### Re: My Number is Bigger! (Everymen Needed!)

Posted:

**Fri Jan 01, 2016 4:54 pm UTC**What number do you think we'll reach at the end of this page? 10^^3? A googolplex?

10^256

10^256

Page **3** of **45**

Posted: **Fri Jan 01, 2016 4:54 pm UTC**

What number do you think we'll reach at the end of this page? 10^^3? A googolplex?

10^256

10^256

Posted: **Fri Jan 01, 2016 8:55 pm UTC**

Posted: **Fri Jan 01, 2016 9:47 pm UTC**

I've made a blogpost on Googology Wiki for people to edit in their notations. See it here:

http://googology.wikia.com/wiki/User_Bl ... _Notations

I believe you should be able to edit it if you have a registered Wikia account.

Anyway...

1 Centillion = 10^303

http://googology.wikia.com/wiki/User_Bl ... _Notations

I believe you should be able to edit it if you have a registered Wikia account.

Anyway...

1 Centillion = 10^303

Posted: **Sat Jan 02, 2016 12:19 pm UTC**

That was only three...

Posted: **Sat Jan 02, 2016 3:32 pm UTC**

2.5F2 = 10^10^2.5 ~ 10^316

Posted: **Mon Jan 04, 2016 8:27 am UTC**

That was four.

Now we await an everyman to keep the game going.

Now we await an everyman to keep the game going.

Posted: **Mon Jan 04, 2016 9:46 am UTC**

10^(101*pi) ~ 1.999E317

Posted: **Mon Jan 04, 2016 1:17 pm UTC**

(8/3)F2 ~ 1.44E464

Posted: **Mon Jan 04, 2016 2:34 pm UTC**

{2}{2}{2}2 = {2}{2}(2x2^{2}) = {2}{2}8 = {2}(8x2^{8}) = {2}2048 = 2048x2^{2048} ~ 6.62x10^{619}

Posted: **Mon Jan 04, 2016 3:09 pm UTC**

That's just past username's 2|()|()|() < 10^{617}.

(-1/3)F4 = EEEE(-1/3) ~ EEE0.46416 ~ EE2.91178 ~ E816.1729 ~ 1.489*10^{816}

(-1/3)F4 = EEEE(-1/3) ~ EEE0.46416 ~ EE2.91178 ~ E816.1729 ~ 1.489*10

Posted: **Mon Jan 04, 2016 4:11 pm UTC**

10^10^3 = 10^1000

Posted: **Mon Jan 04, 2016 4:43 pm UTC**

Define:

[x,y,z] = x^x^x^...^x^z (with y x's)

So:

~~[x,y,1] = x^y~~

[x,y,2] = x^x^y

[x,y,3] = x^x^x^y

[x,y,4] = x^x^x^x^y <- thanks to Emlightened for spotting this to be an error

[x,1,z] = x^z

[x,2,z] = x^x^z

[x,3,z] = x^x^x^z

[x,4,z] = x^x^x^x^z

and so on.

Send:

[3,2,7] = 3^3^7 = 3^2187 ~ 3x10^{1043}

(also, note the following identities always hold:

a^^b = [a,b,1]

aFb = [10,b,a]

)

[x,y,z] = x^x^x^...^x^z (with y x's)

So:

[x,y,2] = x^x^y

[x,y,3] = x^x^x^y

[x,y,4] = x^x^x^x^y

[x,1,z] = x^z

[x,2,z] = x^x^z

[x,3,z] = x^x^x^z

[x,4,z] = x^x^x^x^z

and so on.

Send:

[3,2,7] = 3^3^7 = 3^2187 ~ 3x10

(also, note the following identities always hold:

a^^b = [a,b,1]

aFb = [10,b,a]

)

Posted: **Mon Jan 04, 2016 4:50 pm UTC**

I think I read that the notation ^{y}x^{#z} being the same as that, but I only ever saw it used once.

