Count with PSS expressions of limited length

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PsiCubed2
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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Mon Jun 03, 2019 4:13 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = {2}{1}14 = 28×228 = 7 516 192 768 ≈ 7.5×109 ≈ E9.88

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Mon Jun 03, 2019 6:17 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)[10] = {2}{1}{1}10 = 40 * 240 ~ 4.4 * 1013 ~ EE1.135

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Mon Jun 03, 2019 7:47 pm UTC

EE1.135

You mean F2.055 :wink:

At any rate, here are the full decimal form of the last few numbers:
1 048 576
20 971 520
92 274 688
402 653 184
1 744 830 464
7 516 192 768
43 980 465 111 040

(of-course I won't be able to keep that going for much longer :))

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)[10] = {2}{1}{1}{0}10 = 44×244 = 774 056 185 954 304 ≈ 7.74×1014 ≈ E14.89 ≈ F2.069

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Re: Count with PSS expressions of limited length

Postby Sabrar » Mon Jun 03, 2019 8:37 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = {2}{1}{1}12 = 48 * 248 ~ 1.35×1016 ~ F2.082

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Mon Jun 03, 2019 9:10 pm UTC

That was 13 510 798 882 111 488

And now:

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)[10] = {2}{1}{1}{1}10 = {2}80 = 80×280 =
= 96 714 065 569 170 333 976 494 080 ≈ 9.67×1025 ≈ E25.985 ≈ F2.151

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Tue Jun 04, 2019 5:52 am UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)[10] = {2}{2}10 = 10240 * 210240 ~ 3.61 * 103086 ~ EEE0.543 = F2.543

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Tue Jun 04, 2019 2:23 pm UTC

What? No decimal expansion? It's just:

360957072599898384090605039595548830355368154562216663658407757882686159634992633498094902235890317951559353911247855349486172666633657483467656588012709208748182598356636384334399545815230549708652121305780944094607687051610452806545332430422287685702892997010122003214335508082605878763758507043556411021131882440599131490156862577439953257503054733504014811497383642697438777679619716430794701581449591679689093164315539356126569695803705455174307071116910866061066273442321577547102316976919997191759869752309980772581503326976891743409084980526800964685721621214255852907801943482545725312697422823198175086555263549935207147167916759294630345216830044858194310497469585643388793214920683723361413054467883378587623186830388026709606288329594972154022376673485434803813697555172939568214923110437549731129443639611126859852738744989346253589445842275335089709057738999107362704933446827567946264837762487245074658161789298542273863070037238565810037387697951019945803617801257085790876437756594716845933151615975229353907304463152808968984861025962180160659334221678013914000051722776924934944720336623246536410349671563495582840474975535382609467326519059383049608782570197824570156088538044006148380691681254757288694690489481905743711844181842852717743112266019072319762196328421011040303954345027726782941861882418169484142610931473711982979105056933005658672562620510427959810381446455477346563284940509928834430832475909325288857376410879046828625223279640651753897335374339878482047069769788820613993720652178330715228866933918010269286022686582658559288492858064139725178753096222327775043137725961893906665695825694925440377077920743638601960837791117641595877476961307332573071625842682355292133107640045483246956029669323559670280750415727846152425844774850423021498101974337867585724795113066625866510161810993557764641190807624788591668574563114300977843921380910589544188835912484131356109086668225057069880775806656833318513865539755205915006035377306676434213935959195464072399104805532814456133113489977561952607282641277077600317528478934753582181943660217291661472783057094767504962805464905449619694327203383583442885922628270574456545002062883179784228297808225974006038595728828620619879518842170350704751184638636449217039082062288188714696175654411105225993151788599812851415654264842837422304284010164509705982693049731471323109170157173297417942125904467996480337653121462763247430028011322300406163756468357136334325433159782564464868518326705844968870557751649823818717425250349192461845919023290072836614736734772217186495720772586702700364156747687912174603961965388970385558595245890211448129278043639109041815070649122062005392558094379119925775750476208325991144162271899491238942998087295383317029962614619645583924749228199962075775515843940323113975912791834585885958615365879524094710625916041626044389326756510449876143147527025058780988709896628161732334426568002262550585352954698777482484398843216975680153506461202889572423805010320468864660836910084980649358090891381521055951158218869435922794230497899829410896991347525797191088330506240 :mrgreen:

But let's stop with that before it gets out of hand...

