User talk:Hyp cos

Can i ask who are you? JiawhienIsBackNoEvadeBlockOnceAgainIComeBack (talk) 09:10, October 1, 2013 (UTC)


 * Hyp cos. FB100Z &bull; talk &bull; contribs 19:02, October 1, 2013 (UTC)


 * The average of ex and its reciprocal. -- ☁ I want more clouds! ⛅ 03:18, October 3, 2013 (UTC)

SGH
Please look here. I gave reason why SGH is weaker than you think Wythagoras (talk) 19:00, October 16, 2013 (UTC)

SGH grows from outside --- \(g_{\theta(\Omega^\omega,1)}(n)\) stands above \(g_{p(SVO)}(n)\) for any function p which has lower level than SVO. So it's easy for SGH to grow from outside and hard to grow from inside. For example, \(g_{\theta(\Omega^{\omega+1})}(n)\) expands to \(g_{\theta(\Omega^\omega\theta(\Omega^\omega...\theta(\Omega^\omega)))}(n)\), \(g_{\theta(\Omega^\omega+1)}(n)\) expands to \(g_{\theta(\Omega^\omega,\theta(\Omega^\omega,...\theta(\Omega^\omega)))}(n)\), and even \(g_{\theta(\Omega^\omega,1)}(n)\) can expand to \(g_{\theta(\Omega^n,\theta(\Omega^\omega)+1)}(n)\).

However, many notations grow from inside, such as BEAF, BAN, HH, etc. .For example, \(H_{\omega^{\omega^\omega}+1}(n)=H_{\omega^{\omega^\omega}}(n+1)\), and {3,3,1,2}={3,3,{3,3,3}}. It's hard to compare them with SGH.

FGH can grows from outside or inside. \(f_{\alpha+1}(n)=f_\alpha^n(n)\). And there're 2 very different ways to get \(f_\alpha^2(n)\) from \(f_\alpha(n)\). One way is \(f_\alpha(n+1)\), \(f_\alpha(2n)\), ..., \(f_\alpha(p(n))\), ..., \(f_\alpha(f_\alpha(n))\), and thet other way is \(f_\alpha(n)+1\), \(2f_\alpha(n)\), ..., \(p(f_\alpha(n))\), ..., \(f_\alpha(f_\alpha(n))\). Thus, FGH becomes a general way to compare other notations. Notations growing from inside or outside suit. For comparison between FGH and SGH, we should grow FGH from outside and not inside. &#123;hyp&#60;hyp··cos&#62;cos&#125; (talk) 01:36, October 17, 2013 (UTC)
 * \(f_\alpha^2(n)\) just means \(f_\alpha(f_\alpha(n))\) and generally \(f_\alpha^n(n)\) is \(f_\alpha(f_\alpha(\cdots(f_\alpha(n))\cdots))\) with n f's. Ikosarakt1 (talk ^ contribs) 07:36, October 17, 2013 (UTC)

google site
DO you have your own google site about the large numbers? JiawhienIsBackNoEvadeBlockOnceAgainIComeBack (talk) 07:05, November 17, 2013 (UTC)

No. Actually I'm green to googology. hyp$hyp?cos&#38;cos (talk) 10:31, November 17, 2013 (UTC)

For Aarex
Make an extension of latest extension of Extended Up-arrow Notation. AarexTiao 01:50, December 1, 2013 (UTC)

Number navigator
Can you explain this a little more? FB100Z &bull; talk &bull; contribs 01:46, December 5, 2013 (UTC)

Number navigators in numbers such as Hyper-E numbers, Bowers' gongulus series, and lower hyperfactorial array numbers can be deleted because we can use the templetes as our new and powerful "number navigator". We can see other similar numbers in templetes. But, in nucleaxul we cannot delete it because there's no templete for nucleaxul numbers. If we delete it, we can't find numbers similar to nucleaxul on this page. hyp$hyp?cos&#38;cos (talk) 01:53, December 5, 2013 (UTC)
 * So why not just add a template? FB100Z &bull; talk &bull; contribs 02:44, December 5, 2013 (UTC)
 * Done! But there's still something wrong. In Extremexul template, the "Extremexul" in the first row under the title is bold, and it doesn't link to anywhere. However, in gigantixul template I newly add, the "Gigantixul" in the first row under the title links to page Gigantixul. But I don't know how to fix it. Okey now. Nothing serious in the new templates. hyp$hyp?cos&#38;cos (talk) 11:59, December 5, 2013 (UTC)

Extension of Nested Array Notation
Make an extension of Nested Array Notation
 * I found \(nR\{0\{1\text{^}\}*\} = nR\{0\{0\underbrace{**...**}_n\}\), \(nR\{0\{1\text{^}\}**\} = nR\{0\{0\underbrace{**...**}_n\{1\text{^}\}*\}*\}\), etc.


 * And \(nR\{0\{2\text{^}\}*\} = nR\{0\underbrace{\{1\text{^}\}...\{1\text{^}\}}_n*\}\), \(nR\{0\{3\text{^}\}*\} = nR\{0\underbrace{\{2\text{^}\}...\{2\text{^}\}}_n*\}\), etc.


 * \(\{0\{0\{0*\text{^}\}1\text{^}\}*\} = \{0\{0\{0\{...\text{^}\}1\text{^}\}1\text{^}\}*\}\)


 * The extension of NAN name is HNAN (Hyper-Nested Array Notation.) AarexTiao 14:47, December 29, 2013 (UTC)

Extra: Growth rate of PNAN = psi(X(e(0))) AarexTiao 02:08, December 27, 2013 (UTC)
 * I don't think so. First, we have different ideas from {0{0,1*}1} on. We know the nR{0{0,1*}1} and the nR {0{{0{0,1*}1}*}1} both have growth rate \(\psi(\psi_{I(1,0,0)}(0))=\psi(\psi_{\chi(M)}(0))\), but what's the equvalence? We know {0{1*}1} is equvalent to \(\psi_{I(\omega,0)}(0)\) (not \(\psi(\psi_{I(\omega,0)}(0))\) or \(I(\omega,0)\)), and {0{1*}2} is equvalent to \(\psi_{I(\omega,0)}(1)\), the {0{1*}0,1} is equvalent to \(I(\omega,0)\).
 * More, the {0{{0{1*}1}*}1} is equvalent to \(\psi_{I(\psi_{I(\omega,0)}(0),0)}(0)\),
 * the {0{{0{1*}1}*}0,1} is equvalent to \(I(\psi_{I(\omega,0)}(0),0)\),
 * the {0{{0{1*}0,1}*}1} is equvalent to \(\psi_{I(I(\omega,0),0)}(0)\),
 * and the {0{{0{1*}0,1}*}0,1} is equvalent to \(I(I(\omega,0),0)\).
 * The {0{{0{{0{1*}1}*}1}*}1} is equvalent to \(\psi_{I(\psi_{I(\psi_{I(\omega,0)}(0),0)}(0),0)}(0)\), and the {0{{0{{0{1*}0,1}*}0,1}*}0,1} is equvalent to\(I(I(I(\omega,0),0),0)\).
 * Now think of this: {0{0,1*}1} is the diagonalizer of {0{ ____ *}1}, so {0{{0{0,1*}1}*}1} is equvalent to \(\psi_{I(1,0,0)}(0)\), the {0{{0{{0{0,1*}1}*}1}*}1{{0{0,1*}1}*}1} is equvalent to \(\psi_{I(\psi_{I(1,0,0)}(0),1)}(0)\), the {0{{0{0,1*}1}*}2} is equvalent to \(\psi_{I(1,0,0)}(1)\), and the {0{{0{0,1*}1}*}0,1} is equvalent to \(I(1,0,0)\).
 * And the {0{0,1*}1} is equvalent to M, then.
 * So nR{0,1{0,1*}1} has growth rate \(\psi(\psi_{\chi(\varepsilon_{M+1})}(0))\),
 * nR{0,0,1{0,1*}1} has growth rate \(\psi(\psi_{\chi(\psi_{I_{M+1}}(0))}(0))\),
 * nR{0,0,0,1{0,1*}1} has growth rate \(\psi(\psi_{\chi(\psi_{I(1,M+1)}(0))}(0))\),
 * nR{0{1*}1{0,1*}1} has growth rate \(\psi(\psi_{\chi(\psi_{I(\omega,M+1)}(0))}(0))\),
 * nR {0{{0{{0{0,1*}1}*}1}*}1{0,1*}1} has growth rate \(\psi(\psi_{\chi(\psi_{I(\psi_{I(1,0,0)}(0),M+1)}(0))}(0))\),
 * nR {0{{0{0,1*}1}*}1{0,1*}1} has growth rate \(\psi(\psi_{\chi(\psi_{I(M,1)}(0))}(0))\),
 * nR{0{0,1*}2} has growth rate \(\psi(\psi_{\chi(M_2)}(0))=\psi(\psi_{\chi(\psi_{I(1,0,M+1)}(0))}(0))\),
 * nR{0{0,1*}3} has growth rate \(\psi(\psi_{\chi(M_3)}(0))\),
 * nR{0{0,1*}{0{0,1*}1}} has growth rate \(\psi(\psi_{\chi(M_M)}(0))\),
 * So next to a comma means inaccessible, next to a {0,1*} means mahlo, and then next to a {0,2*} means compact. we can use {0{0,A*}1} for stage function of A, and the diagonalizer, {0{0,0,1*}1}, is the stage cardinal. Next to a {0,0,1*} means stage, so {0{0,0,1*}0,1} is inaccessible stage, {0{0,0,1*}0{0,1*}1} is mahlo stage, {0{0,0,1*}0{0,2*}1} is compact stage, {0{0,0,1*}0{0,0,1*}1} is stage stage. etc. hyp$hyp?cos&#38;cos (talk) 02:36, December 27, 2013 (UTC)
 * Well? {0{1,0,1*}1} is w-ex-stage, {0{{0,1},0,1*}1} is W-ex-stage, {0{{0,0,1},0,1*}1} is I-ex-stage, {0{{0{1*}1},0,1*}1} is I(w,0)-ex-stage, {0{{0{{0{0,1}1}*}1},0,1*}1} is chi(M)-ex-stage, {0{{0{0,1*}1},0,1*}1} is M-ex-stage, {0{{0{0,2*}1},0,1*}1} is compact-ex-stage, {0{{0{0,0,1*}1},0,1*}1}, and {0{0,1,1*}1} is stage-ex-...-ex-stage AarexTiao 02:50, December 27, 2013 (UTC)

