I get exit code -1073741571 (0xC00000FD) when I calculating D(n)
In order to calculate D(n), I created a class named CompressInt, and I modified Naruyoko's D^6(9)（https://codegolf.stackexchange.com/a/197173
نیسٹڈ ہائپر فیکٹوریل (چھوٹے گام2 فنکشن نوٹیشن میں تمام بنیادی باتوں کی تمام نظرثانی پر نظر ثانی جس کے بارے میں مجھے بہت افسوس ہے
(0|0|1) = psi(Ω_2)
(1|0|1) = psi(Ω_2)+1
(2|0|1) = psi(Ω_2)+2
(3|0|1) = psi(Ω_2)+3
((|0)|0|1) = psi(Ω_2)+w
(((|0)|0)|0|1) = psi(Ω_2)+w^w
((((|0)|0)|0)|0|1) = psi(Ω_2)+w^w^w
(((((|0)|0)|0)|0)|0|1) = psi(Ω_2)+w^w^w^w
((|1)|0|1) = psi(Ω_2)+psi(Ω)
((|2)|0|1) = psi(Ω_2)+psi(Ω^2)
((|3)|0|1) = psi(Ω_2)+psi(Ω^3)
((|{1}1)|0|1) = psi(Ω_2)+psi(Ω^Ω)
(0||1) = psi(Ω_2^psi1(Ω_2))
stats of 100 randomly chosen numbers on the wiki
WIP
- 1 By notation
- 2 By coiner
- 3 By class
- 4 Generated numbers
- Hyper-E - 5
- Hyperfactorial Array - 2
- Scientific - 2
- Bird's Array - 1
- Sbiis Saibian - 4
- Lawrence Hollom - 2
- Conway and Guy - 2
- XRQ CORPORATION - 1
- Douglas Shamlin Jr. - 1
- Binary phi - 3
- Bachmann's collapsing - 2
- Class 2 - 2
- Quadratic omega - 2
- N/A - 1
- Duodecinongentillion
- Eneninto-tethracross
- Terrisecunded dustaculated-tethracross
- Bisuperior Grand Kiloenormabixul
- Flastretchemorgulus
- Thruelohgolpeta
- Kilo-BIGG
- Quattuordecitrecentillion
- Thraatagolhepta
- Tethrathoth-turreted-tethratopothoth
نیسٹڈ ہائپر فیکٹوریل (چھوٹے گاما1 فنکشن نوٹیشن میں تمام بنیادی باتوں کی تمام نظرثانی پر نظر ثانی جس کے بارے میں مجھے بہت افسوس ہے
(0|0|1) = psi(Ω_2)
(1|0|1) = psi(Ω_2)+1
(2|0|1) = psi(Ω_2)+2
(3|0|1) = psi(Ω_2)+3
((0|0|1)|0|1) = psi(Ω_2)2
((0|0|1)(0|0|1)|0|1) = psi(Ω_2)3
((0|0|1)(0|0|1)(0|0|1)|0|1) = psi(Ω_2)4
((0|0|1)!^(|0)|0|1) = psi(Ω_2)w
((0|0|1)!^((|0)|0)|0|1) = psi(Ω_2)w^w
((0|0|1)!^(((|0)|0)|0)|0|1) = psi(Ω_2)w^w^w
((0|0|1)!^((((|0)|0)|0)|0)|0|1) = psi(Ω_2)w^w^w^w
((0|0|1)!^(|1)|0|1) = psi(Ω_2)psi(Ω)
((0|0|1)!^(|2)|0|1) = psi(Ω_2)psi(Ω^2)
((0|0|1)!^(|{1}1)|0|1) = psi(Ω_2)psi(Ω^Ω)
({(0|0|1)}|0|1) = psi(Ω_2*psi(Ω_2))
({(0|0|1)}|0|1) = psi(Ω_2*psi(Ω_2*psi(Ω_2)))
({({(0|0|1)}|0|1)}|0|1) = psi(Ω_2*psi(Ω_2*psi(Ω_2*psi(Ω_2))))
({}|0|1) = psi(Ω_2*Ω)
Ascending Operator
I would like to introduce this AO(Ascending Operator) { }, vs. long established hyper-operator.