Also, Psi, your examples are screwed up. They should be:

[3,2,8] = 3^3^8 = 3^6561 ~ 10^{3130}

Also, Psi, your examples are screwed up. They should be:

PsiSquared wrote:[x,1,z] = x^z

[x,2,z] = x^x^z

[x,3,z] = x^x^x^z

[x,4,z] = x^x^x^x^z

[3,2,8] = 3^3^8 = 3^6561 ~ 10

Posted: **Mon Jan 04, 2016 4:58 pm UTC**

emlightened wrote:Also, Psi, your examples are screwed up. They should be:PsiSquared wrote:[x,1,z] = x^z

[x,2,z] = x^x^z

[x,3,z] = x^x^x^z

[x,4,z] = x^x^x^x^z

Ouch. How did that happen?

You're right of-course.

[3,3,2] = 3^3^3^2 = 3^19683 ~ 10

Posted: **Mon Jan 04, 2016 5:05 pm UTC**

(7!)! = 5040! ~ E16476 ~ 4.217F2

Four.

Four.

Posted: **Wed Jan 06, 2016 9:33 am UTC**

It seems that the everymen lost interest in the game, causing it to stall.

(I'm pretty sure it isn't because of a comprehension issue, given that we haven't even reached a googolplex yet)

So what do we do now?

(I'm pretty sure it isn't because of a comprehension issue, given that we haven't even reached a googolplex yet)

So what do we do now?

Posted: **Wed Jan 06, 2016 9:56 am UTC**

5F2

I'm an everyman (although with a PhD in Cryptography). Hit me with your most crazy notations.

I'm an everyman (although with a PhD in Cryptography). Hit me with your most crazy notations.

Posted: **Wed Jan 06, 2016 9:58 am UTC**

6^^3 ~ 2.659E36305

Posted: **Wed Jan 06, 2016 12:13 pm UTC**

5F2 is larger than 6^^3

5F2 = 10^10^5 = 1E100000 which is larger than 2.7E36305 (because 100000 > 36305)

Or alternatively:

6^^3 ~ 2.7E36305 ~ E36305 = EE(log 36305) ~ EE4.56 = 4.56F2 < 5F2 (because 4.56 < 5)

Anyway:

[8,2,8-2] = [8,2,6] = 8^8^6 ~ 4.2E236739 ~ (log 236739)F2 ~ 5.374F2

5F2 = 10^10^5 = 1E100000 which is larger than 2.7E36305 (because 100000 > 36305)

Or alternatively:

6^^3 ~ 2.7E36305 ~ E36305 = EE(log 36305) ~ EE4.56 = 4.56F2 < 5F2 (because 4.56 < 5)

Anyway:

[8,2,8-2] = [8,2,6] = 8^8^6 ~ 4.2E236739 ~ (log 236739)F2 ~ 5.374F2

Posted: **Wed Jan 06, 2016 1:22 pm UTC**

PsiSquared wrote:5F2 is larger than 6^^3

I'm aware of that, I was ninja'd.

5.5F2

Posted: **Wed Jan 06, 2016 3:13 pm UTC**

Edit: Fixed number.

Posted: **Wed Jan 06, 2016 3:58 pm UTC**

That's actually (9!)!.

And how are you doing the factorial approximations? I'm curious, because your results are almost accurate yet ever-so-slightly off (Wolfram-alpha gives (7!)! as 4x10^{16473} and (9!)! as 10^{1 859 933})

Anyway, since the actual number you've posted is (10!)! ~ 7.35F2, I'll post:

{3}3 = {2}{2}{2}3 = {2}{2}24 = {2}402653184 = 402653184x2^{402653184} ~ 10^{121 210 695} = E(121 210 695) = EE[log(121 210 695)] ~ EE8.08 = 8.08F2

Also notice that {3}3 is not very far off from 1F3 (or 10F2). Make note of this fact, because it will be important later.

And how are you doing the factorial approximations? I'm curious, because your results are almost accurate yet ever-so-slightly off (Wolfram-alpha gives (7!)! as 4x10

Anyway, since the actual number you've posted is (10!)! ~ 7.35F2, I'll post:

{3}3 = {2}{2}{2}3 = {2}{2}24 = {2}402653184 = 402653184x2

Also notice that {3}3 is not very far off from 1F3 (or 10F2). Make note of this fact, because it will be important later.

Posted: **Wed Jan 06, 2016 5:09 pm UTC**

Hypercalc. (It gives 3.633825810665 × 10^{16476} for (7!)!, by the way.)

