Count with PSS expressions of limited length
Moderators: jestingrabbit, Moderators General, Prelates
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = {2}{1}14 = 28×2^{28} = 7 516 192 768 ≈ 7.5×10^{9} ≈ E9.88
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)[10] = {2}{1}{1}10 = 40 * 2^{40} ~ 4.4 * 10^{13} ~ EE1.135
Re: Count with PSS expressions of limited length
EE1.135
You mean F2.055
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
(ofcourse 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 = 44×2^{44} = 774 056 185 954 304 ≈ 7.74×10^{14} ≈ E14.89 ≈ F2.069
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = {2}{1}{1}12 = 48 * 2^{48} ~ 1.35×10^{16} ~ F2.082
Re: Count with PSS expressions of limited length
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 = 80×2^{80} =
= 96 714 065 569 170 333 976 494 080 ≈ 9.67×10^{25} ≈ E25.985 ≈ F2.151
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 = 80×2^{80} =
= 96 714 065 569 170 333 976 494 080 ≈ 9.67×10^{25} ≈ E25.985 ≈ F2.151
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)[10] = {2}{2}10 = 10240 * 2^{10240} ~ 3.61 * 10^{3086} ~ EEE0.543 = F2.543
Re: Count with PSS expressions of limited length
What? No decimal expansion? It's just:
360957072599898384090605039595548830355368154562216663658407757882686159634992633498094902235890317951559353911247855349486172666633657483467656588012709208748182598356636384334399545815230549708652121305780944094607687051610452806545332430422287685702892997010122003214335508082605878763758507043556411021131882440599131490156862577439953257503054733504014811497383642697438777679619716430794701581449591679689093164315539356126569695803705455174307071116910866061066273442321577547102316976919997191759869752309980772581503326976891743409084980526800964685721621214255852907801943482545725312697422823198175086555263549935207147167916759294630345216830044858194310497469585643388793214920683723361413054467883378587623186830388026709606288329594972154022376673485434803813697555172939568214923110437549731129443639611126859852738744989346253589445842275335089709057738999107362704933446827567946264837762487245074658161789298542273863070037238565810037387697951019945803617801257085790876437756594716845933151615975229353907304463152808968984861025962180160659334221678013914000051722776924934944720336623246536410349671563495582840474975535382609467326519059383049608782570197824570156088538044006148380691681254757288694690489481905743711844181842852717743112266019072319762196328421011040303954345027726782941861882418169484142610931473711982979105056933005658672562620510427959810381446455477346563284940509928834430832475909325288857376410879046828625223279640651753897335374339878482047069769788820613993720652178330715228866933918010269286022686582658559288492858064139725178753096222327775043137725961893906665695825694925440377077920743638601960837791117641595877476961307332573071625842682355292133107640045483246956029669323559670280750415727846152425844774850423021498101974337867585724795113066625866510161810993557764641190807624788591668574563114300977843921380910589544188835912484131356109086668225057069880775806656833318513865539755205915006035377306676434213935959195464072399104805532814456133113489977561952607282641277077600317528478934753582181943660217291661472783057094767504962805464905449619694327203383583442885922628270574456545002062883179784228297808225974006038595728828620619879518842170350704751184638636449217039082062288188714696175654411105225993151788599812851415654264842837422304284010164509705982693049731471323109170157173297417942125904467996480337653121462763247430028011322300406163756468357136334325433159782564464868518326705844968870557751649823818717425250349192461845919023290072836614736734772217186495720772586702700364156747687912174603961965388970385558595245890211448129278043639109041815070649122062005392558094379119925775750476208325991144162271899491238942998087295383317029962614619645583924749228199962075775515843940323113975912791834585885958615365879524094710625916041626044389326756510449876143147527025058780988709896628161732334426568002262550585352954698777482484398843216975680153506461202889572423805010320468864660836910084980649358090891381521055951158218869435922794230497899829410896991347525797191088330506240
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}(11×2^{11}) = {2}22528 = 22528×2^{22528} = 10^{22528×log(2)+log(22528)} ≈ 10^{6786} = E6786 ≈ F2.583
360957072599898384090605039595548830355368154562216663658407757882686159634992633498094902235890317951559353911247855349486172666633657483467656588012709208748182598356636384334399545815230549708652121305780944094607687051610452806545332430422287685702892997010122003214335508082605878763758507043556411021131882440599131490156862577439953257503054733504014811497383642697438777679619716430794701581449591679689093164315539356126569695803705455174307071116910866061066273442321577547102316976919997191759869752309980772581503326976891743409084980526800964685721621214255852907801943482545725312697422823198175086555263549935207147167916759294630345216830044858194310497469585643388793214920683723361413054467883378587623186830388026709606288329594972154022376673485434803813697555172939568214923110437549731129443639611126859852738744989346253589445842275335089709057738999107362704933446827567946264837762487245074658161789298542273863070037238565810037387697951019945803617801257085790876437756594716845933151615975229353907304463152808968984861025962180160659334221678013914000051722776924934944720336623246536410349671563495582840474975535382609467326519059383049608782570197824570156088