+2]},WorkingPrecision -> 20],{n,3,20}]
$$ \begin{array}{r:l:l} n & \alpha_n & \alpha_n^{asym}\\ \hline 3 & 1.2247448713915890491 & 1.22\color{gray}{5163277484645}\\ 4 & 1.6506801238857845559 & 1.650\color{gray}{876782405881}\\ 5 & 2.0201828704560856329 & 2.020\color{gray}{291127247794}\\ 6 & 2.3506049736744922228 & 2.3506\color{gray}{71049151207}\\ 7 & 2.6519613568352334925 & 2.65\color{gray}{200473184185}\\ 8 & 2.9306374202572440192 & 2.9306\color{gray}{67476213962}\\ 9 & 3.1909932017815276072 & 3.19\color{gray}{1014917109888}\\ 10 & 3.4361591188377376033 & 3.4361\color{gray}{75338417365}\\ 11 & 3.6684708465595825188 & 3.6684\color{gray}{83293817732}\\ 12 & 3.8897248978697819193 & 3.8897\color{gray}{34667392081}\\ 13 & 4.1013375961786396412 & 4.1013\color{gray}{45410857248}\\ 14 & 4.3044485704736318126 & 4.3044\color{gray}{54923665302}\\ 15 & 4.4999907073093915537 & 4.49999\color{gray}{5945280707}\\ 16 & 4.6887389393058183647 & 4.6887\color{gray}{43311031201}\\ 17 & 4.8713451936744030925 & 4.87134\color{gray}{8881984195}\\ 18 & 5.0483640088744667684 & 5.04836\color{gray}{7150538351}\\ 19 & 5.2202716905374821646 & 5.22027\color{gray}{4389563317}\\ 20 & 5.3874808900112328620 & 5.38748\color{gray}{322666128}\\ \end{array} $$
For large $n$, $\alpha_n$ is known to have following asymptotic expansion${}^{\color{blue}{[1]}}$:
$$\alpha_n \asymp \sqrt{2n+1}\left\{ 1 - az - \frac{a^2}{10}z^2 + \left[\frac{9}{280} - \frac{11}{350}a^3\right]z^3 + \left[\frac{277}{12600}a - \frac{823}{63000}a^4\right]z^4 + \cdots\right\} $$ where $a \approx 1.855757081489239$ is the smallest positive zero of the Airy function $\mathrm{Ai}(-2^{1/3} x)$ and $z = (2n+1)^{-2/3}$. The second column $\alpha_n^{asym}$ in above table is obtained by keeping up to the $z^4$ term in this asymptotic expansion.
As one can see, at $n = 6$, the first $n$ where we don't know the exact value, the relative error of this approximation $\frac{\alpha_n^{asym}}{\alpha_n} - 1$ is around $2.81\times 10^{-5}$. At $n = 20$, the relative error drop to $4.34\times 10^{-7}$. If one only need $c$ within $10^{-6}$ relative accuracy, one can use above table of $\alpha_n$ for $n \le 20$ and the asymptotic formula above for $n > 20$.
Notes/References
I don't see a simple form for the problem as given. The polynomials $n!x_{n+1}(c)$ alternate between having either only odd or only even powers of $c$, with coefficients of alternating signs, and seem to have all real roots, but I haven't found an explicit formula for the largest root. Here are the numerical results for the first few $n$ (note that $c = x_2$):
\begin{array} {r | c c c c c c c c c c} \hline n & x_0 & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} \\ \hline 1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 & 0\\ 3 & 0 & 1 & 2.45 & 2 & 0 \\ 4 & 0 & 1 & 4.04 & 6.67 & 4.95 & 0 \\ 5 & 0 & 1 & 5.71 & 14.32 & 19.66 & 13.77 & 0 \\ 6 & 0 & 1 & 7.43 & 25.13 & 49.87 & 61.26 & 41.21 & 0 \\ 7 & 0 & 1 & 9.19 & 39.20 & 101.66 & 174.68 & 198.95 & 129.94 & 0 \\ 8 & 0 & 1 & 10.97 & 56.62 & 181.37 & 398.11 & 619.18 & 667.13 & 425.88 & 0 \\ 9 & 0 & 1 & 12.76 & 77.46 & 295.53 & 788.10 & 1539.02 & 2223.21 & 2294.96 & 1438.40 & 0 \\ 10 & 0 & 1 & 14.58 & 101.76 & 450.79 & 1413.97 & 3311.25 & 5924.50 & 8081.20 & 8061.30 & 5976.66 & 0 \\ \hline \end{array}