[e,3,π] ~ 10^{4871463233.4747} ~ 9.68765F2 (accurate in both Hypercalc and Wolfram)

[e,3,π] ~ 10

Posted: **Wed Jan 06, 2016 5:48 pm UTC**

Hypercalc is known for having a bug when it comes to factorials.

You could probably trust it for everything else. But the truth is that for nearly all practical purposes, WolframAlpha can do anything that Hypercalc can do (although W-A has a really annoying bug which causes it to crash when given certain power towers, regardless of the actual size of the number involved: 2^2^2^5 crashes it, yet 2^2^2^6 works just fine).

Anyway:

10^^3 = 10^10^10 = 1F3 = 10F2 = (10 000 000 000)F1 = 10^{10 000 000 000}

(Major Milestone)

And that's four.

(that went quickly... maybe Username should consider increasing this limit a bit? What does everybody else think about this suggestion?)

You could probably trust it for everything else. But the truth is that for nearly all practical purposes, WolframAlpha can do anything that Hypercalc can do (although W-A has a really annoying bug which causes it to crash when given certain power towers, regardless of the actual size of the number involved: 2^2^2^5 crashes it, yet 2^2^2^6 works just fine).

Anyway:

10^^3 = 10^10^10 = 1F3 = 10F2 = (10 000 000 000)F1 = 10

(Major Milestone)

And that's four.

(that went quickly... maybe Username should consider increasing this limit a bit? What does everybody else think about this suggestion?)

Posted: **Wed Jan 06, 2016 7:31 pm UTC**

10^10987654321

Posted: **Wed Jan 06, 2016 9:02 pm UTC**

I'd be fine with it, but it's probably best left < 15 or < 10.

[2,3,2*3] = 2^2^2^6 = 2^2^64 = 2^18446744073709551616 ~ EE18.744

[2,3,2*3] = 2^2^2^6 = 2^2^64 = 2^18446744073709551616 ~ EE18.744

Posted: **Wed Jan 06, 2016 10:22 pm UTC**

I agree.

As for your number:

That's 18.744F2, or in "standard super scientific notation": 1.273F3

So I submit:

{2}{2}{2}4 = {2}{2}64 = {2}(64x2^{64}) = {2}(2^{70}) = 2^{70} x (2^2^70) =

= 2^(70+(2^70)) = 10^((70+(2^70)) x log 2) ~ 10^{355 393 490 465 494 856 487} ~ EE20.5507 ~ EEE1.313 = 1.313F3

As for your number:

That's 18.744F2, or in "standard super scientific notation": 1.273F3

So I submit:

{2}{2}{2}4 = {2}{2}64 = {2}(64x2

= 2^(70+(2^70)) = 10^((70+(2^70)) x log 2) ~ 10

Posted: **Wed Jan 06, 2016 11:23 pm UTC**

17^^3 ~ 10^10^21.008

Posted: **Thu Jan 07, 2016 12:56 am UTC**

I'd be willing to increase it to 7 or even 10. Probably no more than that.

Anyway...

Define a:b as aaa...aaa with b a's (so 6:5 = 66666).

Define a::b as a:(a:(...(a:(a:a)))...))) with b a's (So 2::3 = 2:(2:2) = 2:22 = 2.222E21).

I send:

2::4 = 2:2222222222222222222222 ~ 10^(2.222*10^21)

Anyway...

Define a:b as aaa...aaa with b a's (so 6:5 = 66666).

Define a::b as a:(a:(...(a:(a:a)))...))) with b a's (So 2::3 = 2:(2:2) = 2:22 = 2.222E21).

I send:

2::4 = 2:2222222222222222222222 ~ 10^(2.222*10^21)

Posted: **Thu Jan 07, 2016 8:10 am UTC**

22.2F2 = 10^10^22.2 ~ 10^{15 848 931 924 611 134 852 021}

Posted: **Thu Jan 07, 2016 9:50 pm UTC**

10^10^25

Also:

I hereby declare that only one out of every ten numbers needs to be made by an everyman. That is, if nine consecutive numbers are submitted by googologists, the next one must be made by an everyman.

If you don't agree, state so now, because this will be official soon.

Also:

I hereby declare that only one out of every ten numbers needs to be made by an everyman. That is, if nine consecutive numbers are submitted by googologists, the next one must be made by an everyman.