538044006148380691681254757288694690489481905743711844181842852717743112266019072319762196328421011040303954345027726782941861882418169484142610931473711982979105056933005658672562620510427959810381446455477346563284940509928834430832475909325288857376410879046828625223279640651753897335374339878482047069769788820613993720652178330715228866933918010269286022686582658559288492858064139725178753096222327775043137725961893906665695825694925440377077920743638601960837791117641595877476961307332573071625842682355292133107640045483246956029669323559670280750415727846152425844774850423021498101974337867585724795113066625866510161810993557764641190807624788591668574563114300977843921380910589544188835912484131356109086668225057069880775806656833318513865539755205915006035377306676434213935959195464072399104805532814456133113489977561952607282641277077600317528478934753582181943660217291661472783057094767504962805464905449619694327203383583442885922628270574456545002062883179784228297808225974006038595728828620619879518842170350704751184638636449217039082062288188714696175654411105225993151788599812851415654264842837422304284010164509705982693049731471323109170157173297417942125904467996480337653121462763247430028011322300406163756468357136334325433159782564464868518326705844968870557751649823818717425250349192461845919023290072836614736734772217186495720772586702700364156747687912174603961965388970385558595245890211448129278043639109041815070649122062005392558094379119925775750476208325991144162271899491238942998087295383317029962614619645583924749228199962075775515843940323113975912791834585885958615365879524094710625916041626044389326756510449876143147527025058780988709896628161732334426568002262550585352954698777482484398843216975680153506461202889572423805010320468864660836910084980649358090891381521055951158218869435922794230497899829410896991347525797191088330506240
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}(11×2^{11}) = {2}22528 = 22528×2^{22528} = 10^{22528×log(2)+log(22528)} ≈ 10^{6786} = E6786 ≈ F2.583
Re: Count with PSS expressions of limited length
I leave the crazy stuff to you.
(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(0,0)[10] = {2}{2}12 = 49152 * 2^{49152} ~ 8.3 * 10^{14800} ~ EEE0,62 = F2.62
(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(0,0)[10] = {2}{2}12 = 49152 * 2^{49152} ~ 8.3 * 10^{14800} ~ EEE0,62 = F2.62
Re: Count with PSS expressions of limited length
I leave the crazy stuff to you.
Good call
(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}(13×2^{13}) = {2}106496 = 106496×2^{106496} = 10^{106496×log(2)+log(106496)} ≈ 10^{32063.52} ≈ 3.3×10^{32063} ≈ E32064 ≈ EEE0.654 = F2.654
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(1,0)[10] = {2}{2}20 = 20971520 * 2^{20971520} ~ 7.876 * 10^{6313063} ~ EEE0.83 = F2.83
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)[10] = {2}{2}22 ≈ 10^{27777457} ≈ F2.87
Re: Count with PSS expressions of limited length
(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 * 10^{3086}) ~ EE3086 ~ F3.543
Re: Count with PSS expressions of limited length
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}_{10}10 ≈ 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)
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}_{10}10 ≈ 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)
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(0,0)[10] = {3}11 = {2}_{11}[11] ~ 10pt6786 ~ F11.58 ~ G2.01
I need to refamiliarize myself with the letter notation, that took me way too long to doublecheck.
I need to refamiliarize myself with the letter notation, that took me way too long to doublecheck.
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(0,0)(0,0)[10] = {3}12 = {2}_{12}12 ≈ 11pt14801
(≈ F12.62 ≈ FEE0.0418 ≈ FF1.0418 = FFE0.0178 = G2.0178)
(≈ 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.
Re: Count with PSS expressions of limited length
(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
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = {3}14 = {2}_{14}14 ≈ 13pt69054
(≈ F14.68 ≈ FEE0.0670 ≈ FF1.0670 = FFE0.0282 = G2.0282)
(≈ F14.68 ≈ FEE0.0670 ≈ FF1.0670 = FFE0.0282 = G2.0282)
Re: Count with PSS expressions of limited length
(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
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)[10] = {3}20 = {2}_{20}20 ≈ 19pt6313063
(≈ F20.83 ≈ FEE0.1202 ≈ FF1.1202 = FFE0.0493 = G2.0493)
(≈ F20.83 ≈ FEE0.1202 ≈ FF1.1202 = FFE0.0493 = G2.0493)
Re: Count with PSS expressions of limited length
(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
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = {3}24 = {2}_{24}24 ≈ 23pt121 210 694)
(≈ F24.91 ≈ FEE0.1450 = FF1.1450 = FFE0.059 = G2.059)
(≈ F25 = 10↑↑25)
(≈ F24.91 ≈ FEE0.1450 = FF1.1450 = FFE0.059 = G2.059)
(≈ F25 = 10↑↑25)
Re: Count with PSS expressions of limited length
(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
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(0,0)(1,0)(0,0)(1,0)[10] = {3}40 = {2}_{40}40 ≈ 39pt13239439221689 ≈ 40pt13
(≈ F41.05 ≈ FEE0.208 = FF1.208 ≈ FFE0.082 = G2.082)
(≈ F41 = 10↑↑41)
(≈ F41.05 ≈ FEE0.208 = FF1.208 ≈ FFE0.082 = G2.082)
(≈ F41 = 10↑↑41)
Re: Count with PSS expressions of limited length
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*10^{14}) ~ F45.06 ~ G2.086
(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*10^{14}) ~ F45.06 ~ G2.086
Re: Count with PSS expressions of limited length
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)
(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)
Re: Count with PSS expressions of limited length
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?
