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.
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.
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.
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...
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.
I'm pretty sure there was a project Euler problem premised on this property but I can't find it at the moment.
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.
I'm a big qntm fan. I highly recommend their "antimemetics" SCP stories and articles.
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
I loved this book. The audiobook is available on spotify and was a great listen.
I really enjoyed one of their other stories - Ra https://qntm.org/ra
I'll add that Lena/MMAcevedo[0] is both a wonderful story and terrifying
[0] https://qntm.org/mmacevedo
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?
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...
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.
Hey he does good scifi too