[HN Gopher] Qntm's Power Tower Toy
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Qntm's Power Tower Toy
Author : ravenical
Score : 34 points
Date : 2025-12-20 11:19 UTC (4 days ago)
(HTM) web link (qntm.org)
(TXT) w3m dump (qntm.org)
| konmok wrote:
| I'm a big qntm fan. I highly recommend their "antimemetics" SCP
| stories and articles.
| jdpage wrote:
| There's a new, professionally-published book version of "There
| Is No Antimemetics Division" out as well[1], if you want to
| support Sam's work that way. I have print copies of both the
| self-published V1 and the new V2. I'm very excited about the
| latter, though I haven't finished it yet.
|
| [1]: https://qntm.org/antimemetics
| patleeman wrote:
| I loved this book. The audiobook is available on spotify and
| was a great listen.
| solid_fuel wrote:
| I really enjoyed one of their other stories - Ra
| https://qntm.org/ra
| riffraff wrote:
| I'll add that Lena/MMAcevedo[0] is both a wonderful story and
| terrifying
|
| [0] https://qntm.org/mmacevedo
| analog8374 wrote:
| Hey he does good scifi too
| AnotherGoodName wrote:
| Fun fact with arrow notation, if you put it under a modulus it
| quickly converges to the same value no matter how high in
| exponents you go!
|
| Eg. 2^2^2 = 2^4 mod 35 = 16
|
| Let's go one higher
|
| 2^2^2^2 = 2^16 mod 35 = 16 too!
|
| and once more for the record
|
| 2^2^2^2^2 = 2^65536 mod 35 = 16 as well. It'll keep giving this
| result no matter how high you go.
|
| https://www.wolframalpha.com/input?i=2%5E2%5E2%5E2+mod+35 for a
| link of this to play with.
|
| I could do this with any modulus and any exponent too.
|
| 2^3^3 = 2^3^3^3 = 7 mod 11 etc.
|
| The reason is that the orders of powers are effected by the
| totient recursively and since totients always reduce, eventually
| the totient converges to 1. This is where the powers no longer
| matter under modulus. Eg. the totient of 35 is 12 (the effective
| modulo of the first order power), the totient of 12 is 2 (the
| effective modulo of the second order power), the totient of 2 is
| 1 (the effective modulo of the third order power) and so after 3
| powers under mod 35 it converges.
| ashivkum wrote:
| I'm pretty sure there was a project Euler problem premised on
| this property but I can't find it at the moment.
| AnotherGoodName wrote:
| A classic would be quickly computing such big numbers under a
| modulus. You just compute the carmichael totient recursively
| till it hits 1, disregard higher orders and then going
| backwards calculate the powers, reducing by the modulo of the
| current order (this way it never gets large enough to be a
| pain to calculate). The totients reduce in logn time and each
| step is logn so it's merely logn^2 to calculate.
| 112233 wrote:
| As someone from time to time peeking into googology.fandom.com ,
| my favorite big number device probably still is loader.c, simply
| because of how concrete and unreachable it feels.
|
| Too bad most Friedman's work has linkrotted by now...
| piskov wrote:
| Ah, for a second I hoped it is another novel.
|
| If you haven't read "There is no antimemetics division", do it
| now. Easily one of the top science fiction out there.
|
| However buy the Penguin books 2025 edition, not the self-
| published free one -- that version has a meh ending and suffers
| from not having an editor.
| Sharlin wrote:
| Is it buggy for at least 2^(n)^2? It gives 4 for any n, but
| surely for example 2^^2 = 2^(2^2) != 4?
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