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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).

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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

Threads: 4

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).