If you don't agree, state so now, because this will be official soon.

Posted: **Fri Jan 08, 2016 6:16 am UTC**

No objection.

[3,3,4] = 3^3^3^4 = 3^3^81 ~ 10^10^38.325 = 38.325F2 ~ 1.583F3

[3,3,4] = 3^3^3^4 = 3^3^81 ~ 10^10^38.325 = 38.325F2 ~ 1.583F3

Posted: **Fri Jan 08, 2016 5:19 pm UTC**

Also no objection.

@user (not related to this game, just posted here because he'll more likely see it):**Spoiler:**

4↑↑↑2 = 4↑↑4 = 4↑4↑4↑4 = 4↑4↑256 ~ EE153.907 ~ 2.187F3

I don't think anyone's actually explained ↑ arrows fully yet.

a↑b = a^{b}

a↑↑b = a↑a↑...a (for b a's)

a↑↑↑b = a↑↑a↑↑...a (for b a's)

a↑↑↑↑b = a↑↑↑a↑↑↑...a (for b a's)

etc.

We use ↑^{k} as shorthand for ↑↑...↑↑, with k arrows.

Formally, we say that:

a↑b = a^{b}

a↑^{k}1 = a

a↑^{k}b = a↑^{k-1}(a↑^{k}(b-1)).

So the full way to expand 4↑↑↑2 is:

4↑↑↑2 = 4↑↑(4↑↑↑1) = 4↑↑(4) = 4↑4↑↑3 = 4↑4↑4↑↑2 = 4↑4↑4↑4↑↑1 = 4↑4↑4↑4↑ = 4↑4↑(4^{4}) = 4↑4↑256

= 4↑(4^{256}) ≈ 4↑(10^{154.127}) = 4^{10^154.127} ≈ 10^{(10^154.127)*log(4)} = 10^{10^(154.127+log(log(4)))} ≈ 10^{10^153.907} = EE153.907

@user (not related to this game, just posted here because he'll more likely see it):

4↑↑↑2 = 4↑↑4 = 4↑4↑4↑4 = 4↑4↑256 ~ EE153.907 ~ 2.187F3

I don't think anyone's actually explained ↑ arrows fully yet.

a↑b = a

a↑↑b = a↑a↑...a (for b a's)

a↑↑↑b = a↑↑a↑↑...a (for b a's)

a↑↑↑↑b = a↑↑↑a↑↑↑...a (for b a's)

etc.

We use ↑

Formally, we say that:

a↑b = a

a↑

a↑

So the full way to expand 4↑↑↑2 is:

4↑↑↑2 = 4↑↑(4↑↑↑1) = 4↑↑(4) = 4↑4↑↑3 = 4↑4↑4↑↑2 = 4↑4↑4↑4↑↑1 = 4↑4↑4↑4↑ = 4↑4↑(4

= 4↑(4

Posted: **Sat Jan 09, 2016 11:36 am UTC**

And I wanted to use 3|()() (which is about 40F2)...

10^10^200

10^10^200

Posted: **Sat Jan 09, 2016 3:50 pm UTC**

username5243 wrote:And I wanted to use 3|()() (which is about 40F2)...

I thought you might. That's exactly why I posted [3,3,4], which is a little bit less than that.

Anyway:

2.4F3= 10^10^10^2.4 ~ 10^10^251

P.S.

You should update the OP to reflect that change in the rules.

Posted: **Mon Jan 11, 2016 2:14 pm UTC**

According to Wolfram Alpha

4.2↑↑4 ~ 10^10^258

4.2↑↑4 ~ 10^10^258

Posted: **Mon Jan 11, 2016 5:46 pm UTC**

2.468F3 = 10^10^10^2.468 ~ 10^10^294

Posted: **Tue Jan 12, 2016 10:45 pm UTC**

2{2}1000 = {2}(1000*2^1000) ~ 10^10^303

Posted: **Wed Jan 13, 2016 12:58 am UTC**

10^10^400

Posted: **Wed Jan 13, 2016 1:12 am UTC**

username5243 wrote:What number do you think we'll reach at the end of this page? 10^^3? A googolplex?

10^256

emlightened wrote:2{2}7 ~ 4.7335E272

Probably around .

10^10^680