General solution to a certain simple recurrence relation

Question

This is follow up to a question answered here https://math.stackexchange.com/a/4224037/168758

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Fix $\lambda \in [0, 1)$. Define $m_{-1}(\lambda) = m_0(\lambda) = 1$, and for $k \ge 1$, define $m_k(\lambda) \in \mathbb R$ by

$$ m_{k}(\lambda) = \frac{1}{k(1-\lambda)^2} \Big((2k-3)(1+\lambda) m_{k-1}(\lambda) - (k-3)m_{k-2}(\lambda)\Big). \quad (\star) $$

It is easy to deduce that $m_1(\lambda)=\dfrac{1}{(1-\lambda)^2}(-(1+\lambda)+2) = \dfrac{1}{1-\lambda}$ and $m_2(\lambda) = \dfrac{1}{2(1-\lambda)^2} = \dfrac{1}{2(1-\lambda)^2}((1+\lambda)(1-\lambda)^{-1}+1)=\dfrac{1}{(1-\lambda)^3}$.

Question. What is a general formula for $m_k(\lambda)$, perhaps in terms of well-known (special ?) functions, valid for all $k \ge 1$ ?

Answer 1

Computing several iterations gives \begin{align} m_2&=\frac{1}{(1-\lambda)^3}\\ m_3&=\frac{1+\lambda}{(1-\lambda)^5}\\ m_4&=\frac{1+3\lambda+\lambda^2}{(1-\lambda)^7}\\ m_5&=\frac{1+6\lambda+6\lambda^2+\lambda^3}{(1-\lambda)^9}\\ m_6&=\frac{1+10\lambda+20\lambda^2+10\lambda^3+\lambda^4}{(1-\lambda)^{11}}\\ m_7&=\frac{1+15\lambda+50\lambda^2+50\lambda^3+15\lambda^4+\lambda^5}{(1-\lambda)^{13}} \end{align} and so on. The denominator increases by power of 2 each iteration, and the coefficients of the numerator follow a nice Pascal-triangle like pattern, so I [looked it up on the OEIS ][1] and it turns out they are the Narayana numbers, [which follow the formula ][2]

\begin{align} \mathrm N(n, k)=\frac{1}{n} {n\choose k}{n\choose k-1}. \end{align} It is not hard (translation: I don’t have a legitimate proof) to then arrive at the result \begin{align} m_k(\lambda)=\frac{1}{(1-\lambda)^{2k-1}}\sum_{i=0}^{k-2}\mathrm N(k-1, i+1)\lambda^i, \quad k\geq 2. \end{align}

Hopefully this was without error, let me know if you find one. [1]: https://oeis.org/search?q=1%2C%201%2C%201%2C%201%2C%203%2C%201%2C%201%2C%206%2C%206%2C%201&language=english&go=Search [2]: https://en.m.wikipedia.org/wiki/Narayana_number

Answer 2

Eli's answer seems correct already.

I have an observation to add: the moment generating function $S_{\lambda}(z) = \sum \limits_{k=0}^{\infty} m_k(\lambda) z^k$ has a closed form. Let us prove that $S_{\lambda}(z) = \frac{1+\lambda-z}{2\lambda} - \frac{1}{2\lambda}\sqrt{(1-\lambda)^2-2(1+\lambda)z + z^2}$.

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Define $T_{\lambda}(z) = (1+\lambda)z + \sum \limits_{k=2}^{\infty} \frac{(1+\lambda) m_{k-1}(\lambda) - m_{k-2}(\lambda)}{k}{z^k}$. It is such that $$T_{\lambda}'(z) = 1+\lambda + \sum \limits_{k=2}^{\infty} \big((1+\lambda) m_{k-1}(\lambda) - m_{k-2}(\lambda)\big){z^{k-1}} = (1+\lambda)S_{\lambda}(z) - zS_{\lambda}(z)$$

Now use the recurrence relation (and that $m_0=1$, $m_1(\lambda)=\frac{1}{1-\lambda}$) to replace terms in $S_{\lambda}$:

\begin{align*} S_{\lambda}(z) & = 1 + \frac{z}{1-\lambda} + \frac{\sum \limits_{k=2}^{\infty} \Big(2(1+\lambda)m_{k-1}(\lambda) - \frac{3(1+\lambda)}{k}m_{k-1}(\lambda) + \frac{-k+3}{k}m_{k-2}(\lambda)\Big)z^k}{(1-\lambda)^2} \\ & = 1 + \frac{z}{1-\lambda} + \frac{2(1+\lambda)}{(1-\lambda)^2}(zS_{\lambda}(z)-z) - \frac{3}{(1-\lambda)^2}(T_{\lambda}(z)-z) - \frac{1}{(1-\lambda)^2}z^2S_{\lambda}(z) \end{align*}

Now that allows us to get an ODE for $T_{\lambda}$:

$$\frac{3}{(1-\lambda)^2}T_{\lambda}(z) = 1 + \frac{2}{(1-\lambda)^2}z - \Big(1 - \frac{2(1+\lambda)}{(1-\lambda)^2}z + \frac{1}{(1-\lambda)^2} z^2\Big)S_{\lambda}(z)$$

so $$3 T_{\lambda}(z) = (1-\lambda)^2 + 2z + \Big(-(1-\lambda)^2 + 2(1+\lambda)z - z^2\Big) \frac{T_{\lambda}'(z)}{1+\lambda - z}$$

Note already that the second degree polynomial is the opposite of the classic polynomial $L=(\lambda^+-x)(x-\lambda^-)$ with $\lambda^{\pm} = (1\pm\sqrt{\lambda})^2$

i.e. $$3(1+\lambda - z)T_{\lambda}(z) = (1-\lambda)(1-\lambda^2) + (1+4\lambda-\lambda^2)z - 2z^2 + (\lambda^+-z)(z-\lambda^-)T_{\lambda}'(z) $$

Searching a solution as a linear combination of $(-L)^{3/2}$ and a third degree polynomial, such that $T_{\lambda}(0)=0$, we find

$$T_{\lambda}(z) = \frac{1}{6\lambda}z^3 - \frac{1}{2}\big(1+\frac{1}{\lambda}\big)z^2 + \frac{1}{2}\big(L + 2 + \frac{1}{L}\big)z + \frac{1}{\lambda^3} (\lambda^+-z)^{3/2}(\lambda^- -z)^{3/2} - \big(\frac{1}{\lambda}-1)^3$$

Finally, recall that \begin{align*} S_{\lambda}(z) & = \frac{T_{\lambda}'(z)}{1+\lambda-z} \\ & = \frac{z^2 - 2(1+\lambda)z + (z - 1 - \lambda)\sqrt{(\lambda^+-z)(\lambda^- - z)} + (1+\lambda)^2}{2\lambda(1+\lambda - z)} \\ & = \frac{1+\lambda-z}{2\lambda} - \frac{1}{2\lambda}\sqrt{(\lambda^+-z)(\lambda^- -z)} \end{align*}