Recurrence relation help

Question

The function $\psi_k(n)$ satisfies the recurrence relation: $$\sum_{j=0}^k\binom{k}{j}(-1)^j\psi_j(n)\ln(n)^{k-j}=\psi_k(n)$$ Using this, is there a general way I can re-write

the function $ \psi_k(n)$, when $k$ is odd, in terms of other $\psi_i(n)'s$ where $i$ is even?

The first few odd ones I solved for, in terms of there even counter parts are:

$\psi_1(n)=\frac{1}{2}\psi_0(n)\ln(n)$

$\psi_3(n)=\frac{3}{2}\psi_2(n)\ln(n)-\frac{1}{4}\psi_0(n)\ln(n)^3$

$\psi_5(n)=\frac{5}{2}\psi_4(n)\ln(n)-\frac{5}{2}\psi_2(n)\ln(n)^3+\frac{1}{2}\psi_0(n)\ln(n)^5$

I know in practice this can be done for all of them, but its very tedious to solve for them, so is there a general way I can re-write the functions of odd subscript in terms of other functions of even subscribt?

I would appreciate any help

Answer 1

Define the exponential generating function $\Psi(z)$ by $$\Psi(z):=\sum_{k\ge 0} \psi_k(n) \frac{z^n}{n!}.$$ Then from the recurrence, $$ \Psi(-z) e^{z\ln n} = \Psi(z).$$ Setting $\Psi(z)=A(z^2)+z C(z^2)$ and solving for $C$ gives \begin{eqnarray*} C(z^2)&=&\frac{A(z^2)}{z}\frac{e^{z\ln n}-1}{e^{z\ln n}+1}\\ &=&\frac{A(z^2)}{z}(1+\frac{4}{e^{2z\ln n}-1}-\frac{2}{e^{z \ln n}-1}). \end{eqnarray*} Using the generating function of the Bernoulli numbers $B_k$, $$ \frac{z}{e^z-1}=\sum_{k\ge 0} z^k \frac{B_k}{k!},$$ expanding the coefficients of $z^{2k}$ on both sides gives $$ \psi_{2k+1}(n)=\sum_{0\le j\le k} \frac{B_{2k-2j+2}}{2k-2j+2} \binom{2k+1}{2j} (2^{2k-2j+3}-2) (\ln n)^{2k-2j+1} \psi_{2j}(n). $$ This is a general expression for the $\psi_j(n)$s with odd subscript in terms of the $\psi_j(n)$s with even subscript.