(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)[10] = {2}{2}{0}10 = {2}(11×211) = {2}22528 = 22528×222528 = 1022528×log(2)+log(22528) ≈ 106786 = E6786 ≈ F2.583

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Tue Jun 04, 2019 6:07 pm UTC

I leave the crazy stuff to you. :D

(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(0,0)[10] = {2}{2}12 = 49152 * 249152 ~ 8.3 * 1014800 ~ EEE0,62 = F2.62

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Tue Jun 04, 2019 6:35 pm UTC

I leave the crazy stuff to you. :D


Good call :mrgreen:


(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(0,0)(0,0)[10] = {2}{2}{0}{0}{0} = {2}(13×213) = {2}106496 = 106496×2106496 = 10106496×log(2)+log(106496) ≈ 1032063.52 ≈ 3.3×1032063 ≈ E32064 ≈ EEE0.654 = F2.654

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Re: Count with PSS expressions of limited length

Postby Sabrar » Tue Jun 04, 2019 6:56 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(1,0)[10] = {2}{2}20 = 20971520 * 220971520 ~ 7.876 * 106313063 ~ EEE0.83 = F2.83

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Tue Jun 04, 2019 7:03 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)[10] = {2}{2}22 ≈ 1027777457 ≈ F2.87

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Re: Count with PSS expressions of limited length

Postby Sabrar » Tue Jun 04, 2019 7:32 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)[10] = {2}{2}{2}10 ~ {2}(3.61 * 103086) ~ EE3086 ~ F3.543

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Tue Jun 04, 2019 9:53 pm UTC

We can use Robert's Munafo's power tower notation to write the approximations now:

N pt x = a power tower of N 10's with an x on top.

(0,0)(1,0)(1,0)(1,0)[10]= {3}10 = {2}1010 ≈ 9pt3086
(that's 10^10^10^10^10^10^10^10^10^3086)

This notation should remain useful for the next dozen entries or so.

(the above can also be approximated as F10.54 or as G2.0043, the latter being the standard [letter]x form with 2<x<10)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 4:22 am UTC

(0,0)(1,0)(1,0)(1,0)(0,0)[10] = {3}11 = {2}11[11] ~ 10pt6786 ~ F11.58 ~ G2.01

I need to re-familiarize myself with the letter notation, that took me way too long to double-check.

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 9:38 am UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(0,0)[10] = {3}12 = {2}1212 ≈ 11pt14801
(≈ F12.62 ≈ FEE0.0418 ≈ FF1.0418 = FFE0.0178 = G2.0178)
Last edited by PsiCubed2 on Wed Jun 05, 2019 10:54 am UTC, edited 1 time in total.

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 10:37 am UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)[10] = {3}13 = {2}13[13] ~ 12pt32064 ~ F13.65 ~ G2.023

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 10:56 am UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = {3}14 = {2}1414 ≈ 13pt69054
(≈ F14.68 ≈ FEE0.0670 ≈ FF1.0670 = FFE0.0282 = G2.0282)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 11:37 am UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = {3}15 = {2}15[15] ~ 14pt147968 ~ F15.71 ~ G2.0325

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 1:03 pm UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)[10] = {3}20 = {2}2020 ≈ 19pt6313063
(≈ F20.83 ≈ FEE0.1202 ≈ FF1.1202 = FFE0.0493 = G2.0493)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 2:28 pm UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)[10] = {3}22 = {2}22[22] ~ 21pt27777457 ~ F22.87 ~ G2.054

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 2:38 pm UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = {3}24 = {2}2424 ≈ 23pt121 210 694)
(≈ F24.91 ≈ FEE0.1450 = FF1.1450 = FFE0.059 = G2.059)
(≈ F25 = 10↑↑25)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 3:01 pm UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)[10] = {3}26 = {2}26[26] ~ 25pt525246316 ~ F26.94 ~ G2.063

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 3:38 pm UTC

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)[10] = {3}40 = {2}4040 ≈ 39pt13239439221689 ≈ 40pt13
(≈ F41.05 ≈ FEE0.208 = FF1.208 ≈ FFE0.082 = G2.082)
(≈ F41 = 10↑↑41)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 6:13 pm UTC

If I didn't screw it up

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)[10] = {3}44 = {2}44[44] ~ 43pt(2.33*1014) ~ F45.06 ~ G2.086

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 7:16 pm UTC

Looks good to me.

