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Ratio limits of fibonacci-like series

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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.

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3 answers

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JavaScript (Node.js), 27 22 bytes

-5 bytes thanks to Hakerh400‭

Direct computation

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

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

Try it online!

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4 comments

Nice, interested in how you figured out the explicit formula. rak1507‭ 7 days ago

@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. Moshi‭ 7 days ago

Math.sqrt -> **.5 Hakerh400‭ 7 days ago

@Hakerh400 Thanks! Moshi‭ 7 days ago

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APL (Dyalog Classic), 7 bytes

⎕+∘÷⍣=1

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

Try it online!

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Stax, 5 bytes

òP^↓Φ

Run and debug it

Same idea as Quintec's answer.

Input as a floating point number.

Explanation

1gpun+
1        start with 1
 gp      apply the following till a fixpoint:
   u     reciprocal
    n+   add the input
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