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Let m and n be positive integers.

Let $f(m,0)=m$

Let $f(m,n)= e \ln(f(m,n-1))$

$$\lim_{m\to\infty} \ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) = 163^{1/3}+C$$

Where $C$ is a constant.

It seems $0.005 > C > 0$

Is this true ? Why is this so ? Is $C = 0$ ?

Is this an analogue to the computation of the Paris constant ?

Can we give a closed form for $C$ ?

EDIT :

Conjecture :

$$\lim_{m\to\infty} \ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) = A$$

$A > 0$

Is this true ? How to prove this ?

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    Are you just making this up, or do you have a source?2012-09-07
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    What does it have to do with tetration or dynamical systems?2012-09-07
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    repeatedly doing exp or ln is clearly tetration related. If you dont agree , plz give a definition of tetration and show the evidence of that definition.2012-09-07
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    You still haven't given us even the remotest reason why you think this is ture.2012-09-07
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    Your changing subject somewhat. It is clearly tetration related.2012-09-07
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    I didn't change the subject, @mick, because the tetration comment was not mine. Why do you say this is true? Seems clear that $g(x)=e$ for all $x$, and hence your limit does not exist.2012-09-07
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    @mick What does this new sentence even mean? "The limit is taken in the sense that ln(x) > n" does not mean anything, as far as my mathematical experience goes.2012-09-07
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    Are you really trying to say that $g(x)=f(k)$ where $k=\lfloor \ln x\rfloor$?2012-09-07
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    Yes thats it. Maybe another edit.2012-09-07
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    Ok i will make an edit , but i need to think about it a bit more ...2012-09-08
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    i made the edit.2012-09-08
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    I started a new question because the edit of the older version is too severe , and i could not close it.2012-09-08
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    Thanks for the edit Brian. No big difference but looks much nicer now.2012-09-08
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    When $m=10^{500}$, I get $\ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) < 163^{1/3}$.2012-09-08
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    Really ? i got a different result , although i must admit i dont have confidence in my computation. What did you use ? Maple Mathematica ... ? or did you use an acceleration method ?( p-adic ? ) Do you think C will be - 163^(1/3) or -163^(1/3)+1 or do you have another value in thought ? Did you use theory to compute the value for m = 10^500 ?2012-09-08
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    I used Maple computing with 50 decimals. No acceleration. Perhaps the limit is zero, but it decreases very slowly. I could do $m=10^{2000}$, but $m=10^{3000}$ had too many levels of recursion for Maple.2012-09-08

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