I am eager about fixing the next distinction equation: $x[k-1]+(okay^2+okay+a)/x[k]=b$, $okay=1,2,ldots$, the place $a,b$ are fastened optimistic numbers; for instance $x[1]=c>0$. Mathematica’s `RSolve`

operate does not yield the final answer.

```
RSolve[{x[k - 1] + (okay^2 + okay + a)/x[k] == b}, x, okay]
```

I attempted `RecurrenceTable`

. It produced one thing that appears like a cont fraction. Does anybody acknowledge it?

```
RecurrenceTable[{x[k - 1] + (okay^2 + okay + a)/x[k] == b,
x[1] == c}, x, {okay, 10}]
```

```
{c, (6 + a)/(b - c), (12 + a)/(b - (6 + a)/(b - c)), (20 + a)/(
b - (12 + a)/(b - (6 + a)/(b - c))), (30 + a)/(
b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))), (42 + a)/(
b - (30 + a)/(b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c))))), (
56 + a)/(b - (42 + a)/(
b - (30 + a)/(
b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))))), (72 + a)/(
b - (56 + a)/(
b - (42 + a)/(
b - (30 + a)/(
b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c))))))), (90 + a)/(
b - (72 + a)/(
b - (56 + a)/(
b - (42 + a)/(
b - (30 + a)/(
b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))))))), (
110 + a)/(
b - (90 + a)/(
b - (72 + a)/(
b - (56 + a)/(
b - (42 + a)/(
b - (30 + a)/(
b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))))))))}
```

I think the continued fraction might be a hypergeometric operate of some kind.