# Spherical Bessel functions. Sum of squares

In (1) there is a property of spherical Bessel functions, which's derivation I can not find in the literature.

$${\mathsf{j}_{n}^{2}}\left(z\right)+{\mathsf{y}_{n}^{2}}\left(z\right)=\sum_{k=% 0}^{n}\frac{s_{k}(n+\frac{1}{2})}{z^{2k+2}},$$ where $$s_{k}(n+\tfrac{1}{2})=\frac{(2k)!(n+k)!}{2^{2k}(k!)^{2}(n-k)!}.$$

I tried to obtain it by myself with the use of 10.49.1

$$a_{k}(n+\tfrac{1}{2})=\begin{cases}\dfrac{(n+k)!}{2^{k}k!(n-k)!},& k=0,1,\dotsc% ,n,\\ 0,&k=n+1,n+2,\dotsc.\end{cases}$$

and 10.49.6

$$h_{n}(z)\displaystyle=e^{iz}\sum_{k=0}^{n}i^{k-n-1}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}},$$

but I haven't managed with sums rearrangement. Can someone tell me where to find the derivation?

From the definition 10.47.10, it follows that $$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = h_n^{(1)}(z)\cdot h_n^{(2)}(z).$$ So, by the expansions 10.49.6 and 10.49.7, $$\begin{split} \mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) &= \sum_{k=0}^n I^{k-n-1} \frac{a_k(n+\frac{1}{2})}{z^{k+1}}\cdot \sum_{l=0}^n (-I)^{l-n-1} \frac{a_l(n+\frac{1}{2})}{z^{l+1}} \\ &= \sum_{s=0}^{2n} \frac{(-I)^s}{z^{s+2}} \sum_{k=\max\{0,s-n\}}^{\min\{n,s\}} (-1)^k a_k(n+\frac{1}{2})a_{s-k}(n+\frac{1}{2}). \end{split}$$ From the definition 10.49.1, the inner term can be restated as $$(-1)^k\frac{s!}{2^s} \binom{s}{k}\binom{n+k}{s}\binom{n+s-k}{s}$$ and it naturally nullifies when $$k$$ is outside the summation range. Furthermore, if $$s$$ is odd then the terms for $$k=k'$$ and $$k=s-k'$$ cancel each other. So, we can set $$s=2t$$ and obtain $$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = \sum_{t=0}^{n} \frac{(-1)^t (2t)!}{z^{2t+2}2^{2t}} \sum_{k\geq 0} (-1)^k \binom{2t}{k}\binom{n+k}{2t}\binom{n+2t-k}{2t}.$$
It remains to show that $$\sum_{k\geq 0} (-1)^k \binom{2t}{k}\binom{n+k}{2t}\binom{n+2t-k}{2t} = (-1)^t\frac{(n+t)!}{t!^2(n-t)!},$$ which at very least can be done with the WZ method. But perhaps it's just a consequence from something well-known.