This can be done with help of Maple, realizing the Brendan McKay's suggestion, in such a way:

```
a := convert(expand(convert(-sqrt(1-4*z), FPS, z, j)+1+2*z^2), FPS, z);
```

$$\sum _{j=0}^{\infty }{\frac { \left( 2\,j \right) !\,{z}^{j}}{ \left(
j! \right) ^{2} \left( -1+2\,j \right) }}+1+2\,{z}^{2}
$$

```
b := b := allvalues(convert(1/(2*z*(-z^3-z+1)), FPS, z));
```

$$1/2\,{z}^{-1}+\sum _{k=0}^{\infty } \left( {\frac {4\, \left( 1/6\,
\sqrt [3]{108+12\,\sqrt {93}}-2\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {
93}}}} \right) ^{2}+\sqrt [3]{108+12\,\sqrt {93}}-12\,{\frac {1}{
\sqrt [3]{108+12\,\sqrt {93}}}}+13}{62\, \left( 1/6\,\sqrt [3]{108+12
\,\sqrt {93}}-2\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}} \right) ^{
k+1}}}+{\frac {4\, \left( -1/12\,\sqrt [3]{108+12\,\sqrt {93}}+{\frac
{1}{\sqrt [3]{108+12\,\sqrt {93}}}}+i/2\sqrt {3} \left( 1/6\,\sqrt [3]
{108+12\,\sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}}
\right) \right) ^{2}-1/2\,\sqrt [3]{108+12\,\sqrt {93}}+6\,{\frac {1
}{\sqrt [3]{108+12\,\sqrt {93}}}}+3\,i\sqrt {3} \left( 1/6\,\sqrt [3]{
108+12\,\sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}}
\right) +13}{62\, \left( -1/12\,\sqrt [3]{108+12\,\sqrt {93}}+{\frac
{1}{\sqrt [3]{108+12\,\sqrt {93}}}}+i/2\sqrt {3} \left( 1/6\,\sqrt [3]
{108+12\,\sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}}
\right) \right) ^{k+1}}}+{\frac {4\, \left( -1/12\,\sqrt [3]{108+12
\,\sqrt {93}}+{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}}-i/2\sqrt {3}
\left( 1/6\,\sqrt [3]{108+12\,\sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108
+12\,\sqrt {93}}}} \right) \right) ^{2}-1/2\,\sqrt [3]{108+12\,\sqrt
{93}}+6\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}}-3\,i\sqrt {3}
\left( 1/6\,\sqrt [3]{108+12\,\sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108
+12\,\sqrt {93}}}} \right) +13}{62\, \left( -1/12\,\sqrt [3]{108+12\,
\sqrt {93}}+{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}}-i/2\sqrt {3}
\left( 1/6\,\sqrt [3]{108+12\,\sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108
+12\,\sqrt {93}}}} \right) \right) ^{k+1}}} \right) {z}^{k}
$$

It remains to multiply these series.