AO is right-associative. For natural numbers a, b and n:
- For n = 1,
- a{n}b = a{1}b = a+b
- For n > 1,
- a{n}1 = a
- a{n}b = a{n-1}[(a+1){n}(b-1)]
- 2{1}2 = 2+2 = 4
- 2{2}2 = 2{1}(3{2}1) = 2{1}3 = 2+3 = 5
- 2{3}3
- = 2{2}(3{3}2)
- = 2{2}(3{2}(4{3}1))
- = 2{2}3{2}4
- = 2{2}(3{1}4{1}5{1}6)
- = 2{2}(3+4+5+6)
- = 2{2}18
- = 2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19
- = 189
- 3{3}3
- = 3{2}(4{3}2)
- = 3{2}(4{2}(5{3}1))
- = 3{2}4{2}5
- = 3{2}(4{1}5{1}6{1}7{1}8)
- = 3{2}30
- = 3+4+ ... +31+32
- = 525
- 3{4}3
- = 3{3}4{3}5
- = 3{3}(4{2}5{2}6{2}7{2}8)
- = 3{3}(4{2}5{2}6{2}(7+8+9+10+11+12+13+14))
- = 3{3}(4{2}5{2}6{2}84)
- = 3{3}(4{2}5{2}(6+7+ ... +88+89))
- = 3{3}(4{2}5{2}3990)
- = 3{3}(4{2}(5+6+ ... +3993+3994))
- = 3{3}(4{2}7978005)
- ≈ 3{3}(3.182…
Hierarchical Array Notation
Hierarchical Array Notation (HAN) is a notation created by AndyShow1000000. It is notated as {a,b,c}n.
=== Fundamental Hierarchical Array Notation (FHAN) (
New method
You need to start with the basic massif:
§(n)=n
§(n,2)=n^2
§(n,3)=n^3
§(n,1,2)=n^n
§(n,2,2)=n^n^2
§(n,1,3)=n^n^n
§(n,2,3)=n^n^n^2
§(n,1,4)=n^n^n^n
§(n,1,1,2)=n^^n
§(n,1,1,3)=n^^n^^n
§(n,1,1,1,2)=n^^^n
§(n,1,1,1,1,2)=n^^^^n
And now the most interesting β structures. All further extensions will be based on them.
Please note that β is an amplifical effect on the usual numbers.
β/n=§(n)=n
β^2/n=§(n,2)
β^3/n=§(n,3)
β^β/n=§(n,1,2)
β^β^2/n=§(n,1,3)
β^β^3/n=§(n,1,4)
β^β^β/n=§(n,1,1,2)=f3(n)
β^β^β^2/n=§(n,1,1,3)=f3(f3(n))
β^β^β^β/n=§(n,1,1,1,2)=f4(n)
β^β^β^β^β/n=§(n,1,1,1,1,2)=f5(n)
β^^β/n=fω(n)
β^^β^2/n=fω(fω(n))
β^^β^3/n=fω(fω(fω(n)))
β^^β^β/n=fω+1(n)
β^^β^β^β/n=fω+2(n)
β^^β^^β/n=fω*2(n)
β^^β^^β^β/n=fω*2+1(n)
β^^β^^β^^β/n=fω*3(n)
β^^^β/n=fω^2(n)
β^^^β^^^β/n=fω^2*2(n)
β^^^^β/n=fω^3…
نیسٹڈ ہائپر فیکٹوریل (چھوٹے گاما) فنکشن نوٹیشن میں تمام بنیادی باتوں کی تمام نظرثانی پر نظر ثانی جس کے بارے میں مجھے بہت افسوس ہے
(|2) = psi(Ω^2)
(1|2) = psi(Ω^2)+1
(2|2) = psi(Ω^2)+2
(3|2) = psi(Ω^2)+3
((|1)|2) = psi(Ω^2)+psi(Ω)
((|2)|2) = psi(Ω^2)+psi(Ω^2)
((|2)(|2)|2) = psi(Ω^2)+psi(Ω^2)+psi(Ω^2)
((|2)!^(|0)|2) = psi(Ω^2)w
((|2)!^((|0)|0)|2) = psi(Ω^2)w^w
((|2)!^(((|0)|0)|0)|2) = psi(Ω^2)w^w^w
((|2)!^((((|0)|0)|0)|0)|2) = psi(Ω^2)w^w^w^w
((|2)!^(|1)|2) = psi(Ω^2)psi(Ω)
((|2)!^(|2)|2) = psi(Ω^2)psi(Ω^2)
((|2)!^(|2)|2) = psi(Ω^2+2)
((|2)!^((|2)!^(|2)|2)|2) = psi(Ω^2+3)
((|2)!^(((|2)!^(|2)|2)!^(|2)|2)|2) = psi(Ω^2+4)
((|2)!^((((|2)!^(|2)|2)!^(|2)|2)!^(|2)|2)|2) = psi(Ω^2+5)
