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

CC BY-SA 4.0

Quintec‭

2356 reputation 17 46 247 49

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

CC BY-SA 4.0

4y ago

Moshi‭

1762 reputation 17 58 208 80

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

CC BY-SA 4.0

4y ago

Razetime‭

2930 reputation 42 65 328 83

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 $n\in\mathbb Z$), 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 $n\in\mathbb Z$. Now, we can write $G(x)$ in recurrence relation form as$$\lambda^2-n\lambda-1=0$$The solution to this equation (let'_s call this $\lambda_+$),$$\lambda_+=\dfrac{n+\sqrt{n^2+4}}2$$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$$\lambda^2+n\lambda-1=0$$and now our ratio between terms is$$\lambda_-=\dfrac{-n+\sqrt{n^2+4}}2$$Finally, we get $\lambda_+-\lambda_-=n$, or equivalently$$\lambda_+=\lambda_-+n$$$$\lambda_-=\lambda_+-n$$

posted 18 days ago

CC BY-SA 4.0

9d ago

CrSb0001‭

164 reputation 6 14 18 22