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Consider the following sequence : Let $a_1 = a_2 = 1.$ For integer $ n > 2 : $

$$a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$$

$$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$

$$T = ??$$

What is the value of $T$ ?

Is there a closed form or integral for $T$?

I get

$$ T = 3.73205080..$$

The convergeance is fast.

Does anyone recognize this ?

——

Edit

So apparantly $ T = 2 + \sqrt 3 $

Let us generalize.

Take $a_1= 1, a_2 > a_1$

And now the whole sequence depends on $y = a_2$.

We thus define

$$ T(y) = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$

We know $T(1) = T(2) = 2 + \sqrt 3 $.

$$T(3) = 4.4415184401122.. $$

Apparantly $T(3) = \frac{7 + 2 \sqrt 10}{3} $ as found ( no proof ) by lhf.

How about a closed form for $T(y)$ ?

Can all of these rational recursions be transformed into a linear recursion ?

——— Update

See also

About $a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}} , t(12,13) = \frac{3}{2}$

mick
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    I recognize it as $2+\sqrt3$ – J. W. Tanner Oct 16 '19 at 22:07
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    $a_n$ is https://oeis.org/A101879, a sequence of integers! – lhf Oct 16 '19 at 22:19
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    See also https://faculty.uml.edu/jpropp/reach/Barton/b-interp.html – leonbloy Oct 16 '19 at 22:49
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    Where does this problem come from? – lhf Oct 16 '19 at 23:26
  • My mentor tommy1729. Mid 90’s but he talked about it and related recursions yesterday. – mick Oct 16 '19 at 23:30
  • One of the reasons was that with a_2 = 1 or 2 the sequence is integer despite the fraction recursion. Somos like but simpler in a sense. He considered The number theoretic things of it. Like divisors and primes. I already posted many recursions and questions originating from him. – mick Oct 16 '19 at 23:39
  • He called them “ artistic sequences “ – mick Oct 16 '19 at 23:42
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    You may be able to cheat your way out of the question by assuming $$a_n=C_0r^n+C_1+C_2r^{-n},~r>1$$though I'm not sure if this is a sufficiently general form or not... – Simply Beautiful Art Oct 17 '19 at 01:48
  • See also https://math.stackexchange.com/questions/3477954/tommy-s-amazing-new-lim-for-the-golden-number – mick Dec 16 '19 at 02:13
  • See also https://math.stackexchange.com/questions/3479854/about-a-n-fraca-n-1a-n-1-ca-n-2-t12-13-frac32 – mick Dec 17 '19 at 17:15
  • see also : https://math.stackexchange.com/questions/4955849/limit-related-to-fn-fracfn-12-fn-112fn-2fn-213 – mick Aug 07 '24 at 23:04

2 Answers2

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EDIT: By already knowing the solution to your recurrence a priori, I am able to solve your problem. However, a more interesting question might be why your recurrence is equivalent to $$a_n=5a_{n-1}-5a_{n-2}+a_{n-3}$$ and how to see this a priori.

Your recurrence is solved by the sequence (as can be checked by direct calculations) $$a_{n+1} = \frac{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^n + (2-\sqrt 3)^n\cdot(3+\sqrt 3)}{12}.$$

So \begin{split}\frac{a_{n+1}}{a_n}&=\frac{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^n + (2-\sqrt 3)^{n}\cdot(3+\sqrt 3)}{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^{n-1} + (2-\sqrt 3)^{n-1}\cdot(3+\sqrt 3)} \\ &= (2+\sqrt 3)\cdot\frac{\frac{6}{(2+\sqrt3)^{n}}+3-\sqrt 3+3+\sqrt 3}{\frac{6}{(2+\sqrt3)^{n-1}}+3-\sqrt 3+3+\sqrt 3}\\ &=(2+\sqrt 3)\cdot\frac{\frac{6}{(2+\sqrt3)^{n}}+6}{\frac{6}{(2+\sqrt3)^{n-1}}+6} \\&\xrightarrow{n\to\infty}(2+\sqrt 3). \end{split}

I got the closed form for $a_{n+1}$ from A101879 of the OEIS.

Maximilian Janisch
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    Your answer is correct ! “ as can easily be checked “, well easily is subjective i would say , but indeed it is true. But how was it found ? Or how should it be found ? In other words is there a nice method ? I assume you just used the OEIS. But there should be a method to find that formula for a_n or at least ( indirectly ) The simpler linear recursion a_n satisfies. How about that ? – mick Oct 16 '19 at 22:37
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    Hi @mick ; Solving the linear recurrence $a_n=5a_{n-1}-5a_{n-2}+a_{n-3}$ relation is standard theory (see here). However, I don't know how one would notice a priori that both recurrence relations are equivalent... – Maximilian Janisch Oct 16 '19 at 22:59
  • See The edit to The OP. Sorry to make it harder after your answered. But maybe this is a better challenge to you ;) – mick Oct 16 '19 at 23:06
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    @lhf Ok (I'll edit it) – Maximilian Janisch Oct 16 '19 at 23:20
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Empirical answer to the last question:

$T(y)$ seems to be $\dfrac{A(y)+\sqrt {B(y+2)}}{y}$, where $A(y)$ is OEIS/A000124 and $B(y)$ is OEIS/A327319.

Tested for $y=1,2,\dots,6$.

lhf
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  • The author of the conjectured B(y) wrote about a “ fibonacci-Lucas experiment “ at The 2017 Ohio mma Fall meeting. Related ??? – mick Oct 16 '19 at 23:24