[Game] My integer is larger

[pzq_alex]

Like My number is larger, but integers only.

To prevent cheating, we enforce the following rules:

1. Integers in standard ZFC set theory only. (This prohibits nonstandard integers. )
2. Your integer must be well-defined, utilizing preexisting notations under (eventually dominated by) f_(w^w) under the standard choices of fundamental sequences only.
3. No trivial variations on previous posts. This includes “the previous number +1” and “sum of all previously posted numbers”.
4. Don’t jump too far. This is less formal, but please don’t jump from e.g. power tower to TREE, or from BEAF to uncomputable functions.
5. Your integer must be greater than the previous one (obviously).

I’ll start: 0.

Edit: New rule:

6. If it is not immediately clear from the definition, you must show that your integer is well-defined.

[yujh]

99^0

[HotWheels9232]

Not CA related

[rowett]

Let ds(n) be the smallest prime number where the digit sums of it when written in bases 2 to n+1 are all prime.

ds(34)

[pzq_alex]

Let ds(n) be the smallest prime number where the digit sums of it when written in bases 2 to n+1 are all prime.

ds(34)

How would you show that such a prime exists?

[rowett]

How would you show that such a prime exists?

ds(1) = 3
3 in base 2 is the binary number 11, the digit sum of 11 is 2, which is prime

ds(2) = 5
5 in base 2 is 101, the digit sum of 101 is 2, which is prime
5 in base 3 is 12, the digit sum of 12 is 3, which is prime

ds(3) = 5
5 in base 2 is 101, the digit sum of 101 is 2, which is prime
5 in base 3 is 12, the digit sum of 12 is 3, which is prime
5 in base 4 is 11, the digit sum of 11 is 2, which is prime

ds(4) = 11
11 in base 2 is 1011, the digit sum of 1011 is 3, which is prime
11 in base 3 is 102, the digit sum of 102 is 3, which is prime
11 in base 4 is 23, the digit sum of 23 is 5, which is prime
11 in base 5 is 21, the digit sum of 21 is 3, which is prime

etc.

[pzq_alex]

How would you show that such a prime exists?

ds(1) = 3
3 in base 2 is the binary number 11, the digit sum of 11 is 2, which is prime

ds(2) = 5
5 in base 2 is 101, the digit sum of 101 is 2, which is prime
5 in base 3 is 12, the digit sum of 12 is 3, which is prime

ds(3) = 5
5 in base 2 is 101, the digit sum of 101 is 2, which is prime
5 in base 3 is 12, the digit sum of 12 is 3, which is prime
5 in base 4 is 11, the digit sum of 11 is 2, which is prime

ds(4) = 11
11 in base 2 is 1011, the digit sum of 1011 is 3, which is prime
11 in base 3 is 102, the digit sum of 102 is 3, which is prime
11 in base 4 is 23, the digit sum of 23 is 5, which is prime
11 in base 5 is 21, the digit sum of 21 is 3, which is prime

etc.

I understand how the function works, but how do you show that ds(34) is well-defined, i.e. there is a prime whose digit sums in base 2..35 are also primes?

Edit: According to this calculation, ds only increases exponentially, so something like graham(with a small g) is going to beat it...

[Wyirm]

This one is going to take a bit of explaining.
Imagine a numerical counting system with a base less than unary, say base .5. Each 1 in the base would be only half the last, and no number could be larger than 2. Now assume we have a function f(x) that represents the number of digits required to reach 2 for a number system base x. This function starts at 2 for base 1, and rockets off to infinity at 2. I do not know the growth rate of this function, but it can be represented as the number of terms in a series of x^n needed to equal 2 as x approaches 0.5.

f(0.5001) should be a good enough start

I would show it with the sum function, but i'm on my phone.

Edit* this isn't rational function, it surpasses any x^-n because the asymptote is at .5 and rational functions are at 0.