In my system, stage-ex-stage-ex-...-ex-stage is just the first ordinal in the second row of n-dimensions, or psi(X(ww+12)) Wythagoras (talk) 07:42, December 27, 2013 (UTC)
 * I stronger than you! Check it out! AarexTiao 01:11, December 29, 2013 (UTC)
 * Well, if you have some strong ordinal notations, can you compare it to nested array notation? hyp$hyp?cos&#38;cos (talk) 02:27, December 29, 2013 (UTC)

HNAN II
Can you add Hyper-Nested Array Notation II on R function series? AarexTiao 15:17, February 1, 2014 (UTC)


 * click here AarexTiao 02:41, February 9, 2014 (UTC)

Thanks, Hyp cos. I see you have more experience with the categorization than I do.

Questions
Hello Hypcos. I would like to ask a few questions about your work with large numbers. I would like to ask a couple of questions about your website called WordPress Steps Toward Infinity. How come parts 8-12 don't work in your Strong Array Notation? Do parts 1-7 work in your Strong Array Notation? Does your R function work? Two people on this website told me that your R function has an infinite loop.
 * Some expressions in NDAN cause infinite loop, such as s(3,3{1(2(1)○△2)△3}2). That makes NDAN and beyond don't work. But I've just know what's the problem, and I'll fix it soon.
 * Parts 1-7 work, and they're as strong as R function.
 * But I don't know what's wrong with R function now. &#123;hyp/^,cos&#125; (talk) 00:42, July 1, 2016 (UTC)
 * iirc it was littlepeng9 who found the error. -- vel! 17:21, July 1, 2016 (UTC)
 * No, it wasn't me. For what I know it was Fluoroantimonic Acid. I have happened to meet him on IRC recently and I've asked him about this. I don't have access to logs right now, but he said that the flaw regarded an old version of the notation and that error isn't there anymore. I'm pretty sure Deedlit was there at the time of the conversation, so he might be able to provide logs. LittlePeng9 (talk) 19:51, July 1, 2016 (UTC)


 * I don't think the error was ever there. Here is the passage that FA thought meant there was a problem with the notation:


 * To solve s(a,b{1{2}1,2}2), we start the process, then meet case B2, then meet case B1. Before case B1 applies, the array is s(a,b{1{2}1,2}2{1{2}1,2}1); after case B1 applies, the array becomes s(a,a{1{2}b,1}2{1{2}1,2}1). Then rule 2a and 2b applies, so s(a,b{1{2}1,2}2) = s(a,a{1{2}b}2). What if we use the case B1 in Linear array notation? s(a,b{1{2}1,2}2) will be s(a,a{a{a}b}2). When a ≥ 3 and b ≥ 2, it’ll reduces to something containing {1{2}1,2} again – it can never be solved. So a change on case B1 is necessary.


 * But what Hyp Cos was saying was that if we used the version of B1 in "Linear array notation" in his "Extended array notation", then we would get an infinite loop, which is why he changed that particular rule for "Extended array notation". Deedlit11 (talk) 00:18, July 2, 2016 (UTC)


 * That cause only LAN will work and since NDAN is fixed, why DEN will have 2 types includes weak and strong? GoogleAarex2001 17:18, July 7, 2016 (UTC)

My feeling is that as a notation becomes more complex, the more necessary it becomes for someone to prove that it terminates. -- vel! 04:49, July 7, 2016 (UTC)

Question from Googleaarex
Is wDEN parts finally completed? GoogleAarex2001 23:32, July 11, 2016 (UTC)
 * It depends on how you define "wDEN parts". My plan for the next part is converting superscripts on droppers into a very special separator - that's a bit similar to "from mDEN to pDAN". &#123;hyp/^,cos&#125; (talk) 13:12, July 12, 2016 (UTC)
 * Since wDmEN is completed, what is your next part plan? AarexWikia04 (talk) 01:35, July 18, 2016 (UTC)
 * The case B3 in wDmEN makes it hard to extend. So I may make a new version of wDmEN before extend it. &#123;hyp/^,cos&#125; (talk) 01:39, July 18, 2016 (UTC)
 * Can you do the new case B3 changing instead? AarexWikia04 (talk) 15:47, July 21, 2016 (UTC)

Will you ever make number names past two-row arrays? Username5243 (talk) 01:43, July 18, 2016 (UTC)
 * Yes, but I'll finish it much later. &#123;hyp/^,cos&#125; (talk) 01:45, July 18, 2016 (UTC)
 * But I have the ideas! AarexWikia04 (talk) 01:46, July 18, 2016 (UTC)

I will beat your biggest number on SAN.
I am nearby passing Pentenlinel = s(3,2,2{2}3,3,3,3,3)!

I am currently on Googolis-exchgain = s(10,100{2}1,1,1,2). AarexWikia04 (talk) 15:44, July 21, 2016 (UTC)

Question about PDAN
Hi Hyp Cos, I had a question about PDAN, in particular the evaluation of {a,b {1 {1,,1,,2} 2} 2}.

I get that A_1 = {1 { 1,,1,,2} 2}, A_2 = {1,,1,,2}. Then B_3 = ",,", B_2 = {1,,1,,2,,1} = {1,,1,,2}, and B_1 = {1 {1,,1,,2} 2 {1,,1,,2} 1} = {1 {1,,1,,2} 2}. (Actually, do we apply rule 2 at this point, or not?) Then X = "{1,,1" and Y = "}", so we compare lvl(A_1) to lvl(X ,, n-1 Y) = lvl({1,,1,,1}) = lvl({1}), and lvl (A_1) is greater.

So we go to case B2.5: P = "{1", Q = "2}", so S_1 = ",", S_2 = {1,2}, S_3 = {1{1,2}2}, S_4 = {1{1{1,2}2}2}, and so on. Is this correct? If so, that seems problematic, as that should be how {1,,2} reduces and not {1,,1,,2}. The problem seems to be that we want {1,,1,,2} to drop down to something like {1,,X}, but in your comparison you compare to {1,,1,,1} = {1}. Deedlit11 (talk) 04:14, August 2, 2016 (UTC)
 * You're right, and s(a,b {1 {1,,1,,2} 2} 2) = s(a,b {1 {1,,2} 2} 2). That seems problematic because it's a "wrong way" of pDAN. Another example for "wrong way" is that (in \(\psi\) function up to \(\psi(\Omega_\omega)\)) \(\psi(\psi_1(\Omega_\omega))=\psi(\Omega_2)<\psi(\Omega_\omega)\), and making \(\alpha\) in \(\psi(\psi_1(\alpha))\) larger is a "wrong way". Since we have many ways to extend the expressions, we can't avoid the appearence of "wrong ways" for notations complicated enough.
 * In pDAN, the main reason for some expressions running into "wrong way" is that the Case B2.6 doesn't work correctly on them. Then, how to find expressions in the "right way" of pDAN? We use the most simple expressions, e.g. s(a,b {1,,1,,2} 2), s(a,b {1,,1,,3} 2), s(a,b {1,,1,,1,,2} 2) and so on. They meet Case B2.6 for the first reduce, and, after that, they'll continue the "right way". &#123;hyp/^,cos&#125; (talk) 05:40, August 2, 2016 (UTC)

Thanks! I'm curious, have you made comparisons with your notation and ordinal collapsing functions beyond \(\psi(\psi_I(0))\)? For the R function, you've said that LAN reaches \(\psi(\psi_{I(\omega,0)}(0))\), and for PNAN you've compared {0{0,1*}1} to M and {0{0,2*}1} to K; so, if my interpretation to the new array notation is correct, that should mean that PDAN reaches \(\psi(\psi_{I(\omega,0)}(0))\) (I'm inclined to agree with this) and that {1 {1,,2,,,2}2} is analogous to M and {1{1,,3,,,2}2} is analogous to K (I'm less sure about this). Do you agree with this, or do you get something different? Deedlit11 (talk) 10:58, August 2, 2016 (UTC)
 * No, PNAN reaches over C(e(W2+1)). AarexWikia04 - 11:00, August 2, 2016 (UTC)


 * I didn't talk about the limit of PNAN in my post. Are you referring to Taranovsky's notation? What makes you say PNAN surpasses C(e(W2+1))? Deedlit11 (talk) 11:11, August 2, 2016 (UTC)


