Posts: 29
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12/12/2011, 12:50 AM
(This post was last modified: 12/12/2011, 12:52 AM by marcokrt.)
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 \(G(1) \notin \{2, 3, 7\}\), then \(^{G(n)}G(n) \pmod {10^d}≡^{G({n+1})}G({n+1}) \pmod {10^d}\), \(\forall n \in \mathbb{N}-\{0\}\)
("La strana coda della serie n^n^...^n", p. 60).
Posts: 1,214
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12/12/2011, 05:20 AM
(This post was last modified: 12/12/2011, 05:21 AM by JmsNxn.)
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 \( ^k 3 \) 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.
Posts: 29
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(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 \( ^k 3 \) 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
M
Let \(G(n)\) be a generic reverse-concatenated sequence. If \(G(1) \notin \{2, 3, 7\}\), then \(^{G(n)}G(n) \pmod {10^d}≡^{G({n+1})}G({n+1}) \pmod {10^d}\), \(\forall n \in \mathbb{N}-\{0\}\)
("La strana coda della serie n^n^...^n", p. 60).
Posts: 98
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12/20/2011, 11:29 PM
(This post was last modified: 12/20/2011, 11:29 PM by nuninho1980.)
I understand that.
2^^3 = 1
6
2^^4 = 655
36
2^^5 = 20035... 19718 digits ...156
736
2^^6 = ???... ?.??x10^19727 digits ...
8736
2^^7 = ???... ?.??x10^(?.??x10^19727) digits ...
48736
4^^2 = 25
6
4^^3 = 13407... 145 digits ...840
96
4^^4 = ???... ?.??x10^153 digits ...
896
4^^5 = ???... 10^^2^153 digits ...
8896
4^^6 = ???... 10^^3^153 digits ...
28896
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12/21/2011, 12:03 AM
(This post was last modified: 12/21/2011, 12:13 AM by marcokrt.)
(12/20/2011, 11:29 PM)nuninho1980 Wrote: I understand that.
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- 97
65625
2- 06435109023004770278930
6640625
3- 9278745586052536964416
50390625
4- 768305384553968906402
587890625
5- 42344429437071084976
1962890625
6- 6498173708096146583
55712890625
7- 838703774847090244
293212890625
8- 12594428006559610
3668212890625
9- 6484958166256546
97418212890625
10- 388659619726240
634918212890625
11- 25514140073210
0009918212890625
12- 4043342107906
93759918212890625
13- 333762311376
631259918212890625
14- 37804331723
6006259918212890625
15- 8208533758
29756259918212890625
16- 248953961
767256259918212890625
… …
0- 5
1- 97
65625
2- 06435109023004770278930
6640625
3- 927874558605253696441
650390625
4- 7683053845539689064
02587890625
5- 42344429437071084
9761962890625
6- 649817370809614
658355712890625
7- 8387037748470
90244293212890625
8- 12594428006
5596103668212890625
9- 648495816
625654697418212890625
10- 3886596
19726240634918212890625
11- 255141
400732100009918212890625
12- 4043
34210790693759918212890625
13- 33
3762311376631259918212890625
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 \(G(1) \notin \{2, 3, 7\}\), then \(^{G(n)}G(n) \pmod {10^d}≡^{G({n+1})}G({n+1}) \pmod {10^d}\), \(\forall n \in \mathbb{N}-\{0\}\)
("La strana coda della serie n^n^...^n", p. 60).
Posts: 29
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Joined: Dec 2011
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 \(G(1) \notin \{2, 3, 7\}\), then \(^{G(n)}G(n) \pmod {10^d}≡^{G({n+1})}G({n+1}) \pmod {10^d}\), \(\forall n \in \mathbb{N}-\{0\}\)
("La strana coda della serie n^n^...^n", p. 60).
