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Consider the function $t(x)$ defined as :

$$ x_1 = x $$ $$x_2 = x $$ $$ x_3 = 2 x^2 $$ $$ x_4 = 4 x^4 + 2 x^2 $$

and for $n > 4 $

$$ x_{n} = \frac { x_{n-1}^2 + x_{n-2}^2 + x_{n-3}^2}{x_{n-2} + x_{n-3} + x_{n-4} } $$

If the sequence converges to a constant then we define $t(x) = \lim x_n $.

This function is not so well understood by me.

I assume it is analytic for instance. But i have no formal proof.

The function grows fast. But I am not sure how fast exactly. ( ofcourse 10 iterations give a good asymptotic and then notice it goes double exponentially fast to its lim )

I have no series expansion , integral representation , differential equation , continued fraction or such for this. Nor do I have a way to compute these values in a different way such as different iterations , a koenigs type formula , combinatorical methods , fractals , cellular automatons , a bifurcation point , the area of a filled julia set etc etc.

practically there is not a big problem for small imput since it converges rapidly. But the values are mysterious to me.

In particular is $t(1)$ transcendental ?? Can we even prove it to be irrational ??

Do we have a closed form for $t(1)$ ? [ main question ! ]

This all reminds me of the Somos sequences and the Somos constant ( which has a closed form as the derivative of a lerch form ! )

Assuming it is analytic , how does its analytic continuation look like ? Is analytic continuation possible to any complex number or its neighbourhood , or is there a natural boundary ?

numerically we have

$$t(1) = 35.2850889...$$

Do you recognize this number ??

Is there a closed form ???

Has this been studied before ??

Edit

Some remark or motivation :

Notice that when you naively try to analyse this , after you know it converges fast you get

Set all $x_n = y$ for large $ n $ and some $ y > x > 1.$

$$ y = \frac{y^2 + y^2 + y^2}{y + y + y} $$

So

$$ y = 3 y^2 / (3y) $$

$$ y = y^2 / y $$

This is a tautology like $ z = z $. Useless.

It seems all naive methods lead to such tautologies.

This is the one of the simplest possible with sum of squares ; rational function iterations of degree $2.$

This only informally suggests the dependance of the starting value $x $ ofcourse.

Also there seems no link to iterations that solves equations like Newton iterations.

Therefore the recursion fascinates me.

Also notice $ t(x) $ is strictly increasing for $ x > 1.$

mick
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    Is there some motivation behind this question? This looks like a "recurrence of cluster-type," which generalize Somos sequences. The caveat is that they must also yield integer values, whereas yours does not. – Alex R. Sep 25 '19 at 22:48
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    Wolfram suggests: $-1 + e - 2 \sqrt{1 + e} + 6 \sqrt{1 + e^2} + \pi^2 + 5 \sqrt{1 + \pi}≈35.285088900886$ – Surb Sep 25 '19 at 22:54
  • Alex R , I added some motivations. Recurrence of cluster type does not sound familiar to me. If there is some theory about that plz inform us thanks. – mick Sep 26 '19 at 07:36
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    A nice one (among many other) : $\frac{867}{22 \log (\pi )}+\frac{3 \log (\pi )}{4}=35.2850889035603$. Can you provide more digits ? – Claude Leibovici Sep 26 '19 at 10:05
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    $\frac{463986 e^e}{199273}=35.2850889000266$ is not bad – Claude Leibovici Sep 26 '19 at 10:11
  • It appears that $ t(1) = 35.28508892852.. $ the Precision of my programs is insufficient to even make a conjecture for many more digits. But you know how it is computed so .. – mick Sep 26 '19 at 11:17
  • Is it on purpose that $x_4$ does not follow the pattern for $x_n$, $n>4$? –  Oct 10 '19 at 10:02
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    Based on computations in C++, I'm reasonably sure the value is $35.2850889285276615170428016199\dots$. – YiFan Tey Oct 10 '19 at 10:11
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    35.2850889285276615207652307264286692165640411569825581698414274720345996961636628369523260781462773487233897716541224894028653927496201757257239628652739126303264607314379854344783657207592433512446939836904309061986181584445582701842520656297136055578242086656685910819241234589915119952884871871455836385451305511968027345598378895550617584307050005261737969962713278718510530653467923189770181268049356579281058857417513652270481787930777146173840897413414293153707668060549844730768295099102778387454143886965495027043307047763137440562129240122196130975494994015741277121015923067889150106873

     – Raymond Manzoni Oct 12 '19 at 11:24
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    f(x,n)=local(x4=x,x3=x,x2=2x^2,x1=4x^4+2*x^2,xk);for(k=5,n,xk=(x1^2+x2^2+x3^2)/(x2+x3+x4);x4=x3;x3=x2;x2=x1;x1=xk);xk (pari/gp's script used to get more digits for @Claude and OP!). No closed form found yet...

     – Raymond Manzoni Oct 12 '19 at 11:26
  • Im beginning to believe this number has a minimum polynomial. But it will take some time to work out the details. – mick Mar 22 '23 at 22:01

1 Answers1

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This isn't an answer because it is doubtful that an exact analytical answer can be found. Allow me just to summarize what is already known from the comments.

A good approximate numerical value from mick : $$t(1)\simeq 35.28508892852..$$ and from my own computation : $$t(1)\simeq 35.28508892852\color{red}{76}.. \tag 1$$ I am not sure of the two last digits. So, my approximate value is the same as mick's value.

Since no analytical solution was found, one look for formulas giving approximate results.

The ISC Inverse Symbolic Calculator http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html gives only a rough framework :

enter image description here

From Wolfram, reported by Surb : $-1+e-2\sqrt{1+e}+6\sqrt{1+e^2}+\pi^2+5\sqrt{1+\pi}$ $$\simeq 35.285088900886$$ From Claude Leibovici : $\frac{867}{22\ln(\pi)}+\frac{3\ln(\pi)}{4}$ $$\simeq 35.2850889035603$$ From Claude Leibovici : $\frac{463986e^e}{199273}$ $$\simeq 35.2850889000266$$ From my home-made ISC : $-\frac{\left(\ln(5)-\sinh(e)\right)^2}{\sin\left(25/\ln(\gamma) \right)}\quad$ , $\gamma$ is the Euler-Mascheroni constant. $$\simeq 35.285088928836$$ Even more accurately : $-\frac{\left(\ln(5)-\sinh(e)\right)^2}{\sin\left(25/\ln(\gamma) \right)}-\frac{\gamma^3\pi}{\cosh\left(e^3/\sin(2) \right)}$ $$\simeq 35.2850889285276$$ This is exactly the goal $(1)$ with 15 digits. Of course all this is not analytically true. An infinity of similar results can easily be obtained thanks to ISC as pointed out in the paper : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales .

JJacquelin
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