Let $F:\mathbb{R}^2\to\mathbb{R}$ be a harmonic function (it can be thought as one component of a holomorphic function), take $m$ odd and let $f:\mathbb{R}^{m+1}\to\mathbb{R}$, $$f(a,x_1,\dots,x_m)=\frac{1}{x_1^2+\dots x_m^2}F\left(a,\sqrt{x_1^2+\dots x_m^2}\right).$$ Prove that for any $k\in\mathbb{N}$ it holds
$$
\Delta_{m+1}^kf(a,x_1,\dots,x_m)=(m-3)\cdot...\cdot(m-2k-1)\sum_{j=1}^{k+1}a_j^{(k+1)}b^{j-2k-2}\partial_b^{(j-1)}F(a,b(x_1,\dots,x_m)),
$$
where $\Delta_{m+1}=\partial^2_a+\sum_{j=1}^m\partial^2_{x_j}$ is the Laplacian of $\mathbb{R}^{m+1}$ and $a_j^{(k)}:=\frac{(2k-j-1)!}{(j-1)!\,(k-j)!\,(-2)^{k-j}}$ satisfy $a_j^{(k+1)}=a_{j-1}^{(k)}+(j-2k)a_j^{(k)}$ for any $j=1,\dots,k+1$.
In particular, $f$ is polyharmonic of degree $\frac{m-1}{2}$.
A proof by induction is what I am trying to get.
I think I solved the problem.
Suppose by induction that for some $k$ it holds $$ \Delta_{m+1}^{k-1}f=(m-3)\cdot...\cdot(m-2k+1)\sum_{j=1}^ka_j^{(k)}b^{j-2k}\partial_b^{(j-1)}F $$ and let us compute $\Delta_{m+1}(\Delta_{m+1}^{k-1}f)$. First compute $\partial^2_{x_i}(b^{j-2k}\partial_b^{(j-1)}F)$ for $i=1,...,n$: $$ \partial_{x_i}(b^{j-2k}\partial_b^{(j-1)}F)=(j-2k)x_ib^{j-2k-2}\partial_b^{(j-1)}F+x_ib^{j-2k-1}\partial_b^{(j)}F, $$ since it holds $$ \partial_{x_i}(b^n)=nx_ib^{n-2},\qquad \partial_{x_i}F=x_ib^{-1}\partial_bF. $$ Then, $$ \begin{split} \partial^2_{x_i}(b^{j-2k}\partial_b^{(j-1)}F)&=(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+(j-2k)(j-2k-2)x_i^2b^{j-2k-4}\partial_b^{(j-1)}F+\\ &+(j-2k)x_i^2b^{j-2k-3}\partial_b^{(j)}F+b^{j-2k-1}\partial_b^{(j)}F+\\ &+(j-2k-1)x_i^2b^{j-2k-3}\partial_b^{(j)}F+x_i^2b^{j-2k-2}\partial_b^{(j+1)}F. \end{split} $$ Now, recalling that $b^2=\sum_{i=1}^mx_i^2$, we have $$ \begin{split} \sum_{i=1}^m\partial^2_{x_i}(b^{j-2k}\partial_b^{(j-1)}F)&=m(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+(j-2k)(j-2k-2)b^{j-2k-2}\partial_b^{(j-1)}F+\\ &+(j-2k)b^{j-2k-1}\partial_b^{(j)}F+mb^{j-2k-1}\partial_b^{(j)}F+\\ &+(j-2k-1)b^{j-2k-1}\partial_b^{(j)}F+b^{j-2k}\partial_b^{(j+1)}F\\ &=(m+j-2k-2)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+(m+2j-4k-1)b^{j-2k-1}\partial_b^{(j)}F+\\ &+b^{j-2k}\partial_b^{(j+1)}F \end{split} $$ and, since $F$ is harmonic, $$ \begin{split} \Delta_{m+1}(b^{j-2k}\partial_b^{(j-1)}F)=\left(\partial_a^2+\sum_{i=1}^m\partial^2_{x_i}\right)(b^{j-2k}\partial_b^{(j-1)}F)=(m+j-2k-2)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+(m+2j-4k-1)b^{j-2k-1}\partial_b^{(j)}F +b^{j-2k}\partial_b^{(j-1)}(\partial_b^2F+\partial_a^2F)\\ =(m+j-2k-2)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+(m+2j-4k-1)b^{j-2k-1}\partial_b^{(j)}F \end{split} $$ Let us force to pick up the factor $m-2k-1$, indeed $m+j-2k-2=m-2k-1+j-1$ and $m+2j-4k-1=m-2k-1+2j-2k$, so we have $$ \begin{split} \Delta_{m+1}(b^{j-2k}\partial_b^{(j-1)}F)&=(m-2k-1)[(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+b^{j-2k-1}\partial_b^{(j)}F]+\\ &+(j-1)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+2(j-k)b^{j-2k-1}\partial_b^{(j)}F \end{split} $$ So let us consider the whole Laplacian $$ \begin{split} \Delta_{m+1}^{k}f&=\Delta_{m+1}(\Delta_{m+1}^{k-1}f)=(m-3)\cdot...\cdot(m-2k+1)\sum_{j=1}^ka_j^{(k)}\Delta_{m+1}(b^{j-2k}\partial_b^{(j-1)}F)\\ &=(m-3)\cdot...\cdot(m-2k+1)(m-2k-1)\sum_{j=1}^ka_j^{(k)}[(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+b^{j-2k-1}\partial_b^{(j)}F]+\\ &+(m-3)\cdot...\cdot(m-2k+1)\sum_{j=1}^ka_j^{(k)}[(j-1)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+2(j-k)b^{j-2k-1}\partial_b^{(j)}F]. \end{split} $$ Let us prove that the first summation actually is $\Delta^k_{m+1}f$: $$ \begin{split} &(m-3)\cdot...\cdot(m-2k+1)(m-2k-1)\sum_{j=1}^ka_j^{(k)}[(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+b^{j-2k-1}\partial_b^{(j)}F]=\\ &=(m-3)\cdot...\cdot(m-2k-1)\left[\sum_{j=1}^{k+1}a_{j}^{(k)}(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+\sum_{j=1}^{k+1}a_{j-1}^{(k)}b^{j-2k-2}\partial_b^{(j-1)}F\right]\\ &=(m-3)\cdot...\cdot(m-2k-1)\sum_{j=1}^ka_j^{(k+1)}b^{j-2k-2}\partial_b^{(j-1)}F=\Delta_{m+1}^kf, \end{split} $$ where we have used $a_j^{(k+1)}=a_{j-1}^{(k)}+(j-2k)a_j^{(k)}$ and that $a_j^{(k)}=0$ if $j\notin\{1,\dots, k\}$. Now, let us prove that the second sum is actually zero. Indeed, rearranging that sum, we have $$ \sum_{j=1}^ka_j^{(k)}(j-1)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+\sum_{j=1}^ka_j^{(k)}2(j-k)b^{j-2k-1}\partial_b^{(j)}F\\ =\sum_{j=2}^ka_j^{(k)}(j-1)(j-2k)b^{j-2k-2}\partial_b^{(j-1)}F+\sum_{j=2}^ka_{j-1}^{(k)}2(j-k-1)b^{j-2k-2}\partial_b^{(j-1)}F\\ =\sum_{j=2}^k[a_{j-1}^{(k)}2(j-k-1)+a_{j-1}^{(k)}2(j-k-1)]b^{j-2k-2}\partial_b^{(j-1)}F. $$ Then we see that $a_{j-1}^{(k)}2(j-k-1)+a_{j-1}^{(k)}2(j-k-1)=0$, indeed by definition $$ a_{j-1}^{(k)}2(j-k-1)+a_{j-1}^{(k)}2(j-k-1)\\=\frac{(2k-j-1)!}{(j-1)!(k-j)!(-2)^{k-j}}(j-1)(j-2k)+\frac{(2k-j)!}{(j-2)!(k-j+1)!(-2)^{k-j+1}}2(j-k-1)\\ =\frac{-(2k-j)!}{(j-2)!(k-j)!(-2)^{k-j}}+\frac{(2k-j)!}{(j-2)!(k-j)!(-2)^{k-j}}=0. $$