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)[10] = {3}{2}10 = {3}{2}10 ≈ {3}10240 ≈ 10240pt3086
(≈ F10241.54 ≈ FE4.01 ≈ FEE0.603 = FF1.603 ≈ G2.205)
(≈ F10242 = 10↑↑10242)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Wed Jun 05, 2019 7:38 pm UTC

Help me out please. How is this step calculated?

{3}10240 ≈ 10240pt3086

Are you just substituting 10 as the base instead of 2 in the chain?

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Wed Jun 05, 2019 9:48 pm UTC

It's exactly the same calculation we've done for the previous dozen numbers or so:

{3}44 = {2}4444 ≈ 43pt(2.33×1014) ≈ 43pt(1014) = 44pt14

{3}10240 = {2}1024010240 ≈ 10239pt103186 = 10240pt3186

The trickiest part in both calculations, is the second step. But since you clearly know how to do it for n=44, I'm not sure why you're having a hard time with n=10240.

At any rate, here is a general approximation for {2}kn when k>2,n>3:

{2}kn ≈ (k-1)pt(n×2n×log(2)+log(n)+n×log(2)+log(log(2)))

For k=3, the righthand formula is simply the result of calculating {2}{2}{2}n the hard way and using the rules of logarithms to convert all the 2's into 10's.

For k>3, we use the fact that for a large enough n (say n>F4), 10n is a very good approximation of {2}n=n×2n. Thanks to this fact, the expression to the right of the "pt" (in the approximation) doesn't depend on k.

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Re: Count with PSS expressions of limited length

Postby Sabrar » Thu Jun 06, 2019 7:34 am UTC

PsiCubed2 wrote:But since you clearly know how to do it for n=44, I'm not sure why you're having a hard time with n=10240.
Because I'm lazy and use wolfram alpha and one of the steps was too much for it.

PsiCubed2 wrote:For k>3, we use the fact that for a large enough n (say n>F4), 10n is a very good approximation of {2}n=n×2n.
That's what I guessed, thanks for confirming.

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)[10] = {3}22528 = {2}22528[22528] ~ 22528pt6785.435 ~ F22529.58 ~ G2.2145

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Thu Jun 06, 2019 12:03 pm UTC

Well, I'm even lazier.

I no longer do any calculations at all. I just take a peek at the previous values and adjust them as needed.

For example, {3}10240 and {3}10 are both equal to {2}....{2}10 (with 10241 and 10 {2}'s respectively). We've already calculated that {3}10 is about 9pt3086. So we can write immediately, without any further calculation, that {3}10240 = 10240pt3086.

Using this trick, I'll do the next one:
(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(0,0)[10] = {3}{2}12

Looking up the thread, I see that:
(a) {2}12 = 49152
(b) {3}12 ≈ 11pt14801 ≈ F12.62

So I can write immediately:

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(0,0)[10] = {3}49152 = 49152pt14801 ≈ F49153.62
(≈G2.223, which is the only point I've used a calculator)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Thu Jun 06, 2019 12:32 pm UTC

Okay, let me try that one:

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(1,0)[10] = {3}{2}20 ~ 20971520pt6313063 ~ F20971521.83 ~ G2.27

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Thu Jun 06, 2019 1:02 pm UTC

Yup.

Now for a pretty big step:
(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)(1,0)[10] = {3}{3}10 ≈ {3}(9pt3086) ≈ 10↑↑(9pt3086) = F(9pt3086) = FF10.54 ≈ FF10 = G3.00
( ≈ 10↑↑↑3)

And I propose another way to write the numbers form now on:

[ a,b ; x ] = FaEbx with 0<=a,b<=8 integers and 10<=x<1010

Any number between 10 and G10 can be written uniquely as such an expression. Note that:

[ 0,b ; n ] = b pt n
[ 1,0 ; n ] = n pt 1 = Fn

So the previous number could be written as [ 1,0 ; 20971521.83 ]

FF10.54 = [ 2,0 ; 10.54 ]

(funny how we've jumped entirely over all the numbers of the form [ 1,b ; n ] for 1<=b<=8)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Thu Jun 06, 2019 1:22 pm UTC

Next step's gonna be even bigger, no? No room left at {3}, you have to move to {4} level.

