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Integer tetration and convergence speed rules
#1
Hi all,
This is my first post on this wonderful place and I have to confess I'm feeling like a kid in the playground.

I've just published a quite simple book (in Italian) about integer tetration. It's mainly focused on its ending digits (perfect p-adic convergence, regular displacements of the most important constrained digits, and so on), plus some related topics.
It's possible to read the first (introductive) chapter online: http://www.uni-service.it/images/stories...review.pdf

The required skill level is quite low (high school proficiency), but I hope it could be a nice amusement for recreational math lovers.

All the best,
Marco

Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60).
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#2
Sadly, I don't speak Italian, but looking at the math that seems very wild and interesting. The fact that the same digits reoccur in is very... weird?

I wonder, does this happen exclusively for base ten? Would this happen in binary or hexadecimal? I think it would... right?

That's very weird. This has to tell us something. Too bad I don't speak Italian to see what you've concluded.

Nonetheless; this takes the cake for "mathematical beauty of the day", maybe week, for me.

very cool find! I don't even know how you evaluate such large values of integer tetration of three.
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#3
(12/12/2011, 05:20 AM)JmsNxn Wrote: Sadly, I don't speak Italian, but looking at the math that seems very wild and interesting. The fact that the same digits reoccur in is very... weird?

I wonder, does this happen exclusively for base ten? Would this happen in binary or hexadecimal? I think it would... right?

That's very weird. This has to tell us something. Too bad I don't speak Italian to see what you've concluded.

Nonetheless; this takes the cake for "mathematical beauty of the day", maybe week, for me.

very cool find! I don't even know how you evaluate such large values of integer tetration of three.

Thank you. Yes... it's a general outcome.
BTW this represents only the beginning, I've explained every single rule concerning the convergence speed of a generic base. I've also found a lot of laws about the digits at the left of the convergent ones...

To calculate this digits, it's sufficient to use Wolfram|Alpha or others free programs Wink

M
Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60).
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#4
I understand that. Smile

2^^3 = 16
2^^4 = 65536
2^^5 = 20035... 19718 digits ...156736
2^^6 = ???... ?.??x10^19727 digits ...8736
2^^7 = ???... ?.??x10^(?.??x10^19727) digits ...48736

4^^2 = 256
4^^3 = 13407... 145 digits ...84096
4^^4 = ???... ?.??x10^153 digits ...896
4^^5 = ???... 10^^2^153 digits ...8896
4^^6 = ???... 10^^3^153 digits ...28896

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#5
(12/20/2011, 11:29 PM)nuninho1980 Wrote: I understand that. Smile

2^^3 = 16
2^^4 = 65536
2^^5 = 20035... 19718 digits ...156736
2^^6 = ???... ?.??x10^19727 digits ...8736
2^^7 = ???... ?.??x10^(?.??x10^19727) digits ...48736

4^^2 = 256
4^^3 = 13407... 145 digits ...84096
4^^4 = ???... ?.??x10^153 digits ...896
4^^5 = ???... 10^^2^153 digits ...8896
4^^6 = ???... 10^^3^153 digits ...28896


Every base is characterized by this kind of convergence... (sometimes) there are only a few steps without convergence.
The asymptotic convergence speed is constant for every base (the proof is in my book)!

If you like a little more fun, you can take a look at this (in the book I have called it "sfasamento"):

[5^10^i](mod 10^30):


0- 5
1- 9765625
2- 064351090230047702789306640625
3- 927874558605253696441650390625
4- 768305384553968906402587890625
5- 423444294370710849761962890625
6- 649817370809614658355712890625
7- 838703774847090244293212890625
8- 125944280065596103668212890625
9- 648495816625654697418212890625
10- 388659619726240634918212890625
11- 255141400732100009918212890625
12- 404334210790693759918212890625
13- 333762311376631259918212890625
14- 378043317236006259918212890625
15- 820853375829756259918212890625
16- 248953961767256259918212890625
… …


0- 5
1- 9765625
2- 064351090230047702789306640625
3- 927874558605253696441650390625
4- 768305384553968906402587890625
5- 423444294370710849761962890625
6- 649817370809614658355712890625
7- 838703774847090244293212890625
8- 125944280065596103668212890625
9- 648495816625654697418212890625
10- 388659619726240634918212890625
11- 255141400732100009918212890625
12- 404334210790693759918212890625
13- 333762311376631259918212890625
14- 378043317236006259918212890625
15- 820853375829756259918212890625
16- 248953961767256259918212890625
… …

... I've discovered different kinds of convergence/pseudo-convergence... it's related to caos theory too (the underlying mathematics is group theory by Galois).

Marco
Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60).
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#6
We can also construct some bases with an unlimited convergence speed, for example, 999...9. The number of "9" (the lenght in digits of the base) gives us an equal "convergence speed in a single step": i.e. [9999999^^n](mod 10^(7*n))==[9999999^^(n+1)](mod 10^(7*n)).

Marco
Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60).
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