 * My guess (originating from some comparisions between the R function and strong array notation) is that \(\psi(\psi_{I(\omega,0)}(0))\) is just the growth rate of s(a,b {1,,1 ... ,,1,,2} 2), and that {1{1,,2,,}2} is analogous to M and {1{1,,3,,}2} is analogous to K. -- ☁ I want more clouds! ⛅ 11:39, August 2, 2016 (UTC)
 * Oh wait, {1{1,,2,,}2} and {1{1,,3,,}2} is equivalent to {1{1,,2,,,2}2} and {1{1,,3,,,2}2}, respectively. -- ☁ I want more clouds! ⛅ 11:42, August 2, 2016 (UTC)
 * pDAN is as strong as PNAN in R function and Taranovsky's "Degrees of Reflection" notation, but they're much weaker than \(C(C(\varepsilon_{\Omega_2+1},0),0)\). &#123;hyp/^,cos&#125; (talk) 12:24, August 2, 2016 (UTC)
 * s(n,n{1{2,,,2}2}2) reaches \(\psi(\psi_{I(\omega,0)}(0))\), {1{1,,2,,,2}2} is analogous to M, but {1{1,,3,,,2}2} is just analogous to \(\Xi[1]\) (the least 1-weakly Mahlo), and {1{1,,1,,2,,,2}2} is analogous to K. &#123;hyp/^,cos&#125; (talk) 12:24, August 2, 2016 (UTC)
 * But this is sDAN. AarexWikia04 - 12:30, August 2, 2016 (UTC)
 * {whatever,,,2} is equivalent to {whatever,,}, so this could still be considered a part of pDAN. -- ☁ I want more clouds! ⛅ 16:09, August 2, 2016 (UTC)

Thanks again... according to your PNAN calculations, you would have

{1{1,,2,,,2}3} is the second Mahlo cardinal {1{1,,2,,,2}a} is the ath Mahlo cardinal {1{1,,2,,,2}1,,2} is the first inaccessibly Mahlo cardinal {1{1,,2,,,2}2,,2} is the second inaccessibly Mahlo cardinal {1{1,,2,,,2}1,,3} is the first 2-inaccessibly Mahlo cardinal {1{1,,2,,,2}1,,1,,2} is the first hyper-inaccessibly Mahlo cardinal {1{1,,2,,,2}1{1,,2,,,2}2} is the first 2-Mahlo cardinal

but apparently this is incorrect... where does it go wrong?

Also, would it be correct to say that {1{1,,1,,2,,,2}a} is analogous to the ath weakly compact cardinal? Deedlit11 (talk) 16:28, August 2, 2016 (UTC)
 * I might make analysis soon but starting with pDAN. AarexWikia04 - 16:37, August 2, 2016 (UTC)
 * I still haven't fully understand ordinal collapsing notation on K, so the comparisons are done between my array notation and Taranovsky's notation. He claimed that \(C(\Omega_2+C(\Omega_2,C(\Omega_22,0)),0)\) is the least recursive inaccessible, \(C(\Omega_2+\omega^{C(\Omega_2,C(\Omega_22,0))2},0)\) is the least recursive Mahlo, and \(C(\Omega_2+\omega^{\omega^{C(\Omega_2,C(\Omega_22,0))2}},0)\) is the least recursive \(\Pi_3\)-reflecting, and I get the results from here. As a result, my PNAN calculations are wrong beyond M. &#123;hyp/^,cos&#125; (talk) 00:56, August 3, 2016 (UTC)

Ah, I see. I believe I more or less understand the ordinal collapsing function with K, but unfortunately I don't quite grasp your notation or Taranovsky's notation, so we are at an impasse. How far have you compared Taranovsky's notation with your own? Do you know if Taranovsky's main notation is equal to, say, Dropping Array Notation, or if one is stronger than the other? Deedlit11 (talk) 04:50, August 3, 2016 (UTC)
 * Limit of DAN is \(C(C(\Omega_22+C(\Omega_2+C(\Omega_2+1,C(\Omega_22,0)),0),0),0)\) (also limit of R function), limit of NDAN is \(C(C(\Omega_22+C(\Omega_2+C(\Omega_22,C(\Omega_22,0)),0),0),0)\), and limit of WDmEN is \(C(C(\Omega_22+C(\Omega_2+C(\Omega_22+1,0),0),0),0)\) (but I'm not very sure about the last one). So Taranovsky's notation is much stronger. &#123;hyp/^,cos&#125; (talk) 08:43, August 3, 2016 (UTC)
 * He made new part, so my limit ordinal of pDDN is C(C(W_2*2+C(W_2*2,0),0)). AarexWikia04 - 17:25, August 31, 2016 (UTC)

Woah, really?! The notations for Taranovsky's system barely budge even all the way up your notation! I guess Taranovsky's notation is really that strong. Deedlit11 (talk) 17:55, August 3, 2016 (UTC)


 * Hyp cos is wrong. I check and C(1,1) is stronger, which is 2nd fixed point of x -> C(0,x). AarexWikia04 - 11:33, August 3, 2016 (UTC)
 * NVM AarexWikia04 - 15:10, August 3, 2016 (UTC)

After WDmEN?
How about soWDmEN (Second order weak dropper-expanding notation?) AarexWikia04 - 14:13, August 2, 2016 (UTC)
 * According to pDDN after WDEN, will you make like sDDN, DDN, and nDDN? AarexWikia04 - 15:43, August 11, 2016 (UTC)
 * I pause making parts now, and go to naming numbers. &#123;hyp/^,cos&#125; (talk) 03:50, August 12, 2016 (UTC)
 * When will you make parts again? AarexWikia04 - 00:28, August 13, 2016 (UTC)
 * He made the new part, pDDN. Will you make sDDN after or before EAN numbers? AarexWikia04 - 15:07, August 23, 2016 (UTC)
 * Ummm, hyp cos will analyze entire SAN first; then he will make EAN numbers or extensions included:

* Secondary Dropper-Dropping Notation * Dropper-Dropping Notation * Nested Dropper-Dropping Notation * Second-Order Weak Dropper-Expanding Notation * Higher-Order Weak Dropper-Expanding Notation * Strong Dropper-Expanding Notation
 * Googleaarex (talk) 02:25, February 20, 2017 (UTC)

Hello, I'm back in here again, and I have revamped idea for sDDN.
 * Imagine A without grave accents in the end is first mark of the definition of A.
 * Define A with adding 1 grave accent in the end is the next mark of the definition of B, where B is like A but removing grave accents in the end.
 * Define A with adding comma in the end is dropper of the definition of A.
 * ,, is a simplest dropper and (n+1)-tuple comma is a simplest n-dropper, which is dropper of (n-1)-dropper.
 * Define A with adding colon in the end is weak dropper-expanding of the definition of A.
 * Define A with adding colon then n grave accents in the end is weak (n+1)th mark of dropper-expanding of the definition of A.
 * Let dropper-(nth mark of A) separators are same as (dropper-A)-(nth mark of A).
 * Finally, we could define 1-dropper-2-dropper separators, which is a strongest separator of sDDN.

We can easily extend this:
 * DDN defines separators up to the limit of dropper-dropper separators
 * nDDN defines separators up to the limit of the fixed point A, which is dropper-A separators

Good luck for defining 3 extensions using my idea! Googleaarex (talk) 22:23, August 5, 2017 (UTC)

Can you define ntepAAN?
First, read the incomplete definition right here:. AarexWikia04 - 00:45, August 4, 2016 (UTC)

Is your site resumed for more analysis?
Please answer me... AarexWikia04 - 00:42, November 19, 2016 (UTC)
 * It continued up to sDAN analysis. Googleaarex (talk) 02:23, February 20, 2017 (UTC)

Adding strong array notation approximation
Can I add approximations using strong array notation on non-SAN numbers? For example, number pages that coined and defined using other notations especially Saibian's ExE (e.g. grangol, greagol, etc.) or BEAF. ARsygo (talk) 16:03, June 29, 2017 (UTC)
 * Yes. &#123;hyp/^,cos&#125; (talk) 00:10, June 30, 2017 (UTC)

My ideas up to SDEN
Nishada 00:33, August 6, 2017 (UTC) EDIT: I fixed a mistake. Nishada 00:34, August 6, 2017 (UTC) EDIT 2: I fixed another mistake. Nishada 00:35, August 6, 2017 (UTC)
 * Grave accent is 1-separator.
 * ,, is 2-separator.
 * ,,, is 2-separator over 2-separator = 3-separator.
 * ,,,, is 2-separator over 2-separator over 2-separator = 4-separator.
 * : is 2-separator over x-separators = 1-separator-separator.
 * + is 2-separator-separator.
 * Now, here is the ideas beyond pDDN:
 * ++ is 2-separator-separator over 2-separator-separator = 3-separator-separator.
 * +++ is 2-separator-separator over 2-separator-separator over 2-separator-sepator = 4-separator-separator.
 * [1]□□◊ is 2-separator-separator over x-separator-separators = 1-separator-separator-separator.
 * [2]□□◊ is 2-separator-separator-separator.
 * [3]□□◊ is 2-separator-separator-separator over 2-separator-separator-separator = 3-separator-separator-separator.
 * [1]□□□◊ is 2-separator-separator-separator over x-separator-separator-separators = 1-separator-separator-separator-separator.
 * [x]□y◊ is x-separatory, where ab = aaa... b times (concatenation of strings, not multiplication)
 * sDDN (Secondary Dropper-Dropping Notation) goes up to ++
 * DDN (Dropper-Dropping Notation) goes up to [1,2]□◊
 * NDDN (Nested Dropper-Dropping Notation) goes up to {1[1]□□◊2}
 * SNDN (Strong Nested Dropping Notation) goes up to [1]□□□...◊
 * You are wrong, 2-separator-separator is stronger than you thought. Googleaarex (talk) 00:36, August 6, 2017 (UTC)
 * Well, my ideas for future parts seems like the logical extension to the current parts.
 * Forgot the signature Nishada 00:38, August 6, 2017 (UTC)
 * Also, your 3-separator-separator corresponds to my 2-dropper-(1-dropper) separators. That's why my sDDN is stronger than yours. Googleaarex (talk) 00:57, August 6, 2017 (UTC)

OCF vs DAN
From http://googology.wikia.com/wiki/User:Hyp_cos/OCF_vs_Array_Notation_p2

...fitting Taranovsky's correspondences (x,x{1{1{2,,}2,,}2}2) - the growth rate limit in KP + $$\Pi_n$$ -reflection

...not fitting Taranovsky's correspondences (x,x{1{1`,,2,,}2}2) the growth rate limit in KP + $$\Pi_n$$ -reflection

KP + $$\Pi_n$$ -reflection equivalent KP + $$\Pi_\omega$$ ?