{3}10240 ≈ 10240pt3086
Are you just substituting 10 as the base instead of 2 in the chain?
Re: Count with PSS expressions of limited length
It's exactly the same calculation we've done for the previous dozen numbers or so:
{3}44 = {2}_{44}44 ≈ 43pt(2.33×10^{14}) ≈ 43pt(10^{14}) = 44pt14
{3}10240 = {2}_{10240}10240 ≈ 10239pt10^{3186} = 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}_{k}n when k>2,n>3:
{2}_{k}n ≈ (k1)pt(n×2^{n}×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), 10^{n} is a very good approximation of {2}n=n×2^{n}. Thanks to this fact, the expression to the right of the "pt" (in the approximation) doesn't depend on k.
{3}44 = {2}_{44}44 ≈ 43pt(2.33×10^{14}) ≈ 43pt(10^{14}) = 44pt14
{3}10240 = {2}_{10240}10240 ≈ 10239pt10^{3186} = 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}_{k}n when k>2,n>3:
{2}_{k}n ≈ (k1)pt(n×2^{n}×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), 10^{n} is a very good approximation of {2}n=n×2^{n}. Thanks to this fact, the expression to the right of the "pt" (in the approximation) doesn't depend on k.
Re: Count with PSS expressions of limited length
Because I'm lazy and use wolfram alpha and one of the steps was too much for it.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.
That's what I guessed, thanks for confirming.PsiCubed2 wrote:For k>3, we use the fact that for a large enough n (say n>F4), 10^{n} is a very good approximation of {2}n=n×2^{n}.
(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
Re: Count with PSS expressions of limited length
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)
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)
Re: Count with PSS expressions of limited length
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
(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
Re: Count with PSS expressions of limited length
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 ] = F_{a}E_{b}x with 0<=a,b<=8 integers and 10<=x<10^{10}
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)
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 ] = F_{a}E_{b}x with 0<=a,b<=8 integers and 10<=x<10^{10}
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)
Re: Count with PSS expressions of limited length
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)
(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)
Re: Count with PSS expressions of limited length
Exactly.
We already need to add another number to our approximation array:
[ a,b,c ; x ] = G_{a}F_{b}E_{c}x
(0,0)(1,0)(1,0)(1,0)(1,0)[10] = {4}10 = {3}_{10}10 = {3}_{9}{3}10 ≈ F_{9}({3)}10 ≈ F_{9}F10.54 = F_{10}10.54 ≈ F_{11}1.01 ≈ G11.00 = 10↑↑↑11 = [ 1,0,0 ; 11 ]
( ≈ H2.003)
We already need to add another number to our approximation array:
[ a,b,c ; x ] = G_{a}F_{b}E_{c}x
(0,0)(1,0)(1,0)(1,0)(1,0)[10] = {4}10 = {3}_{10}10 = {3}_{9}{3}10 ≈ F_{9}({3)}10 ≈ F_{9}F10.54 = F_{10}10.54 ≈ F_{11}1.01 ≈ G11.00 = 10↑↑↑11 = [ 1,0,0 ; 11 ]
( ≈ H2.003)
Re: Count with PSS expressions of limited length
(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)[10] = {4}11 = {3}_{11}11 ~ F_{10}({3}11) ~ F_{11}11.58 ~ G12.0115 ~ [ 1,0,0 ; 12 ]
Re: Count with PSS expressions of limited length
(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 ≈ F_{11}({3}12) ≈ F_{11}(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 ≈ [ n1,n1 ; (the value after the 'pt' we've shamelessly stolen from the previous {3}n calculation) ]
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 ≈ F_{11}({3}12) ≈ F_{11}(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 ≈ [ n1,n1 ; (the value after the 'pt' we've shamelessly stolen from the previous {3}n calculation) ]
Re: Count with PSS expressions of limited length
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 ]
(0,0)(1,0)(1,0)(1,0)(1,0)(0,0)(0,0)(0,0)[10] = {4}13 ~ [ 12,12 ; 32064 ]
Re: Count with PSS expressions of limited length
(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 cutoff 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.
Hmmm... You're right.
To make this variation unique, we need to define an exact cutoff 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
I semiarbitrarily 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 ]
(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.

 Posts: 5
 Joined: Fri Jun 07, 2019 1:43 am UTC
Re: Count with PSS expressions of limited length
It seems still OK to post here (in order not to post on a dead tenyearold 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.
(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.
Re: Count with PSS expressions of limited length
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
Is there a new googology forum I'm not aware of? If so, please do tell
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