I will resolve this market to the biggest number.
21
1.8kṀ550resolved Oct 27
ResolvedN/A
15%
Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number)
12%
A(999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999) where x is the first non-recursive ordinal and A is the xth function in the Grzegorczyk hierarchy.
9%
notation F[n] = composition of func F with itself n times. R=RayosFunc
Extending Graham's-number-like towers
Lyr(1)=R(999), Lyr(n)=R[Lyr(n-1)](999)
Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9)
Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9)
Row[9999](9)
7%Other
6%
A(9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999) where x is the first non-recursive ordinal and A is the xth function in the Grzegorczyk hierarchy
6%
Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number squared)
5%
So let's say you have:
F1(x,y) = addition
F2(x,y) = multiplication
F3(x,y) = factorial
F4 ...
etc.
Then my number is:
Fgrahamsnumber(Graham's number, Graham's number)
4%
T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(9
))))))))))))))))))))))))))))))))))))))))))))))))))))))), where T is Friedman's function in the context of Kruskal's tree theorem
4%
notation F[n] = composition of fun F with itself n times. R=RayosFunc
Extending Graham's-number-like towers
Lyr(1)=R(999), Lyr(n)=R[Lyr(n-1)](999)
Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9)
Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9)
Row[R(9999)](9)
4%
Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Graham's Number passed to the busy beaver function, with the result passed to the busy beaver function - repeat a number of times equal to Graham's Number))))))))))))))
4%
max(all numbers submitted so far) + 1
3%
If you divide the visible universe into cubes of one Planck length. How many ways are there to rearrange them? Rearrange all cubes every Planck time. From the big bang until now, how many different timelines are there possible?
3%
Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number
3%
Uncountable infinity, but just enough subtracted from it so it becomes finite. Right there at the edge. That's my number. So if you add only a little bit, it would become infinity, which would fail.
2%
The largest finite number definable in fewer than 10000 characters in English minus 7
1.1%
Graham's number = G. f(1)=G, f(x+1) when x > 1 = f(x) ↑^f(x) f(x). h(1) = f(G), h(x) when x > 1 = h(x) ↑^h(x) h(x). h(f(G)) = U. i(1) = h(U), i(x) when x > 1 = i(x) ↑^i(x) i(x). i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(U))))))))))))))))))+1 = W.
1.1%
Let n(k) be defined as the length of the longest possible sequence that can be constructed with a k-letter alphabet such that no block of letters x_i,...,x_2i is a subsequence of any later block x_j,...,x_2j. Then, n(n(n(n(n(n(n(n(9)))))))
1.1%
The "High Five" sculpture in Silverwood Park
1.1%
Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number) + 1 (sorry)
Using Scott Aaronson's rules:
"Using standard math notation, English words, or both, name a single whole number—not an infinity—on a [free response answer]. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your [answer] and, if necessary, the published literature."
Largest number wins!
Close date updated to 2022-10-26 10:07 pm
This question is managed and resolved by Manifold.
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