Then:

What the growth rate has $$\Pi_{OCF(\Pi_\omega)}$$ in DAN?

What the growth rate has $$\Pi_{OCF(\Pi_{OCF(\Pi_\omega)})}$$ in DAN?

etc. up to $$\alpha \mapsto \Pi_{OCF(\alpha)}$$ ?

Or what stable ordinals will be equivalent? If I correctly understand the stable ordinals are equivalent to transfinite $$\Pi_n$$ -reflection? Scorcher007 (talk) 10:44, August 6, 2017 (UTC)


 * Here is the answer, and it's not that easy:
 * \(\Pi_{OCF(\Pi_\omega)}\) = \(\Pi_{\psi_{\Omega}(\Pi_{\omega})}\) approx. corresponds to \(\{1,,1\{1',,1\{1\{1,,1,2\}2\}2\}2\}\)
 * \(\Pi_{OCF(\Pi_{OCF(\Pi_\omega)})}\) = \(\Pi_{\psi_{\Omega}(\Pi_{\psi_{\Omega}(\Pi_{\omega})})}\) approx. corresponds to \(\{1,,1\{1',,1\{1\{1,,1\{1',,1\{1\{1,,1,2\}2\}2\}2\}2\}2\}2\}\)
 * \(\alpha \mapsto \Pi_{OCF(\alpha)}\) = \(\Pi_{\psi_{\omega}(\Pi_{\Omega})}\) approx. corresponds to \(\{1,,1\{1',,1\{1\{1,,1\{1',,1\{1,,2\}2\}2\}2\}2\}2\}\)
 * Googleaarex (talk) 09:06, August 6, 2017 (UTC)
 * Fixed. Googleaarex (talk)


 * This expression $$\Pi_\Omega$$ can exist? Then do the following expressions have meaning: $$\Pi_I$$, $$\Pi_M$$, $$\Pi_{\Pi_3}$$, $$\Pi_{\Pi_\omega}$$, $$\Pi_{\Pi_{\Pi_\omega}}$$, etc. up to $$\alpha \mapsto \Pi_\alpha$$ ? And how this the growth rate in DAN? I believe that stable ordinals are equivalent to transfinite $$\Pi_n$$ -reflection (if n - transfinite countable ordinal) and this equivalent totally indescribable cardinal. And $$\Pi_n$$ -reflection (if n - uncountable cardinal and large) equivalent shrewd indescribable cardinal or subtle ordinals (In accordance with Rathjen's work The art of ordinal analysis). Is it so? Scorcher007 (talk) 10:44, August 6, 2017 (UTC)


 * The expression $$\Pi_\alpha$$ doesn't have a meaning when $\alpha$ is transfinite, as far as I know. $$\Pi_n$$ for n finite is defined as the set of formulas of the form $$Q_1 x_{1,1},\ldots, x_{1,n_1} Q_2 x_{2,1},\ldots,x_{2,n_2} \ldots Q_m x_{m,1},\ldots x_{m,n_m} F$$ where $$Q_i$$ is $$\forall$$ if $$i$$ is odd and $$\exists$$ when $$i$$ is even, and $$F$$ is a formula with only bounded quantifiers. In order to talk about $$\Pi_\alpha$$ or $$\Sigma_\alpha$$ for $$\alpha$$ transfinite, you have to make sense of formulas with infinite chains of alternating quantifiers. This would belong to the subject of infinitary logic. Looking at the Wikipedia page on infinitary logic, they do define some infinitary languages, but they do not have infinite chains of alternating quantifiers. (They do have infinite quantification, but only in the sense of $$Q_i x_{i,1},\ldots x_{i,\alpha}$$ for $$\alpha$$ transfinite.) Perhaps someone somewhere has addressed the topic of infinite alternating quantification, but maybe it didn't work out so well or they couldn't come up with a reasonable definition. In any case, we can't talk about $$\Pi_\alpha$$ for transfinite $$\alpha$$ until we define what it means. Deedlit11 (talk) 01:20, August 29, 2017 (UTC)

Analyze SAN with Lambda Notation
Click here for the definition of Lambda Notation Nishada 00:47, August 25, 2017 (UTC)
 * It might be ill-defined. Googleaarex (talk) 10:28, August 25, 2017 (UTC)

Hyp cos, stop ignoring me... Nishada 02:13, September 6, 2017 (UTC)
 * I haven't read that... It's a bit complicated, so I need more time. &#123;hyp/^,cos&#125; (talk) 03:37, September 6, 2017 (UTC)

Will you cut it out?
It is becoming near-impossible to follow the actual new conversations here, when you're editing hundreds of articles per hour.

Yes, I know I did the same thing a few weeks ago, but Cloudy told both of us to stop because it is annoying. And you know something? He is right (especially when 90%+ of those articles will be deleted anyway, once we finish the regiments project). PsiCubed2 (talk) 19:16, August 29, 2017 (UTC)


 * Can confirm it's annoying. LittlePeng9 (talk) 19:32, August 29, 2017 (UTC)


 * Yeah. I goto school on most weakdays, so can't access the wiki from about 6:30 to about 3:00 (and often later, because of homework). Because of that, I want to see what happened while I was gone. And I can't get a complete idea because I have to sroll througha hundred of edits on pages that will get deleted. Username5243 (talk) 19:39, August 29, 2017 (UTC)

Presently the changing of classification of size require more than 10 000 edits. I admire the Hypcos, his intelligence and efficiency, if he would help with the creation of pages for regiments then this process will go faster. Further if we will replace pages for single numbers with pages for regiments, then the changing of size classes will require smaller amount of edits--Denis Maksudov (talk) 21:51, August 29, 2017 (UTC)
 * All "numbers" (including redirects) from "Chained arrow notation level" to "Beyond legiattic array notation level" should be edited, and there are approximately 7000 "numbers". If I slow down it must take more days to finish this. &#123;hyp/^,cos&#125; (talk) 23:24, August 29, 2017 (UTC)
 * Just note: if we will have 86 pages for Saibian's regiments+some amount of saved pages for single Saibian's numbers, which currently contains approximations, then changing of size classes requires much smaller amount of edits than presently when we have 6439 pages for Saibian's numbers. And same way for other naming systems. --Denis Maksudov (talk) 23:39, August 29, 2017 (UTC)


 * So? It will take more time. What's the rush? Meanwhile you are rendering the "wiki activity" page unusable. And three people, all respectful members of this community, have asked you to slow down. I think we deserve a little more consideration than "If I slow down, it will take more time".


 * Moreover, as Denis tried to explain in his always-diplomatic tone, you are editing stuff which won't be relevant in a few months anyway. If you sit back and do nothing, 90%+ of your edits won't be necessary anyway. In short: you're disrupting our ability to have discussions for nothing.


 * And if you will continue to ignore our pleas, I think I'm going to take a break from this wiki until you're finished (with the sole exception of updating my PEGG page). PsiCubed2 (talk) 07:56, August 30, 2017 (UTC)


 * I have a suggestion for a new feature on the "Wiki Activity" page: filtering by user. That way, you can get rid of Hyp cos's edits on the wiki activity page, so they don't clog it up. that way, he can edit the articles, and people can still use the wiki activity page. Username5243 (talk) 23:58, August 29, 2017 (UTC)

is it over yet LittlePeng9 (talk) 10:33, September 9, 2017 (UTC)
 * No. I've just done numbers by Sbiis Saibian. HAN numbers, BEAF numbers, FGH numbers and SAN numbers in "Tetrational array level" and "Higher array level" are not done. &#123;hyp/^,cos&#125; (talk) 12:31, September 9, 2017 (UTC)


 * LittlePeng9, of-course it isn't over.


 * This guy decided he's going to be ranked #1 in the badge ranking, and doesn't give a hoot about the damage he's doing in the process. So why would he stop now?


 * I don't care how smart he is, or how big his past contributions are. NOBODY should be allowed to behave like this. IMO, niether being a googology genius nor being a community veteran should give a person the right to act like a jerk.


 * But what do I know? After all, the big boss Cloudy thinks "it is a cool situation"... (and yes, I'm pissed). PsiCubed2 (talk) 16:58, September 9, 2017 (UTC)


 * I don't think it's correct to assume that Hyp cos is changing the categories just to get badge points. I think he did this to switch from one category system to another as soon as possible, for some reason. -- ☁ I want more clouds! ⛅ 17:04, September 9, 2017 (UTC)
 * Frankly, I don't care why he is doing it.