((1|2)|2) = psi(Ω^2+w)
((2|2)|2) = psi(Ω^2+w+1)
((3|2)|2) = psi(Ω^2+w+2)
(((|0)|2)|2) = psi(Ω^2+w2)
((((|0)|0)|2)|2) = psi(Ω^2+w^2)
(((((|0)|0)|0)|2)|2) = psi(Ω^2+w^w^w)
((|1)|2)|2) = psi(Ω^2+psi(Ω))
((|2)|2)|2) = psi(Ω^…
Hyperotation (help me simplify the rules)
So this is a notation I call Hyperotation
Let me explain it to you as simply as I can but before that, I need to explain some terms:
- [H] is called the Hyper Variable
- [] are the Hyper Brackets
The Hyper Variable is equal to the value between the Hyper Brackets
for example: 8↑[10], here the hyper variable is 10, so if we would have written 8↑[H][10] then the hyper variable would be equal 10 so it would be equal to 8↑1010, however what we can make this even more complex. Because what happens when the Hyper Variable is between the Hyper Brackets, then the Hyper Variable would be still equal to the number between the hyper brackets but when we want to know how much is the Hyper Variable, we replace the Hyper Variable by ten. So [100↑[H]5]=={10,10,1…
Naive Oblivion
Define a function:
o[1]=The largest finite number that can be uniquely defined with no more than Kungulus symbols
o[n]=The largest finite number that can be uniquely defined with no more than o[n-1] symbols if n>1
o[n,m]=o[o[n,m-1],m-1]
o[#,m,k]=o[#,o[#,m-1,k],k-1]
o2[n]=o[n,n,...,n,n] with n entries
o3[n]=o2[n,n,...,n,n] with n entries
and so on
Naive Oblivion = ooo...oooOblivion[Oblivion,Oblivion,...,Oblivion,Oblivion] with Oblivion entries and Oblivion o's
sequence of growth rates
[n] -> 2^(1+n) ~> 2^n
[n, 1] ~> 2^2^n
[n, a] ~> 2^^(1+a);n
[n, 0, 1] ~> 2^^(3+n);n ~> 2^^n
[n, 1, 1] ~> 2^^2^^n
[n, a, 1] ~> 2^^^(1+a);n
[n, 0, b] ~> 2[1+b]n
[n, 0, 0, 1] ~> 2[2+n]n ~> fω(n)
[n, 1, 0, 1] ~> fω+1(n)
[n, a, 0, 1] ~> fω+a(n)
[n, 0, 1, 1] ~> fω2(n)
[n, 1, 1, 1] ~> fω2+ω(n)
[n, a, 1, 1] ~> fω2+ωa(n)
[n, 0, b, 1] ~> fωb(n)
[n, 0, 0, 2] ~> fωω(n)
[n, 1, 0, 2] ~> fωω+1(n)
[n, a, 0, 2] ~> fωω+a(n)
[n, 0, 1, 2] ~> fωω^ω(n)
[n, 1, 1, 2] ~> fωω^(ω+1)(n)
[n, a, 1, 2] ~> fωω^(ω+a)(n)
[n, 0, b, 2] ~> fω^^b(n)
[n, 0, 0, 3] ~> fε_0(n)
[n, 1, 0, 3] ~> fε_1(n)
[n, a, 0, 3] ~> fε_a(n)
[n, 0, 1, 3] ~> fε_ε_0(n)
[n, 1, 1, 3] ~> fε_ε_1(n)
[n, a, 1, 3] ~> fε_ε_a(n)
[n, 0, b, 3] ~> fω^^^b(n)
[n, 0, 0, c] ~> fφ_c(0)(n)
[n, 0, 0, 0, 1] ~> fφ_ω(0)(n)
[n, 1, 0, 0, 1] ~> fφ_(ω+1)(0)(n)
[n, a, 0, 0, 1] ~> fφ…
Concept for notation
This is part of my in Clubsuit Notation & Grangol-tetrational-duadekal in my naming system.
