for me ψ'_1(ω)[3,b] worked out to Ω_1*2^5 which can start the system of creating loads of ω 32 times, and that's just a really small portion of ψ'_0(Ω_2)[3,b] and your number uses ψ'_0(Ω_9)[n,Ω_9] wow I'm getting a sense that it grows much much quicker than TREE(n)
I will start by analyzing $\psi_1$.
Some early values:
$\psi_1(\alpha,0) = \omega^{\omega^{1+\alpha}}$ for $\alpha \le \varepsilon_0$.
$\psi_1(\omega_1 + \alpha, 0) = \omega^{\omega^{\varepsilon_0+\alpha}}$ for $\alpha \le \varepsilon_1$.
$\psi_1(\omega_1(1+ \alpha) + \beta,0) = \omega^{\omeg...
From this, we know that ψ'_0(x) is equivalent to the standard OCF at particular large ordinals.
he defined tau with tree-ordinals (defining ordinals on only their fundamental sequences) α∈T_n if: 1)α=0 2)α=β+1 β ∈T_n 3)α: T_(n-1)→T_n (alpha is a function that takes a member of the previous set to collapse down to a smaller member of the current set)
T_0 is the set of integers T_1 is the set of countable ordinals that is to say functions that take an interger and create a smaller ordinal T_2 are the smallest uncoutable ordinals functions that take an ordinal and create a smaller uncountable ordinal
Let S(n) be all of the binary sequences of length N. How many sequence in S are repeating? Repeating means that it consists only of a subsequence of length N/2 or less. (Partial repeats are OK) @El'endiaStarman @flawr @feersum @anybodyiforgot
but in a mathematical way there is no real thing that means ^ is more fundamental than ↑↑ too me it just seems like +,*,^ was cut of at a random point but sure lets focus on the collapsing function
@fejfo Oh yeah, and feel free to try and produce a code that outputs something larger than TREE(3), or rather its approximation H(3,ω^(ψ'_0(ω^(Ω_1^(Ω_1^(Ω_1))))+1)).
^ The above approximation is crude, and may be invalid. I was just guesstimating.
I assumed a sequence is repeating if and only if there is a prefix of this sequence such that the whole sequence is just that prefix repeated over and over again
Let S(n) be all of the binary sequences of length N. How many sequence in S are repeating? Repeating means that it consists only of a subsequence of length N/2 or less. (Partial repeats are OK) @El'endiaStarman @flawr @feersum @anybodyiforgot
import sympy, itertools
lcm = lambda a, *b: sympy.lcm(a, lcm(*b)) if b else a
def f(n):
a = 2**n
if n < 2: return a
if sympy.isprime(n): return a - 2
factorization = sympy.factorint(n)
ff = [n / p for p in factorization]
for i in range(1, len(factorization) + 1):
a += (-1)**i * sum(2**lcm(*comb) for comb in itertools.combinations(ff, i))
return a
It can probably be made more efficient by using the structure of a factorization more.
Write a Manufactoria program that will accept the empty input tape. But don't do it quickly! I mean, write the program quickly, but don't let it run quickly. The slower the program, the better, as long as it terminates eventually. The example program below takes 3:51 (the "total time" reporte...
I will start by analyzing $\psi_1$.
Some early values:
$\psi_1(\alpha,0) = \omega^{\omega^{1+\alpha}}$ for $\alpha \le \varepsilon_0$.
$\psi_1(\omega_1 + \alpha, 0) = \omega^{\omega^{\varepsilon_0+\alpha}}$ for $\alpha \le \varepsilon_1$.
$\psi_1(\omega_1(1+ \alpha) + \beta,0) = \omega^{\omeg...
Write a Manufactoria program that will accept the empty input tape. But don't do it quickly! I mean, write the program quickly, but don't let it run quickly. The slower the program, the better, as long as it terminates eventually. The example program below takes 3:51 (the "total time" reporte...