 * The point is - it isn't helpful. 90% of the articles he's editing are going to be deleted within a few months, All this "work" he is completely and utterly useless.


 * We've already told him that, and he completely ignored us. I'm sorry, but a person who is editing hundreds of pages per day does not have the luxury of ignoring the discussions that follow his actions. It's not a "cool situation" when a wiki member decides to do his own thing while completely ignoring his peers. It is a f***-ing nuisance. PsiCubed2 (talk) 17:56, September 9, 2017 (UTC)


 * Also, if the impossiblity of following the actual new conversations is your concern, you can hide the edits on the main namespace in Special:RecentChanges. -- ☁ I want more clouds! ⛅ 17:13, September 9, 2017 (UTC)


 * I wish I knew about that two weeks ago. LittlePeng9 (talk) 17:28, September 9, 2017 (UTC)


 * PsiCubed2, does that work for you? Deedlit11 (talk) 17:36, September 9, 2017 (UTC)


 * Deedlit, it works as well as you could expect. Unchecking the "mainspace" box does - indeed - hide all those edits. It doesn't make the general usage of the Special:RecentChanges page any less of a pain-in-the-a*s, though. PsiCubed2 (talk) 18:04, September 9, 2017 (UTC)


 * Cloudy, that's hardly an adequete solution. That page is much less reader-friendly then the usual "Wiki Updates" page and lacks quite a few features. PsiCubed2 (talk) 17:56, September 9, 2017 (UTC)


 * What feature does Special:WikiActivity have that Special:RecentChanges doesn't? You can see who edited what page, exactly at what time, you can dropdown to see individual edits (with descriptions) on each page which you can compare to both a previous version and the current version of a page. I won't comment on reader-friendlyness, since I think this is a subjective choice as to which one is more convenient. LittlePeng9 (talk) 18:21, September 9, 2017 (UTC)
 * Maybe it's due to my own ignorance. I don't know. I just feel lost on that page, when compared to the "Special:WikiActivity" page. PsiCubed2 (talk) 11:31, September 10, 2017 (UTC)

Well, the madness has stopped... finally. Look at the badge leadership board and the two top scores. Look at the difference between those two top scores. Funny coincidence, huh? We've been through hell for nearly a month just for that. Way to go, Cloudy. PsiCubed2 (talk) 11:25, September 17, 2017 (UTC)
 * But later I will start another thing. There will be again large amount of edits. &#123;hyp/^,cos&#125; (talk) 13:07, September 17, 2017 (UTC)

SAN vs Catching function
Will you do a comparison of SAN with Catching function? As I understood from this blog, for relatively small values the Catching function has the following comparisons:

$$C(\varepsilon_0) = KPI = \{1,,2,,2\}$$

$$C(\Omega) = \psi(\psi_{I_\omega}(0)) = \{1,,1,,1,2\}$$

$$C(\Omega^2) = \psi(\psi_{I(1,\omega)}(0)) = \{1,,1,,1,,1,2\}$$

$$C(\Omega^\omega) = \psi(\psi_{I(\omega,0)}(0)) = \{1\{2''\}2\}$$

$$C(\Omega^{\varepsilon_0}) = KPM = \{1,,2\{1,,2''\}2\}$$

$$C(\Omega^\Omega) = \psi(\psi_{\chi(M\omega)}(0)) = \{1\{1,,2''\}1,2\}$$

$$C(\Omega^{\Omega^\omega}) = \psi_K(\omega) = \{1\{1,,1,2''\}2\}$$

$$C(\Omega^{\Omega^{\varepsilon_0}}) = KP+\Pi_3 = \{1,,2\{1,,1,,2''\}2\}$$

$$C(\Omega^{\Omega^{\Omega^{\varepsilon_0}}}) = KP+\Pi_4 = \{1,,2\{1\{1,,2\}2\}2\}$$

$$C(\Omega^{\Omega^{\Omega^{\Omega^{\varepsilon_0}}}}) = KP+\Pi_5 = \{1,,2\{1\{1,,1,,2\}2\}2\}$$

$$C(\varepsilon_{\Omega+1}) = KP+\Pi_n = \{1\{1',,2''\}2\}$$

And, further, if I correctly understand:

$$C(C_1(\Omega)\omega) \approx \{1,,,1,,,2\}$$

$$C(C_1(\Omega)\Omega\omega) \approx \{1,,,1,,,1,,,2\}$$

Is it so? What about the big values? What comparisons will have with SAN? For example, such values of the function: $$C(\Omega_2), C(\Omega_\omega) , C(\psi(\psi_I(0))) , C(KPI) , C(KPM) , C(KP+\Pi_3) , C(KP+\Pi_3) , C(KP+\Pi_n)$$

Will it be stronger, say, $$C(KP+\Pi_n)$$ than SAN? And if not, what expansion of SAN will beat $$C(KP+\Pi_n)$$? DAN, NDAN, WDEN, mWDEN, WDmEN or pDDN? Scorcher007 (talk) 10:27, September 2, 2017 (UTC)
 * Nowadays, I've found the catching hierarchy (from \(C_1\) on) ill-defined, because it depends on very high value of \(\psi\) function, but \(\psi\) functions have not been defined so high. &#123;hyp/^,cos&#125; (talk) 11:39, September 2, 2017 (UTC)


 * But are such values of the function still defined?


 * $$C(\psi_1(\Omega_2))$$


 * $$C(\psi_1(\Omega_\omega))$$


 * $$C(\psi_1(\psi_I(0)))$$


 * $$C(\psi_1(\psi_I(\varepsilon_{I+1})))$$


 * $$C(\psi_1(\psi_{\chi(M)}(0)))$$


 * $$C(\psi_1(\psi_{\chi(M)}(\varepsilon_{M+1})))$$


 * $$C(\psi_1(\psi_K(K)))$$


 * $$C(\psi_1(\psi_K(\varepsilon_{K+1})))$$


 * Scorcher007 (talk) 12:30, September 2, 2017 (UTC)
 * Yes. And one more detail: the last two ordinals you listed are equal, and both equal \(C(\psi_1(K))\). &#123;hyp/^,cos&#125; (talk) 12:44, September 2, 2017 (UTC)
 * Based on the fact that $$C(\varepsilon_{\Omega+1}) = KP+\Pi_n = \{1\{1',,2''\}2\}$$ and that we can work with $$C(\psi_1(...))$$ we can define such expression: $$C(\psi_1(KP+\Pi_n))$$


 * Can you give a comparison with SAN? Scorcher007 (talk) 09:42, October 1, 2017 (UTC)
 * Edit: Clear definition only goes up to \(C(\psi_1(\Omega_{\Omega_{\cdots_{\Omega_{K+1}}}}))\), where \(\Omega_{\Omega_{\cdots_{\Omega_{K+1}}}}\) means the omega-fixed-point after K. &#123;hyp/^,cos&#125; (talk) 13:43, October 1, 2017 (UTC)


 * If I understand the work of DAN correctly, then:


 * PDAN


 * {1,,2} = {1&#39;2} = {1,,,2}


 * {2,,2} = {1{2&#39;}2}2}


 * {1,,1,2}


 * {1,,1,,2}


 * {1,,1,,1,,1,,1,,1 ... 2} = {1{2,,}2} = {2,,,2}


 * {1{1,,2,,}2}


 * {1{1{1,,2,,}2,,}2}


 * SDAN


 * {1{1`,,2,,}2} = {1{1{1{1 ... {1,,2,,} ... 2,,}2,,:}2,,}2} = {1,,`2,,}


 * {1{1`,,1`,,1`,,1`,,1`,, ... 2,,}2} = {1{1{2`,,}2,,:}2}


 * {1{1{1``,,2`,,}2,,}2} = {1,,`3,,}


 * {1{1{1``,,1``,,1``,,1``,,1``,, ... 2`,,}2,,}2} = {1{1{1{2``,,}2`,,}2,,}2}


 * {1{1{1{1```,,2``,,}2`,,}2,,}2} = {1,,`4,,}


 * {1,,`1,,`2,,}


 * {1,,`1,,`1,,`1,,`1,,` ... 2,,} = {1{2,,`}2,,} = :{1,,,3}


 * {1{1,,`1,,`1,,`1,,`1,,` ... 2,,`}2,,} = {1{1{1`,,2,,`}2,,`}2,,}


 * then:


 * {1{1``,,2,,`}2,,} = {1{1{1{1 ... {1`,,2,,`} ... 2,,`}2,,`}2,,`}2,,} = {1,,``2,,`}


 * {1{1``,,1``,,1``,,1``,,1``,, ... 2,,`}2,,} = {1{1{2`,,`}2,,`}2,,}


 * {1{1{1``,,2`,,`}2,,`}2,,} = {1,,``3,,`}


 * {1{1{1``,,1``,,1``,,1``,,1``,, ... 2`,,`}2,,`}2,,} = {1{1{1{2``,,`}2`,,`}2,,`}2,,}


 * {1{1{1{1```,,2``,,`}2`,,`}2,,`}2,,} = {1,,``4,,`}


 * {1,,``1,,``2,,`}


 * {1,,``1,,``1,,``1,,``1,,`` ... 2,,`} = {1{2,,``}2,,`} = {1,,,4}


 * then:


 * 2,,``` = {1,,,5}


 * 2,,```` = {1,,,6}


 * 2,,````` = {1,,,7}


 * {1,,,1,2}


 * {1,,,1,,2}


 * {1,,,1,,,2}


 * The path from {1{1`,,2,,}2} to {1{1``,,2,,`}2,,} is very similar to the path from {1,,2} to {1{1`,,2,,}2}


 * If C(&epsilon;&Omega;+1) = C(&psi;1(&Omega;2)) ~ {1{1`,,2,,}2}


 * Then my guess:


 * C(&psi;1(&Omega;2)) ~ {1{1`,,2,,}2} = {1,,`2,,}


 * C(&psi;1(&psi;2(&Omega;3))) ~ {1{1{1``,,2`,,}2,,}2} = {1,,`3,,}


 * C(&psi;1(&psi;2(&psi;3(&Omega;4)))) ~ {1{1{1{1```,,2``,,}2`,,}2,,}2} = {1,,`4,,}


 * C(&psi;1(&Omega;&omega;)) ~ {1,,`1,2,,}


 * C(&psi;1(&psi;I(0))) ~ {1,,`1,,2,,}


 * C(&psi;1(&psi;I(&omega;,0)(0)))) ~ {1,,`1{2,,}2,,}


 * C(&psi;1(M))) ~ {1,,`1{1,,2,,}2,,}


 * C(&psi;1(K))) ~ {1,,`1{1,,1,,2,,}2,,}


 * If allow ill-defined way, then


 * C(&psi;1(П12))) ~ {1,,`1{1{1,,2,,}2,,}2,,}


 * C(&psi;1(П13))) ~ {1,,`1{1{1,,1,,2,,}2,,}2,,}


 * C(&psi;1(П1n))) ~ {1,,`1{1`,,2,,}2,,}

Scorcher007 (talk) 08:11, October 2, 2017 (UTC)
 * that far from {1,,`1,,`2,,}
 * Something incorrect: {1,,`1,,`1,,`1,,`1,,` ... 2,,} = {1{2,,`}2,,} but not {1,,,3}
 * and {1{1,,`1,,`1,,`1,,`1,,` ... 2,,`}2,,} is not {1{1{1`,,2,,`}2,,`}2,,} - it's {1{1{2,,`}2,,`}2,,}
 * I didn't use separators like `,,` (with grave accents both before and after the double comma), what do you mean by those separators?