AlexFJ defined this really well.
x club^n y = Ex{#^n}y
x club[2]^n y = Ex{#^^n}y
It can be assumed x club[m]^n y = Ex{#{m}n}y
I DONT KNOW ANYTHING AT ALL
Approximation of exponential factorial with tetration
We use the hyperfactorial array notation, where exponential factorial of n is expressed as n!1, and derive the approximation:
\(n!1 \approx 10\uparrow\uparrow (n-2.27876677830609945248)\) for \(n>5\)
By using Hypercalc BASIC
and we can confirm that n!1 has power-tower paradox for n≥6. This result can be rewritten with continuous tetration by
- 1!1 = 1
- 2!1 = 2
- 3!1 = 9
- 4!1 = 262144
- 5!1 = 6.20606987866088 x 10 ^ 183230
- 6!1 = 10^^3.72123322169390054751
- 7!1 = 10^^4.72123322169390054751
- 8!1 = 10^^5.72123322169390054751
- 9!1 = 10^^6.72123322169390054751
- 10!1 = 10^^7.72123322169390054751
and we get the approximation
\(n!1 \approx 10\uparrow\uparrow (n-2.27876677830609945248)\) for \(n>5\)
up arrow notation on crack
Define x↑1 as x+1, so for example 3↑1=4
Define x↑2 as ((...((x↑1)↑1)...)↑1)↑1 x times, so for example 3↑2=((3↑1)↑1)↑1=(4↑1)↑1=5↑1=6=2x
Define x↑3 as ((...((x↑2)↑2)...)↑2)↑2 x times, so for example 3↑3=((3↑2)↑2)↑2=(6↑2)↑2=12↑2=24=2^x*x
The pattern repeats. x↑y=fy(x) and x↑x=fw(x).
All other hyperoperations have the same rules.
x↑↑2=fw(x). 3↑↑3=3↑3↑3=3↑24=f24(3), 3↑↑4=3↑3↑3↑3=3↑3↑24=3↑f24(3)=ff_24(3)(3) and so on and so forth
3↑↑↑2=3↑↑3=f24(3), 3↑↑↑3=3↑↑3↑↑3=3↑↑f24(3)
n↑↑n=n↑nn
more ideas might be added later
5. اگر مساوی کو مساوی میں شامل کیا جائے تو پورے برابر ہوتے ہیں۔5
(!^(2) |0|1) = psi(Ω_2*2+Ω)
(!^(2) |0|1) = psi(Ω_2*2+Ω^2)
(!^(2) |0|1) = psi(Ω_2*2+Ω^3)
(!^(2) |0|1) = psi(Ω_2*2+Ω^Ω)
(0|(0|0|1) |1) = psi(Ω_2^ psi(Ω_2))
(0|(0|(0|0|1)|1) |1) = psi(Ω_2^ psi(Ω_2^ psi(Ω_2)))
(0|(0|(0|(0|0|1)|1)|1) |1) = psi(Ω_2^ psi(Ω_2^ psi(Ω_2^ psi(Ω_2))))
(0||1) = psi(Ω_2^Ω)
4. اگر مساوی کو مساوی میں شامل کیا جائے تو پورے برابر ہوتے ہیں۔4
(0|0|1) = psi(Ω_2)
(1|0|1) = psi(Ω_2)+1
(2|0|1) = psi(Ω_2)+2
(3|0|1) = psi(Ω_2)+3
(4|0|1) = psi(Ω_2)+4
((|0)|0|1) = psi(Ω_2)+w
((0|0|1)|0|1) = psi(Ω_2)2
((0|0|1)(0|0|1)|0|1) = psi(Ω_2)3
((0|0|1)(0|0|1)(0|0|1)|0|1) = psi(Ω_2)4
((0|0|1)!^(|0)|0|1) = psi(Ω_2)w
((0|0|1)!^((|0)|0)|0|1) = psi(Ω_2)w^w
((0|0|1)!^(((|0)|0)|0)|0|1) = psi(Ω_2)w^w^w
((0|0|1)!^((((|0)|0)|0)|0)|0|1) = psi(Ω_2)w^w^w^w
((0|0|1)!^(|1)|0|1) = psi(Ω_2)psi(Ω)
((0|0|1)!^(|2)|0|1) = psi(Ω_2)psi(Ω^2)
((0|0|1)!^(|3)|0|1) = psi(Ω_2)psi(Ω^3)