[HotWheels9232]

This one is going to take a bit of explaining.
Imagine a numerical counting system with a base less than unary, say base .5. Each 1 in the base would be only half the last, and no number could be larger than 2. Now assume we have a function f(x) that represents the number of digits required to reach 2 for a number system base x. This function starts at 2 for base 1, and rockets off to infinity at 2. I do not know the growth rate of this function, but it can be represented as the number of terms in a series of x^n needed to equal 2 as x approaches 0.5.

f(0.5001) should be a good enough start

I would show it with the sum function, but i'm on my phone.

Edit* this isn't rational function, it surpasses any x^-n because the asymptote is at .5 and rational functions are at 0.

Seems like logarithmicly. But I warned you about not being CA-related so now I'm going to ask the moderators. My 509th pot!

[hotdogPi]

What is the point of declaring every single post (or "pot", as you have it) number now that you've reached 500? I can see 512 (2^9), but not 501, 502, 503, or 509.

[pzq_alex]

I’ve added a new rule (see OP), so rowett’s entry is now invalidated.

This one is going to take a bit of explaining.
Imagine a numerical counting system with a base less than unary, say base .5. Each 1 in the base would be only half the last, and no number could be larger than 2. Now assume we have a function f(x) that represents the number of digits required to reach 2 for a number system base x. This function starts at 2 for base 1, and rockets off to infinity at 2. I do not know the growth rate of this function, but it can be represented as the number of terms in a series of x^n needed to equal 2 as x approaches 0.5.

f(0.5001) should be a good enough start

I would show it with the sum function, but i'm on my phone.

Edit* this isn't rational function, it surpasses any x^-n because the asymptote is at .5 and rational functions are at 0.

This entry is actually 13 (https://www.wolframalpha.com/input?i=Su ... 2C13%7D%5D).

My entry: 10^100 (aka Googol)

[get_Snacked]

10^100/(1/(10^100)) (so 10^10000)
(when will mathml be available?)
EDIT: i just learned summation notation! yay for me!
if you have done basic coding before, summation notation is basically just the for(); function.
(now i'll have to learn about log(n) and then i'm done with the basic operations!)

[unname4798]

I will restart.
0

[Colonizor48]

The least Baile-PSW pseudoprime(one probably exists, the known lower bound is 2^64, I would say there is probably one between 2^64 and 2^1000, but I have nothing to back that up)

[Haycat2009]

The least Baile-PSW pseudoprime(one probably exists, the known lower bound is 2^64, I would say there is probably one between 2^64 and 2^1000, but I have nothing to back that up)

The smallest odd perfect number

[unname4798]

The least Baile-PSW pseudoprime(one probably exists, the known lower bound is 2^64, I would say there is probably one between 2^64 and 2^1000, but I have nothing to back that up)

The smallest odd perfect number

There are no odd perfect numbers.

[b3s23love]

There are no odd perfect numbers.

It’s still an unsolved problem.

[MEisSCAMMER]

There are no odd perfect numbers.

It’s still an unsolved problem.

It's possible Haycat has solved it and this is their subtle way of telling the rest of us...

[Colonizor48]

(using BEAF notation)
let f_0(n) = 0
f_1(n) = {n} = 1
f_2(n) = {n, n(1)n, n}
f_3(n) = {n, n, n(1)n, n, n(1)n, n, n(1)(2)n, n, n(1)n, n, n(1)n, n, n(1)(2)n, n, n(1)n, n, n(1)n, n, n}(i think this is right, if it is not correct me please, it is supposed to be a 3x3x3 3d array filled with n
f_k(n) = a kxk k dimensional hypercube array filled with n
f_omega(n) = f_n(n)
f_omega(f_omega(f_omega(3)))
f_omega grows faster then f_epsilon0(i am pretty sure) on the fgh because f_epsilon0 is around a function k(n), where k(n) = {n, n, n, n, n....} n times. and f_omega obviously grows faster, but I don't know how to prove it

[Colonizor48]

The least Baile-PSW pseudoprime(one probably exists, the known lower bound is 2^64, I would say there is probably one between 2^64 and 2^1000, but I have nothing to back that up)

The smallest odd perfect number

This is i'll defined, as odd perfect numbers are not known to exist or not exist.
(But, by that logic mine was also i'll defined, though the probability of a Baille-PSW pseudoprime is far higher then that of an odd perfect number existing.)