In [(1)](https://dlmf.nist.gov/10.49.18) 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?



Spherical Bessel functions. Sum of squares
From the definition [10.47.10](https://dlmf.nist.gov/10.47), 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](https://dlmf.nist.gov/10.49),
\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](https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair). But perhaps it's just a consequence from something well-known.


**P.S.** This identity has a neat representation in terms of hypergeometric functions, which I posted in a [follow-up question](https://mathoverflow.net/questions/334953).