I have compared catching function with R function, and the m-ple comma in R function approximately corresponds to the (m+1)-ple comma in my array notation. We can get "SAN vs. catching function" from those relationship. From "\(C(\psi_1(\psi_I(0)))\) corresponds to {0{0{0**}0{0**}1*}1} in R function" we get "\(C(\psi_1(\psi_I(0)))\) corresponds to {1,,`1,,`2,,} in SAN". &#123;hyp/^,cos&#125; (talk) 10:28, October 2, 2017 (UTC)
 * `,,` I mean `,,2. if it's incorrect, then {1{1``,,1``,,1``,,1``,,1``,, ... 2,,`}2,,} = ?
 * {A,,,3} = {A,,2} = {A,,`}. Ok, my mistake. Then {1{2,,`}2,,} = {1{2,,,3}2,,}?Scorcher007 (talk) 10:59, October 2, 2017 (UTC)
 * {1{1``,,1``,,1``,,1``,,1``,, ... 2,,`}2,,} = {1{1{2``,,}2,,`}2,,} = {1{1{2{1,,,3}3,,,2}2,,,3}2,,,2}, and {1{2,,`}2,,} = {1{2,,,3}2,,} = {1{2,,,3}2,,,2}.
 * Some shorthands:
 * ,, is {1,,,2}
 * {A,,} is {A,,,2}
 * `,, is {1{1,,,3}2,,,2}
 * {A`,,} is {A{1,,,3}2,,,2}
 * ``,, is {1{1,,,3}3,,,2}
 * ```,, is {1{1,,,3}4,,,2}
 * ,,` is {1,,,3}
 * {A,,`} is {A,,,3}
 * Your `,,` is {1{1,,,,4}2,,,3}
 * Your ``,,` is {1{1,,,,4}3,,,3}
 * ,,`` is {1,,,4}
 * {A,,``} is {A,,,4}
 * ,,``` is {1,,,5}
 * The triple comma is 3-separator for { ____ }, the ,,``...`` is 2-separator for { ____ ,,``...`}, the ``...``,,``...`` is 1-separator for { ____ ``...`,,``...``} (notice the change of the amount of grave accents). &#123;hyp/^,cos&#125; (talk) 11:17, October 2, 2017 (UTC)
 * Thanks for the explanation. Here are my guess about the level of C-function if it were defined. http://lihachevss.ru/catchfunction.html If this were so, then I suppose the C(ψI(0)) could beat the NDAN. I believe that, despite its ill-definitely, the Catch-function can be used to demonstrate the level of SAN.Scorcher007 (talk) 14:50, October 4, 2017 (UTC)

The discussion about regiment's project
I've seen you talking about deleting the "article stubs". But I still don't understand this way. "Article stubs" are simple and ugly, but why not improving them instead of avoiding them (I mean, the regiment project)? Do you think there are really so much "numbers" that some of them must be deleted? &#123;hyp/^,cos&#125; (talk) 23:22, September 9, 2017 (UTC)


 * Because there's really no need to have individual articles for thousands upon thousands of numbers, when there isn't anything particularly interesting to say about them. You can say who named it, what numbers it lies between, where it lies in the FGH... But, I think it would actually be more useful to people if they were grouped into "regiments". There would be much fewer pages to navigate, and information would be more consolidated. You could have a nice table for all the numbers saying where they lie in various hierarchies.


 * For that matter, it could reduce many hours of work - I constantly see people making edits by the hundreds, making the same damn edit to hundreds of basically identical articles. Wouldn't it be better not to have to do that? But anyway, I'm not really involved in this... Deedlit11 (talk) 00:01, September 10, 2017 (UTC)
 * But, what's "anything particularly interesting"? For instance, some people may think that gugold, graatagold, greegold, gugolthra and graatagolthra are "interesting"; some may think that graatagold, greegold and graatagolthra are not because the addition of "#100" is already passed in Hyper-E range; some may even think that just the numbers starting to use new rules are "interesting", such as gugold and godgahlah. There's no clear boundary for "anything particularly interesting". &#123;hyp/^,cos&#125; (talk) 03:36, September 10, 2017 (UTC)
 * Anyone has right to create page about a "number" if he/she can say about this "number" something more than "A is equal to B using C. The term was coined by D". All pages containing something more than that form (for example table of approximation) should be saved. Only stubs containing no more than that form should be redirected to corresponding regiment's page since their existence becomes redundant.--Denis Maksudov (talk) 08:10, September 10, 2017 (UTC)
 * I agree with you. But there is some difference between "redirecting a page" and "deleting a page". For the former one, the old size-categories on it should be change; for the latter one, we don't need this because the page will be deleted. &#123;hyp/^,cos&#125; (talk) 08:18, September 10, 2017 (UTC)
 * I only redirect pages not deleting them, and, moreover, in cases if it was necessary, before redirection I changed their categories according new classification of size.--Denis Maksudov (talk) 08:32, September 10, 2017 (UTC)
 * So some people (e.g. Denis Maksudov) want redirection but some (e.g. PsiCubed2) want deletion... &#123;hyp/^,cos&#125; (talk) 08:46, September 10, 2017 (UTC)
 * I think this is just a choice of a word, because redirected pages are essentially "deleted"; the original content is no longer visible. However, the categories can still be used to visit the redirect. -- ☁ I want more clouds! ⛅ 08:49, September 10, 2017 (UTC)
 * Although redirected pages are also "not visible", remaining something in old size-categories is not good. Deleting pages doesn't remain them in old size-categories - that's the difference. &#123;hyp/^,cos&#125; (talk) 08:55, September 10, 2017 (UTC)


 * Hypcos I realized this when you explained to me For example when I was doing Monster-Giant Regiment I was changing categories of size to epsilon level before redirection of stubs.

What do you think about fate of stubs for Saibians numbers? They should be redirected to corresponding regiment's page with pre-changing of categories if it is nessessary They should be just deleted --Denis Maksudov (talk) 09:21, September 10, 2017 (UTC)


 * I don't really care if the articles are redirected or deleted. What I do think, is that these things should be discussed before you embark on massive editing projects.


 * And I also think that you continuing to ignore the requests of other people on the wiki is rude and against the spirit of community work. I find what you're doing to be very disrespectful.


 * And no, I don't think "assuming good faith" applies when people repeatedly ask someone to stop doing something and he repeatedly ignores them. You certainly aren't obliged to "obey" anybody's request, but - when doing editing on this kind of scale - you are obliged to seriously address these complaints in some fashion (and no, saying "If I go slower it will take longer" is not a serious reply).


 * At the very very least, we deserve a candid explanation to why you're doing this.


 * (and don't think - for a moment - that it's only me. Other people have complained right on this page. Just because they are too nice and polite to press the issue, does not mean that they deserve to be ignored) PsiCubed2 (talk) 11:27, September 10, 2017 (UTC)


 * BTW, even if we keep the redirects, I don't see any point of having them listed in any category - old or new. PsiCubed2 (talk) 11:37, September 10, 2017 (UTC)


 * Yes, we have three variants:1) to delete stub at all 2) to change category of size in process of redirection of a stub 3) to delete category of size in process of redirection of a stub. In all of those variants remaining nothing in old size-categories (what Hypcos want to avoid) --Denis Maksudov (talk) 14:30, September 10, 2017 (UTC)

I think I might start this again somewhere else, because I'm not sure how this discussion started on a user's talk page. Username5243 (talk) 15:58, September 10, 2017 (UTC)

List padding
So it seems you are squeezing every name anyone but me has ever come up with for a number into the lists?--L.E./12.144.5.2 15:21, September 17, 2017 (UTC)
 * Now I'm filling the subpages of List of googologisms, and making them full list of numbers. Those lists should contain all number pages, but currently they are incomplete. But only numbers shown in main pages (or their redirects) count. User pages, user blogs are still not included. &#123;hyp/^,cos&#125; (talk) 15:42, September 17, 2017 (UTC)
 * So it's an internal catalogue.Still,I dubbed the exillio-illio-illio-illion (1,000,0001,000,000 1,000,000 1,000,000 1,000,000,000,000,000,000  ) back in the 1980s and it annoys me that something a teenager with a webpage came up with is considered more worthy of mention.12.144.5.2 18:49, September 17, 2017 (UTC)

SAN vs TON
C(C(Ω23,0),0) = How far have you analyzed TON with SAN? And what are the limits of each of the parts of SAN in TON? Nishada 00:21, September 18, 2017 (UTC)
 * {1{1{1,,`1,,2,,}2,,`1{1,,`1{1,,`1,,`2,,}2,,}1,,2{1,,`1,,`2,,}2,,}1,,2} according to your analysis
 * {1,,`1{1,,`1,,`2,,}3,,} according to my approximation method

Fixed the subscript Nishada 00:22, September 18, 2017 (UTC)

superscript* Nishada 00:22, September 18, 2017 (UTC)
 * I haven't done the analysis beyond C(C(Ω23,0),0). And I still haven't found expression in TON really corresponding to the double comma or the ,,` in my array notation. &#123;hyp/^,cos&#125; (talk) 00:36, September 18, 2017 (UTC)


 * Do you still agree with your earlier statement that the limit of WDmEN is C(C(Ω22+C(Ω2+C(Ω22+1,0),0),0),0), which would be less than C(C(Ω23,0),0)? Deedlit11 (talk) 21:50, September 20, 2017 (UTC)
 * No. And I don't think TON is so strong as the "update" states. I guess its limit is still second-order arithmetic, or even lower. &#123;hyp/^,cos&#125; (talk) 22:47, September 20, 2017 (UTC)


 * I see. I would agree with this, however I don't think that is much of a detriment to the notation, because I think second-order arithmetic is ridiculously strong. Do you have any speculation on how SAN vs TON compares, beyond what you have written on SDAN vs TON, or is it all speculation at this point? Deedlit11 (talk) 23:59, September 20, 2017 (UTC)
