# Ratio limits of fibonacci-like series

# Definition

$F_{n}\left(0\right)=0$

$F_{n}\left(1\right)=1$

$F_{n}\left(x\right)=n\cdot F_{n}\left(x-1\right)+F_{n}\left(x-2\right)$

For example:

$F_{1}=\left[0,1,1,2,3,5,8,13,21,34,55,89...\right]$

$F_{2}=\left[0,1,2,5,12,29,70,169,408,985...\right]$

$F_{3}=\left[0,1,3,10,33,109,360,1189,3927...\right]$

# Challenge

Given n, find the limit of the ratio between consecutive terms, correct to at least the first 5 decimal places.

# Test Cases

```
1 -> 1.61803...
2 -> 2.41421...
3 -> 3.30277...
4 -> 4.23606...
```

# Scoring

This is code-golf. Shortest answer in each language wins.

[JavaScript (Node.js)], 27 22 …

3mo ago

[APL (Dyalog Classic)], 7 byte …

3mo ago

Stax, 5 bytes òP^↓Φ Run …

3mo ago

## 3 answers

#
JavaScript (Node.js), ~~27~~ 22 bytes

*-5 bytes thanks to Hakerh400*

Direct computation

```
f=n=>(n+(n*n+4)**.5)/2
```

$$\frac{n+\sqrt{n^2+4}}{2}$$

#### 4 comments

Nice, interested in how you figured out the explicit formula.

@rak1507 lol, does "looking it up" count as an answer to that question? You can derive it pretty easily though by simply assuming that it converges to $F_n(x+1)=rF_n(x)$ (where r is the ratio we want) and plugging it into recurrence equation.

`Math.sqrt`

-> `**.5`

@Hakerh400 Thanks!

# APL (Dyalog Classic), 7 bytes

```
⎕+∘÷⍣=1
```

Well, rak is probably not going to answer this, so I might as well ;p

## 0 comments