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Moosey wrote: November 27th, 2020, 12:11 pm 2^^6 (sorry, quite a jump)

MathAndCode wrote: November 27th, 2020, 8:55 pm

Code: Select all

if the Ackermann function is considered to be past up-arrows:
    2↑↑9
else:
    A(4, 6)

Hunting wrote: November 27th, 2020, 9:40 pm

Moosey wrote: November 27th, 2020, 12:11 pm 2^^6 (sorry, quite a jump)

What does this mean

Moosey wrote: November 27th, 2020, 9:43 pm

MathAndCode wrote: November 27th, 2020, 8:55 pm

Code: Select all

if the Ackermann function is considered to be past up-arrows:
    2↑↑9
else:
    A(4, 6)

they're basically the same thing, so either is fine

Hunting wrote: November 27th, 2020, 9:40 pm

Moosey wrote: November 27th, 2020, 12:11 pm 2^^6 (sorry, quite a jump)

What does this mean

wdym? if you're asking what 2^^6 is, it's 2 tetrated to the 6 = 2^2^2^2^2^2 (associative right; that is, (2^(2^(2^(2^(2^2))))))

f(0)=1
f(x) = 3 ↑^f(x-1) 3

Hunting wrote: November 27th, 2020, 9:53 pm f(0)=1
f(x) = 3 ↑^f(x-1) 3[/code]

f(3856)

// This is my 3856th post

wzkchem5 wrote: November 28th, 2020, 2:20 am

Hunting wrote: November 27th, 2020, 9:53 pm f(0)=1
f(x) = 3 ↑^f(x-1) 3[/code]

f(3856)

// This is my 3856th post

I would say this is too much a jump

Hunting wrote: November 28th, 2020, 3:53 am

wzkchem5 wrote: November 28th, 2020, 2:20 am

Hunting wrote: November 27th, 2020, 9:53 pm f(0)=1
f(x) = 3 ↑^f(x-1) 3[/code]

f(3856)

// This is my 3856th post

I would say this is too much a jump

Well this is my best attempt at something non-trivial (i. e. not just adding arrows)

Moosey wrote: November 29th, 2020, 11:54 am

Hunting wrote: November 28th, 2020, 3:53 am

wzkchem5 wrote: November 28th, 2020, 2:20 am

I would say this is too much a jump

Well this is my best attempt at something non-trivial (i. e. not just adding arrows)

it's to much of a jump
it's roughly g3856

f(0)=1
f(y)=y↑↑f(f(y-1), y)

Hunting wrote: November 29th, 2020, 11:59 am

Moosey wrote: November 29th, 2020, 11:54 am

Hunting wrote: November 28th, 2020, 3:53 am

Well this is my best attempt at something non-trivial (i. e. not just adding arrows)

it's to much of a jump
it's roughly g3856

Well, I don't know any googology or what is g.

So

Code: Select all

f(0)=1
f(y)=y↑↑f(f(y-1), y)

f(3856)

Moosey wrote: November 29th, 2020, 12:14 pm

Hunting wrote: November 29th, 2020, 11:59 am

Moosey wrote: November 29th, 2020, 11:54 am
it's to much of a jump
it's roughly g3856

Well, I don't know any googology or what is g.

So

Code: Select all

f(0)=1
f(y)=y↑↑f(f(y-1), y)

f(3856)

you don't define how f takes two arguments, that's not even defined

Hunting wrote: November 29th, 2020, 12:16 pm

Moosey wrote: November 29th, 2020, 12:14 pm

Hunting wrote: November 29th, 2020, 11:59 am

Well, I don't know any googology or what is g.

So

Code: Select all

f(0)=1
f(y)=y↑↑f(f(y-1), y)

f(3856)

you don't define how f takes two arguments, that's not even defined

Sry ignore the second argument

it was left there when simplifying

Moosey wrote: November 29th, 2020, 12:18 pm

Hunting wrote: November 29th, 2020, 12:16 pm

Moosey wrote: November 29th, 2020, 12:14 pm

you don't define how f takes two arguments, that's not even defined

Sry ignore the second argument

it was left there when simplifying

ok, that's still illdefined though: observe:

f(4) = 4^^f(f(3)) = 4^^f(27) -> eventually references f(4) again, thus f(4) and everything above is illdefined
f(3) = 3^^f(f(2)) = 3^^2 = 27
f(2) = 2^^f(f(1)) = 2
f(1) = 1

I'd recommend you make it f(y+1) = y^^f(y) or something so that it's defined and pentation level (but that's still WAAAAY too big)

Hunting wrote: November 29th, 2020, 1:03 pm

Moosey wrote: November 29th, 2020, 12:18 pm

Hunting wrote: November 29th, 2020, 12:16 pm
Sry ignore the second argument

it was left there when simplifying

ok, that's still illdefined though: observe:

f(4) = 4^^f(f(3)) = 4^^f(27) -> eventually references f(4) again, thus f(4) and everything above is illdefined
f(3) = 3^^f(f(2)) = 3^^2 = 27
f(2) = 2^^f(f(1)) = 2
f(1) = 1

I'd recommend you make it f(y+1) = y^^f(y) or something so that it's defined and pentation level (but that's still WAAAAY too big)

Dang. I still need two arguments

f(x, 0)=1
f(x, y)=y↑↑f(x-1, f(x, y-1))

Isn't pentation too trivial as it's just adding an arrow to tetration?

Hunting wrote: November 30th, 2020, 9:18 pm 2 ↑^(2^^^8) 8

Hdjensofjfnen wrote: November 30th, 2020, 9:44 pm

Hunting wrote: November 30th, 2020, 9:18 pm 2 ↑^(2^^^8) 8

// That looks pretty large in comparison to the last number.

Moosey wrote: November 30th, 2020, 10:47 pm

Hdjensofjfnen wrote: November 30th, 2020, 9:44 pm

Hunting wrote: November 30th, 2020, 9:18 pm 2 ↑^(2^^^8) 8

// That looks pretty large in comparison to the last number.

// yup, 2{2^^^8}8 is a touch large compared to 2^^^8

2^^^12