 * The TON rule "a is (n+1)-built from below from <b iff a doesn't use ordinal above a except in the scope of an ordinal that is n-built from below from <b" means a "rising". In TON, to check whether a is n-built from below from <b, we need to find something included by the a, so it's "going inside", and we need this n times. But in the "dropping" process in SAN, we need to find something including the separator, so it's "going outside". For an n-separator, we need this n times. That's totally different, and it's hard to compare. Possibly TON might be as strong as DAN, if "going inside" and "going outside" have similar strength.
 * Taranovsky claimed that C(C(Ω23,0),0) (the limit using C0(a,b), C1(a,b) and C2(a,b)) reaches \(\Pi^1_2\text{-CA}_0\), but I don't believe that it could correspond to such a complex expression: {1{1{1,,`1,,2,,}2,,`1{1,,`1{1,,`1,,`2,,}2,,}1,,2{1,,`1,,`2,,}2,,}1,,2}. When I first attempt to analyse up to C(C(Ω23,0),0), I guessed C0(a,b) to be non-dropping separators, C1(a,b) to be 1-separators, and C2(a,b) to be 2-separators. If so, C(C(Ω23,0),0) will correspond to s(n,n{1,,,1{1,,,1{1,,,1,,,2,,}2,,}2,,}2). (But it's not so strong)
 * However, I didn't find an ordinal in TON really corresponding to the {1,,,3} (= ,,`), which is the next 2-separator after the double comma. I also didn't find what C(Ω22,0) corresponds to in SAN. Things suddenly become weaker at C(C(Ω22+C(Ω2+C(Ω22,C(Ω22,0)),0),0),0) and C(C(Ω22+C(Ω22,0),0),0), which weaken my guess about the strength of TON. &#123;hyp/^,cos&#125; (talk) 03:46, September 21, 2017 (UTC)

Buchholz Psi vs Feferman Theta vs SAN over Buchholz ordinal
Why &psi;(&Omega;&omega;+&Omega;) = &epsilon;&psi;(&Omega; &omega;)+1 ? &epsilon;&psi;(&Omega; &omega;)+1 must be equal &psi;(&psi;1(&Omega;&omega;)+&Omega;) Then &psi;(&Omega;&omega;+&Omega;) = &psi;(&psi;1(&psi;2(&psi;3(&psi;4(...(&Omega;&omega;+&Omega;)...))))) ? Or &psi;(&Omega;&omega;+&Omega;) = &psi;(&psi;1(&Omega;&omega;)+&Omega;) ?

If first, then: {2{1,,1,2}2} ~ &psi;(&Omega;&omega;)+1 = &theta;(&Omega;&omega;)+1 {1,2{1,,1,2}2} ~ &psi;(&Omega;&omega;)+&omega; = &theta;(&Omega;&omega;)+&omega; {1(1{1,,2}2}2{1,,1,2}2} ~ &psi;(&Omega;&omega;)+&epsilon;0 = &theta;(&Omega;&omega;)+&epsilon;0 {1(1{1,,1,2}2}2{1,,1,2}2} ~ &psi;(&Omega;&omega;)&times;2 = &theta;(&Omega;&omega;)&times;2 {1{1{1,,1,2}2}1,2{1,,1,2}2} ~ &psi;(&Omega;&omega;)&times;&omega; = &psi;(&psi;1(&Omega;&omega;)+1) = &theta;(&Omega;&omega;)&times;&omega; {1{1,,2}2{1,,1,2}2} ~ &epsilon;&psi;(&Omega; &omega;)+1 = &phi;(1,&psi;(&Omega;&omega;)+1) = &psi;(&psi;1(&Omega;&omega;)+&Omega;) = &epsilon;&theta;(&Omega; &omega;)+1 = &phi;(1,&theta;(&Omega;&omega;)+1) {1{1{1,,2}2,,2}2{1,,1,2}2} ~ Г&psi;(&Omega; &omega;)+1 = &psi;(&psi;1(&Omega;&omega;)+&Omega;&Omega;) = Г&theta;(&Omega; &omega;)+1 = &theta;(&Omega;,&theta;(&Omega;&omega;)+1) {1{1,,3}2{1,,1,2}2} ~ &psi;(&psi;1(&Omega;&omega;)+&psi;1(&Omega;2)) = &theta;(&epsilon;&Omega;+1,&theta;(&Omega;&omega;)+1) {1{1,,4}2{1,,1,2}2} ~ &psi;(&psi;1(&Omega;&omega;)+&psi;1(&Omega;3)) = &theta;(&epsilon;&Omega; 2+1 ,&theta;(&Omega;&omega;)+1) {1{1,,1,2}3} ~ &psi;(&psi;1(&Omega;&omega;)+&psi;1(&Omega;&omega;)) = &theta;(&Omega;&omega;,1) {1{1,,2}2{1,,1,2}3} ~ &psi;(&epsilon;&psi; 1(&Omega;&omega;)+1 ) = &psi;(&psi;1(&Omega;&omega;+&Omega;)) = &phi;(1,&theta;(&Omega;&omega;,1)+1)) {1{1{1,,2}2,,2}2{1,,1,2}3} ~ &psi;(Г&psi; 1(&Omega;&omega;)+1 ) = &psi;(&psi;1(&Omega;&omega;+&Omega;&Omega;)) = &theta;(&Omega;,&theta;(&Omega;&omega;,1)+1)) {1{1,,3}2{1,,1,2}3} ~ &psi;(&psi;1(&Omega;&omega;+&psi;1(&Omega;2))) = &theta;(&epsilon;&Omega;+1,&theta;(&Omega;&omega;,1)+1) {1{1,,1,2}4} ~ &psi;(&psi;1(&Omega;&omega;+&psi;1(&Omega;&omega;))) = &theta;(&Omega;&omega;,2) {1{1,,1,2}1,2} ~ &psi;(&psi;1(&Omega;&omega;+&psi;1(&Omega;&omega;)&times;&omega;)) = &psi;(&psi;1(&Omega;&omega;+&psi;1(&Omega;&omega;+1))) = &theta;(&Omega;&omega;,&omega;) {1{1,,1,2}1{1,,2}2} ~ &psi;(&psi;1(&Omega;&omega;+&epsilon;&psi; 1(&Omega;&omega;)+1 )) = &psi;(&psi;1(&Omega;&omega;+&psi;1(&Omega;&omega;+&Omega;)) = &theta;(&Omega;&omega;+1) {1{1,,2}2{1{1,,3}2{1,,1,2}2,,2}2} ~ &psi;(&psi;1(&psi;2(&Omega;&omega;)+&Omega;)) = &psi;(&psi;1(&epsilon;&psi; 2(&Omega;&omega;)+1 ) {1{1,,2}2{1{1,,4}2{1,,1,2}2,,2}2} ~ &psi;(&psi;1(&psi;2(&psi;3(&Omega;&omega;)+&Omega;))) = &psi;(&psi;1(&psi;2(&epsilon;&psi; 2(&Omega;&omega;)+1 )) {1{1,,1,2}2,,1,2} ~ &psi;(&psi;1(&psi;2(&psi;3(&psi;4(...(&Omega;&omega;+&Omega;)...))))) = &psi;(&Omega;&omega;+&Omega;) But {1{1,,1,2}2,,1,2} is &psi;(&Omega;&omega;&times;2) = &theta;(&Omega;&omega;&times;2) It means that  &psi;(&Omega;&omega;+&Omega;) = &psi;(&psi;1(&Omega;&omega;)+&Omega;) ; &psi;(&Omega;&omega;+&Omega;2) = &psi;(&psi;1(&psi;2(&Omega;&omega;)+&Omega;2)) ; &psi;(&Omega;&omega;+&Omega;3) = &psi;(&psi;1(&psi;2(&psi;3(&Omega;&omega;)+&Omega;3)))  This is true?  Scorcher007 (talk) 05:05, November 20, 2017 (UTC)
 * \(\psi_1(\Omega_\omega)\) is an ordinal between \(\psi_1(\Omega_2)\) and \(\Omega_2\).
 * Now consider \(\psi(\psi_1(\Omega_2)+1)\). Ordinals in \(C(\psi_1(\Omega_2)+1)\) are constructed using addition and application of \(\psi_\nu(\xi)\) (with \(\xi<\psi_1(\Omega_2)+1\)) from 0. But we can't construct \(\psi_1(\Omega_2)\) from 0 because \(\Omega_2\ge\psi_1(\Omega_2)+1\), then we can't construct \(\psi(\psi_1(\Omega_2))\) (a.k.a. BHO). So it's still \(\psi(\psi_1(\Omega_2)+1)=\psi(\psi_1(\Omega_2))\).
 * And further, \(\psi(\alpha)=\psi(\psi_1(\Omega_2))\) for \(\psi_1(\Omega_2)\le\alpha\le\Omega_2\) - the \(\psi\) function gets stuck. So \(\psi(\psi_1(\Omega_\omega))=\psi(\psi_1(\Omega_\omega)+\Omega)=BHO\).
 * \(\psi(\Omega_2+1)\) gets unstuck because \(\Omega_2<\Omega_2+1\), then we can apply \(\psi_2(0)=\Omega_2\), \(\psi(\Omega_2)=BHO\), and addition using BHO.
 * \(\psi(\psi_2(\Omega_\omega))=\psi(\psi_2(\Omega_\omega)+\Omega)=\psi(\Omega_3)\), \(\psi_1(\psi_2(\Omega_\omega))=\psi_1(\psi_2(\Omega_\omega)+\Omega)=\psi_1(\Omega_3)\), \(\psi(\psi_3(\Omega_\omega))=\psi(\psi_3(\Omega_\omega)+\Omega)=\psi(\Omega_4)\), \(\psi_2(\psi_3(\Omega_\omega))=\psi_2(\psi_3(\Omega_\omega)+\Omega)=\psi_2(\Omega_4)\), and such expressions get stuck. Only until \(\psi(\Omega_\omega+1)\) can it get unstuck to \(\psi(\Omega_\omega)\times\omega\), and \(\psi(\Omega_\omega+\Omega)=\varepsilon_{\psi(\Omega_\omega)+1}\). &#123;hyp/^,cos&#125; (talk) 09:31, November 20, 2017 (UTC)
 * It means that it's work with singular
 * &psi;(&Omega;&omega;2+1) = &psi;(&Omega;&omega;2)&times;&omega;
 * &psi;(&Omega;&omega;2+&Omega;) = &epsilon;&psi;(&Omega; &omega;2)+1
 * &psi;(&psi;I(0)+1) = &psi;(&psi;I(0))&times;&omega;
 * &psi;(&psi;I(0)+&Omega;) = &epsilon;&psi;(&psi; I(0))+1
 * And with regular too:
 * &psi;(&psi;1(&Omega;2+1)) = &psi;(&psi;1(&Omega;2))&times;&omega; ≠ &psi;(&psi;1(&Omega;2)+1)
 * &psi;(&psi;1(&Omega;2+&Omega;)) = &epsilon;&psi;(&psi; 1(&Omega;2))+1 ≠ &psi;(&psi;1(&Omega;2)+&Omega;)
 * &psi;(&psi;&omega;(&Omega;&omega;+1+1)) = &psi;(&Omega;&omega;+1)&times;&omega; ≠ &psi;(&psi;&omega;(&Omega;&omega;+1)+1)
 * &psi;(&psi;&omega;(&Omega;&omega;+1+&Omega;)) = &epsilon;&psi;(&Omega; &omega;+1)+1 ≠ &psi;(&psi;&omega;(&Omega;&omega;+1)+&Omega;)
 * ?? Scorcher007 (talk) 14:07, November 20, 2017 (UTC)
 * Your last four lines are wrong. \(\psi(\psi_1(\Omega_2+1))=\psi(\psi_1(\Omega_2)+1)=\psi(\psi_1(\Omega_2+\Omega))=\psi(\psi_1(\Omega_2)+\Omega)=\psi(\Omega_2)\), and \(\psi(\psi_\omega(\Omega_{\omega+1}+1))=\psi(\psi_\omega(\Omega_{\omega+1})+1)=\psi(\psi_\omega(\Omega_{\omega+1}+\Omega))=\psi(\psi_\omega(\Omega_{\omega+1})+\Omega)=\psi(\Omega_{\omega+1})\). &#123;hyp/^,cos&#125; (talk) 16:21, November 20, 2017 (UTC)


 * Thanks for the explanation. And one more question about OCF.
 * {1{1{1{1,,2,,}2}2,,}3} ~ ψ(ψI(1,0,0)(0)) = ψ(ψχ(MM)(0))
 * {1{1{1{1,,2,,}2}2,,}1,,2} ~ ψ(I(1,0,0)) = ψ(ψχ(MM)(χ(MM))) = ψ(χ(MM)) = ψ(MM)
 * {1{1{1{1,,2,,}2}3,,}3} ~ ψ(ψI(1,0,0,0)(0)) = ψ(ψχ(MM 2 )(0))
 * {1{1{1{1,,2,,}2}3,,}1,,2} ~ ψ(I(1,0,0,0)) = ψ(ψχ(MM 2 )(χ(MM 2 ))) = ψ(χ(MM 2 )) = ψ(MM 2 )
 * {1{1{1{1,,2,,}2}4,,}3} ~ ψ(ψI(1,0,0,0,0)(0)) = ψ(ψχ(MM 2 )(0))
 * {1{1{1{1,,2,,}2}4,,}1,,2} ~ ψ(I(1,0,0,0,0)) = ψ(ψχ(MM 3 )(χ(MM 3 ))) = ψ(χ(MM 3 )) = ψ(MM 3 )


 * {1{1{1{1,,2,,}2}1,2,,}3} ~ ψ(MM &omega; )
 * Does this mean that
 * ψ(ψI(1,0,0,0,...)(0)) = ψ(ψχ(MM &omega; )(0)) = ψ(I(1,0,0,0,...)) = ψ(ψχ(MM &omega; )(χ(MM &omega; ))) = ψ(χ(MM &omega; )) = ψ(MM &omega; ) ? —Preceding unsigned comment added by Scorcher007 (talk • contribs) 10:54, November 21, 2017 (UTC)
 * No. The notation using inaccessible cardinals (the first notation) and the notation using weakly Mahlo cardinal (the second notation) are different notations. Although \(\chi(M^M)=I(1,0,0)\), \(\psi(\chi(M^M))\) (in the second notation) is not \(\psi(I(1,0,0))\) (in the first notation). Note that the second notation gets stuck at \(\Omega_{\Omega_{\Omega_\cdots}}\le\alpha\le M\), and \(\psi(\chi(M^M))=\psi(\chi(M^{M^\omega}))=\psi(\psi_{\chi(M^{M^\omega})}(0))=\psi(\psi_{\chi(M^{M^\omega})}(\chi(M^{M^\omega})))\) \(=\psi(\psi_{\chi(M^M)}(0))=\psi(\psi_{\chi(M^M)}(\chi(M^M)))=\psi(M)\) - the limit of \(\Pi_1^1\text{ - TR}_0\).
 * But \(\psi(M^M)\) (in the second notation) equals \(\psi(\psi_{I(1,0,0)}(0))\) (in the first notation), also \(\psi(M^{M^\omega})\) (in the second notation) equals the limit of \(\psi(I(1,0,0,\cdots))\) or \(\psi(\psi_{I(1,0,0,\cdots)}(0))\) (in the first notation). &#123;hyp/^,cos&#125; (talk) 00:59, November 22, 2017 (UTC)
 * Ok, but if \(\psi(\psi_{I(1,0,0)}(0))\) = \(\psi(M^{M})\) what is \(\psi(I(1,0,0))\) = ?
 * \(\psi(M^{M}+\chi(M^M))\) ?Scorcher007 (talk) 04:51, November 22, 2017 (UTC)
 * Yes. \(\psi(I(1,0,0))\) in the first notation equals \(\psi(M^M+\chi(M^M))\) in the second notation. &#123;hyp/^,cos&#125; (talk) 08:34, November 22, 2017 (UTC)