4 ms·

That's a lot of words and symbols for some not-very-impressive lower bounds. I present hnuser123456's function: f(n)=n (n Knuth up arrows) n Coming back to Ra

That's a lot of words and symbols for some not-very-impressive lower bounds. I present hnuser123456's function: f(n)=n (n Knuth up arrows) n Coming back to Rayo, my number is instead f(10^100) rather than Rayo(10^100), and given the series of lower bounds presented in that article, this should easily dwarf that, unless Rayo's will end up growing Knuth arrows faster than linear. I'm not smart enough to evaluate that though. If it would, wonder if this would beat it: f(n)=n (n Knuth up arrows) n (n Knuth up arrows) n Or n copies of n Knuth up arrows...

FartyMcFarter 6y ago
Knuth up arrows are computable functions, right? So they can't possibly grow faster than BB(n), let alone even faster-growing functions.
- hnuser123456 6y ago
  Must be. Since Rayo(10000)>2↑↑↑65536 it seemed easy to spam a lot more up arrows there (10k of them), but the tide probably changes after n grows by another 95 orders of magnitude. Wonder where Rayo's function intersects mine.
  - Aardwolf 6y ago
    Rayo's function can describe the idea of "f(n)=n (n Knuth up arrows) n" with relatively few symbols - just like your post did using relatively few characters
Arcorann 6y ago
n (n Knuth up arrows) n (n Knuth up arrows) n = n (n+1 Knuth up arrows) 3, so that's not really impressive either. Your function is comparable in size to f_omega(n) on the fast-growing hierarchy. [1] At any rate, it's known that BB(18) is larger than Graham's number [2]. In fact, the same person has a bunch of long-lasting Turing machines for values around 18 [3], giving the lower bounds: BB(15) > f_omega(2046) > 2046 (2046 up arrows) 2046 BB(16) > f_omega(f_omega(5)) BB(17) > f_(omega+1)(10) BB(18) > f_(omega+1)(7e57) > Graham's number BB(19) > f_(omega+1)^3(9) > f_(omega+2)(3) BB(20) > f_(omega+1)^4(7e57) > f_(omega+2)(4) BB(21) > f_(omega+2)(f_(omega+1)^2(4)) BB(22) > f_(omega+2)^3(f_(omega+1)^2(4)) And for larger values again [2]: BB(38) > f_(omega2)(167) BB(64) > f_(omega^2)(4098) BB(85) > f_(epsilon_0)(1907) Of course, we expect the Busy Beaver function to grow a lot faster than these bounds. [1] https://en.wikipedia.org/wiki/Fast-growing_hierarchy https://en.wikipedia.org/wiki/Fast-growing_hierarchy [2] https://googology.wikia.org/wiki/Busy_beaver_function#Graham.27s_number_vs._.CE.A3.28n.29 https://googology.wikia.org/wiki/Busy_beaver_function#Graham... [3] https://googology.wikia.org/wiki/User_blog:Wythagoras/The_nineteenth_Busy_Beaver_number_is_greater_than_Graham's_Number https://googology.wikia.org/wiki/User_blog:Wythagoras/The_ni...!
- hnuser123456 6y ago
  Thank you. I've read all your links and will continue to improve my googology.

FartyMcFarter 6y ago

Knuth up arrows are computable functions, right? So they can't possibly grow faster than BB(n), let alone even faster-growing functions.

hnuser123456 6y ago
Must be. Since Rayo(10000)>2↑↑↑65536 it seemed easy to spam a lot more up arrows there (10k of them), but the tide probably changes after n grows by another 95 orders of magnitude. Wonder where Rayo's function intersects mine.
- Aardwolf 6y ago
  Rayo's function can describe the idea of "f(n)=n (n Knuth up arrows) n" with relatively few symbols - just like your post did using relatively few characters

Arcorann 6y ago

n (n Knuth up arrows) n (n Knuth up arrows) n = n (n+1 Knuth up arrows) 3, so that's not really impressive either. Your function is comparable in size to f_omega(n) on the fast-growing hierarchy. [1] At any rate, it's known that BB(18) is larger than Graham's number [2]. In fact, the same person has a bunch of long-lasting Turing machines for values around 18 [3], giving the lower bounds: BB(15) > f_omega(2046) > 2046 (2046 up arrows) 2046 BB(16) > f_omega(f_omega(5)) BB(17) > f_(omega+1)(10) BB(18) > f_(omega+1)(7e57) > Graham's number BB(19) > f_(omega+1)^3(9) > f_(omega+2)(3) BB(20) > f_(omega+1)^4(7e57) > f_(omega+2)(4) BB(21) > f_(omega+2)(f_(omega+1)^2(4)) BB(22) > f_(omega+2)^3(f_(omega+1)^2(4)) And for larger values again [2]: BB(38) > f_(omega2)(167) BB(64) > f_(omega^2)(4098) BB(85) > f_(epsilon_0)(1907) Of course, we expect the Busy Beaver function to grow a lot faster than these bounds. [1] https://en.wikipedia.org/wiki/Fast-growing_hierarchy https://en.wikipedia.org/wiki/Fast-growing_hierarchy [2] https://googology.wikia.org/wiki/Busy_beaver_function#Graham.27s_number_vs._.CE.A3.28n.29 https://googology.wikia.org/wiki/Busy_beaver_function#Graham... [3] https://googology.wikia.org/wiki/User_blog:Wythagoras/The_nineteenth_Busy_Beaver_number_is_greater_than_Graham's_Number https://googology.wikia.org/wiki/User_blog:Wythagoras/The_ni...!

hnuser123456 6y ago
Thank you. I've read all your links and will continue to improve my googology.