Consider this expression:
$$A_n(x) =\frac{\pi}{2} \left[\mathrm{\mathbf{H}}_n(x) - (-1)^n \mathrm{\mathbf{H}}_{-n}(x) \right]$$
for $n=1,2,3,...$
Where $\mathrm{\mathbf{H}}_n$ are Struve functions.
For Bessel functions we have $A_n=0$ for all arguments.
For Struve functions, things are not as simple. Wolfram Alpha gives some cases (these are terminating Laurent series):
$$A_1=1 \\ A_2 = \frac{1}{x}+\frac{x}{3} \\ A_3 = \frac{3}{x^{2}}+\frac{1}{3}+\frac{x^{2}}{3\cdot 5}$$
$$A_4 = \frac{3\cdot 5}{x^{3}}+\frac{1}{x}+\frac{x}{3\cdot 5}+\frac{x^{3}}{3\cdot 5 \cdot 7}$$
$$A_5 = \frac{3\cdot 5 \cdot 7}{x^{4}}+\frac{5}{x^{2}}+\frac{1}{5}+\frac{x^{2}}{3\cdot 5 \cdot 7}+\frac{x^{4}}{3\cdot 5 \cdot 7 \cdot 9}$$
There's clearly a pattern here, but it's hard to recognize precisely from such a small number of terms.
We can try to use the series definition:
$$\mathrm{\mathbf{H}}_n(x)= \left( \frac{x}{2} \right)^{n+1} \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{2^{2k}\Gamma \left(k+\frac32 \right) \Gamma \left(k+n+\frac32 \right)}$$
Or the differential equation:
$$\frac{d^2w}{dx^2}+\frac{1}{x} \frac{dw}{dx}+\left(1-\frac{n^2}{x^2} \right)w=\frac{x^{n-1}}{\sqrt{\pi} 2^{n-1}\Gamma \left(n+\frac12 \right)}$$
where $w=\mathrm{\mathbf{H}}_n(x)$.
But despite some apparent symmetry, I don't really see any way to proceed.
Motivation for this question is that I need to evaluate some related integrals and this will help me develop an efficient algorithm.
After some thinking, I figured out the solution by using the series.
$$\mathrm{\mathbf{H}}_{n}(x)=\left(\frac{x}{2}\right)^{n+1}\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+\frac{3}{2}\right)\Gamma\left(k+n+\frac{3}{2}\right)}$$
$$\mathrm{\mathbf{H}}_{-n}(x)=\left(\frac{x}{2}\right)^{-n+1}\left[\sum_{k=0}^{n-1}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+\frac{3}{2}\right)\Gamma\left(k-n+\frac{3}{2}\right)}+\sum_{k=n}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+\frac{3}{2}\right)\Gamma\left(k-n+\frac{3}{2}\right)}\right]$$
$$\mathrm{\mathbf{H}}_{-n}(x)=\left(\frac{x}{2}\right)^{-n+1}\left[\sum_{k=0}^{n-1}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+\frac{3}{2}\right)\Gamma\left(k-n+\frac{3}{2}\right)}+(-1)^{n}\left(\frac{x}{2}\right)^{2n}\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+n+\frac{3}{2}\right)\Gamma\left(k+\frac{3}{2}\right)}\right]$$
$$\mathrm{\mathbf{H}}_{-n}(x)=(-1)^{n}\mathrm{\mathbf{H}}_{n}(x)+\left(\frac{x}{2}\right)^{-n+1}\sum_{k=0}^{n-1}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+\frac{3}{2}\right)\Gamma\left(k-n+\frac{3}{2}\right)}$$
$$\mathrm{\mathbf{H}}_{-n}(x)-(-1)^{n}\mathrm{\mathbf{H}}_{n}(x)=\left(\frac{x}{2}\right)^{-n+1}\sum_{k=0}^{n-1}\frac{(-1)^{k}x^{2k}}{2^{2k}\Gamma\left(k+\frac{3}{2}\right)\Gamma\left(k-n+\frac{3}{2}\right)}$$
$$\frac{1}{\Gamma\left(k-n+\frac{3}{2}\right)}=\frac{\sin\left[\pi\left(n-k-\frac{1}{2}\right)\right]}{\pi}\Gamma\left(n-k-\frac{1}{2}\right)=-\frac{(-1)^{n-k}}{\pi}\Gamma\left(n-k-\frac{1}{2}\right)$$
$$\mathrm{\mathbf{H}}_{-n}(x)-(-1)^{n}\mathrm{\mathbf{H}}_{n}(x)=-\frac{(-1)^{n}}{\pi}\sum_{k=0}^{n-1}\frac{\Gamma\left(n-k-\frac{1}{2}\right)}{\Gamma\left(k+\frac{3}{2}\right)}\left(\frac{x}{2}\right)^{2k-n+1}$$
$$\color{blue}{A_n = \frac{\pi}{2}\left(\mathrm{\mathbf{H}}_{n}(x)-(-1)^{n}\mathrm{\mathbf{H}}_{-n}(x)\right)=\frac{1}{2} \sum_{k=0}^{n-1}\frac{\Gamma\left(n-k-\frac{1}{2}\right)}{ \Gamma\left(k+\frac{3}{2}\right)} \left(\frac{x}{2}\right)^{2k-n+1}\tag{*}}$$