(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(1,0)(1,0)(0,0)[10] = {3}{3}11 ~ F({3}11) ~ FF11.58 = [ 2,0 ; 11.58 ] (~ G3.0115)

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Thu Jun 06, 2019 1:43 pm UTC

Exactly.

We already need to add another number to our approximation array:

[ a,b,c ; x ] = GaFbEcx

(0,0)(1,0)(1,0)(1,0)(1,0)[10] = {4}10 = {3}1010 = {3}9{3}10 ≈ F9({3)}10 ≈ F9F10.54 = F1010.54 ≈ F111.01 ≈ G11.00 = 10↑↑↑11 = [ 1,0,0 ; 11 ]
( ≈ H2.003)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Thu Jun 06, 2019 2:01 pm UTC

(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)[10] = {4}11 = {3}1111 ~ F10({3}11) ~ F1111.58 ~ G12.0115 ~ [ 1,0,0 ; 12 ]

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Thu Jun 06, 2019 3:36 pm UTC

(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)(0,0)[10] = {4}12 ≈ G13 = [ 1,0,0 ; 13 ]

Kinda boring to write the numbers like this... How about allowing the entries of the approximation array to be higher than 8? Then we'll get:

{4}12 ≈ F11({3}12) ≈ F11(11pt14801) = [ 11,11 ; 14801 ]

And the previous few numbers in the same format:
{3}{2}{1}10 ≈ 20971520pt6313063 = [ 20971520 ; 6313063 ]
{3}{3}10 ≈ [ 1,9 ; 3086 ]
{3}{3}{0}10 ≈ [ 1,10 ; 6785 ]
{4}10 ≈ [ 9,9 ; 3086 ]
{4}11 ≈ [ 10,10 ; 6785 ]

And it looks like in general:
{4}n ≈ [ n-1,n-1 ; (the value after the 'pt' we've shamelessly stolen from the previous {3}n calculation) ] :mrgreen:

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Re: Count with PSS expressions of limited length

Postby Sabrar » Thu Jun 06, 2019 5:47 pm UTC

But doesn't that defeat the purpose of having a concise and unique representation?

(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)[10] = {4}13 ~ [ 12,12 ; 32064 ]

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Thu Jun 06, 2019 10:04 pm UTC

(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = {4}14 ~ [ 13,13 ; 69054 ] ~ [ 1,0,0 ; 15 ]

Hmmm... You're right.

To make this variation unique, we need to define an exact cut-off point at which to stop expanding the expression. How about this:

1. We the write the number as <some letter>x for 2<x<10.
2. We expand the number with the usual rules of letter notation.
3. We stop the expansion when either of the following two things happen:
(a) The expansion was completed (i.e. no letters remained in the expression)
(b) The last letter is "E" and the number after it is bigger than 10.
4. We convert the resulting expression into the [ a,b,c,...,n ; x ] format, and that's our new standard form.

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Re: Count with PSS expressions of limited length

Postby username5243 » Thu Jun 06, 2019 10:36 pm UTC

I semi-arbitrarily decide to check in here and see that this exists...

(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)(1,0)[10] = {4}{1}10 = {4}20 ~ G21 ~ [ 1,0,0 ; 21 ]
This is a signature, in case you didn't notice.

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Re: Count with PSS expressions of limited length

Postby Nayuta_Ito » Fri Jun 07, 2019 1:47 am UTC

It seems still OK to post here (in order not to post on a dead ten-year-old thread)

(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)[10] = {4}{1}11 = {4}22 ~ G23 ~ [ 1,0,0 ; 23 ]

I wonder how much will this "sweet time" continues.

PsiCubed2
Posts: 12
Joined: Tue Apr 02, 2019 8:55 pm UTC

Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Fri Jun 07, 2019 2:22 am UTC

How come so many of you guys came here in such a short notice?

Is there a new googology forum I'm not aware of? If so, please do tell :)


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