Ratio limits of fibonacci-like series

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Definition

$F_{n} (0) = 0$

$F_{n} (1) = 1$

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

For example:

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

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

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

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.

code-golf sequence

posted about 4 years ago

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3y ago by General Sebast1an‭

rak1507‭

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is a duplicate

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not constructive

This question cannot be answered in a way that is helpful to anyone. It's not possible to learn something from possible answers, except for the solution for the specific problem of the asker.

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[APL (Dyalog Classic)], 7 byte …

4y ago

+4

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[JavaScript (Node.js)], 27 22 …

4y ago

+2

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Stax, 5 bytes òP^↓Φ Run …

4y ago

+1

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Python 3, 26 bytes ```pytho …

15d ago

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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

posted about 4 years ago

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4y ago

Razetime‭

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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!

posted about 4 years ago

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Quintec‭

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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!

posted about 4 years ago

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4y ago

Moshi‭

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General comments (4 comments)

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Python 3, 26 bytes

lambda n:(n+(n*n+4)**.5)/2

We can actually just compute this directly, although I could have taken an alternate path considering the fact that, given $R_{n}$ as the ratio between any two terms in the $n$ -fibonacci sequence defined above (where $n \in Z$ ), then $R_{- n} = R_{n} - n$ or equivalently, $R_{n} = R_{- n} + n$ although it seems like it would have cost me an additional 21 19 bytes:

f=lambda n:n+f(-n)if n<0else(n+(n*n+4)**.5)/2

^{-2 bytes by rearranging the terms}

Explanation of the above statement:

We can write $G (x) = F_{n} (x)$ for arbitrary $n \in Z$ . Now, we can write $G (x)$ in recurrence relation form as $λ^{2} - n λ - 1 = 0$ The solution to this equation (let'_s call this $λ_{+}$ ), $λ_{+} = \frac{n + \sqrt{n^{2} + 4}}{2}$ gives the ratio of one term to the next as $x \to \infty$ , because $G$ does not depend on parameter $n$ .

Now, if we let $G (x) = F_{- n} (x)$ , the recurrence relation now becomes $λ^{2} + n λ - 1 = 0$ and now our ratio between terms is $λ_{-} = \frac{- n + \sqrt{n^{2} + 4}}{2}$ Finally, we get $λ_{+} - λ_{-} = n$ , or equivalently $λ_{+} = λ_{-} + n$ $λ_{-} = λ_{+} - n$

posted 24 days ago

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15d ago

CrSb0001‭

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Ratio limits of fibonacci-like series

Definition

Challenge

Test Cases

Scoring

0 comment threads

4 answers

Stax, 5 bytes

Explanation

0 comment threads

APL (Dyalog Classic), 7 bytes

0 comment threads

JavaScript (Node.js), 27 22 bytes

1 comment thread

Python 3, 26 bytes

0 comment threads