My number is larger [GAME]

[Moosey]

that's way too big a jump imo

we're not even out of tetration yet, let alone into expansion

[MathAndCode]

Code: Select all

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

[Hunting]

2^^6 (sorry, quite a jump)

What does this mean

[Moosey]

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

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))))))

[Hunting]

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

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))))))

Code: Select all

f(0)=1
f(x) = 3 ↑^f(x-1) 3

f(3856)

// This is my 3856th post

[wzkchem5]

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]

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]

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)

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

anyway 2^^15

[Hunting]

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

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]

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)

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

[Hunting]

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

Sry ignore the second argument

it was left there when simplifying

[Moosey]

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

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]

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)

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?

[Moosey]

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?

not at this point, no
still illdefined since x will go to 0 and you don't have a condition for that

[Moosey]

2^^32

[EvinZL]

2^^^5

[Moosey]

2^^^6 // thx

[NoAyeBeardo]

2^^^8

[Hunting]

2 ↑^(2^^^8) 8

[Hdjensofjfnen]

2 ↑^(2^^^8) 8

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

[Moosey]

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

[Hunting]

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

It's getting boring
2^^^(2^^6)

[Moosey]

k
2^^^(2^^^4)

[EvinZL]

2^^^^4

[Moosey]

that's exactly the same but ok
3^^^^3 = g1
note: inductively, 3^^^^3 > 2^^^^4 = 2^^^2^^^4
3^^^^3 = 3^^^3^^^3

3^^^3 > 2^^^4

2^4 = 16 > 3^3 = 27