((0|0|1)!^(|4)|0|1) = psi(Ω_2)psi(Ω^4)
((0|0|1)!^(|(|0))|0|1) = psi(Ω_2)psi(Ω^w)
((0|0|1)!^(|{1}1)|0|1) = psi(Ω_2)psi(Ω^Ω)
(|0|1) = psi(Ω_2+psi1(Ω_2)^psi(Ω_2))
(|0|1) = psi(Ω_2+psi1(Ω_2)^psi(Ω_2)^psi(Ω_2))
(!^(2)|0|1) = psi(Ω_2*2)
(1 !^(2)|0|1) = psi(Ω_2*2)+1 (tras…
The question of the side attachment
Can you explain with examples how the changing definition and lateral investment works? For example, how ε, S-σ works
گوگل ارتھ پروفیشنل کا استعمال کرتے ہوئے تقریباً زمین کا رداس تلاش کریں۔
فرض کریں کہ مبصر سے پہاڑ کا فاصلہ "a" ہے اور پہاڑ کی اونچائی "o" ہے اور مبصر کا زاویہ \(\theta\) ہے لہذا
یا صرف 44 کلومیٹر (معلوم نہیں ہے)
دوسرے تجربات کرنے کی ضرورت ہے
REMEMBER I AM NOT AN URDU SPEAKER, I JUST TRANSLATE ALL OF THIS USING GOOGLE TRANSLATE
Weak α function
I think it will be easier to analyze
sα[0]=1
sα[1]=ω
sα[2]=ω^ω
sα[3]=ω^ω^ω
sα[ω]=ε0
sα[ω+1]=ω^ε0+1
sα[ω+2]=ω^ω^ε0+1
sα[ω*2]=ε1
sα[ω*2+1]=ω^ε1+1
sα[ω*3]=ε2
sα[ω*4]=ε3
sα[ω^2]=εω
sα[ω^2+1]=ε(ω^ω)
sα[ω^2+2]=ε(ω^ω^ω)
sα[ω^2+ω]=ε(ε0)
sα[ω^2+ω+1]=ε(ω^ε0+1)
sα[ω^2+ω*2]=ε(ε1)
sα[ω^2*2]=ε(εω)
sα[ω^2*2+1]=ε(ε(ω^ω))
sα[ω^2*3]=ε(ε(εω))
sα[ω^2*4]=ε(ε(ε(εω)))
sα[ω^3]=ζ0
sα[ω^3+1]=ω^ζ0+1
sα[ω^3+2]=ω^ω^ζ0+1
sα[ω^3+ω]=ε(ζ0+1)
sα[ω^3+ω+1]=ε(ω^ζ0+1)
sα[ω^3+ω*2]=ε(ε(ζ0+1))
sα[ω^3+ω*3]=ε(ε(ε(ζ0+1)))
sα[ω^3+ω^2]=ζ1
sα[ω^3+ω^2+1]=ω^ζ1+1
sα[ω^3+ω^2+ω]=ε(ζ1+1)
sα[ω^3+ω^2*2]=ζ2
sα[ω^3+ω^2*3]=ζ3
sα[ω^3*2]=ζω
sα[ω^3*2+1]=ζ(ω^ω)
sα[ω^3*3]=ζ(ζω)
sα[ω^3*4]=ζ(ζ(ζω))
sα[ω^4]=η0
sα[ω^4*2]=ηω
sα[ω^5]=ψ(Ω^5)
sα[ω^6]=ψ(Ω^6)
sα[ω^ω]=ψ(Ω^ω)
sα[(ω^ω)+1]=ψ(Ω^ω^ω)
sα[(ω^ω)+2]=ψ(Ω^ω^ω^ω)
sα[(ω^ω)*2]=ψ(Ω^ψ(Ω^ω))
sα[(ω^ω)*3]=ψ(Ω…
List of array notation’s
- Fast array notation
- Graham Array Notation
- Hyperfactorial array notation
- Extended array notation
- Ember Array Notation
- Chained array notation
- Strong array notation
- Hyperdimensional Array Notation
- Jazzy array notation
- Quick array notation
- Galaxy array notation
- Almighty Array Notation
- Hayden's Array Notation
- ThaAwesome's Array Notation
- DeepLineMadom's Array Notation
- Rampant Array Notation
- Stage array notation
- Bird's array notation
what should this function name should be?
1-ag(n)= gag(N)
n-ag(n) = (n-1)ag^n(n)
This function is n-ag(n) therefore stronger for all larger n
F0(n) = Gag(n)
F1(n) = Gagn(n)
FM(N) = M-Ag(H)
Wainer hierarchies and others exist in this Just copy the definition of Fgh and replace F_0(N) = N+1 with F_0(N) = Gag(N)
NooberRevival's Average Notation
- 1 Introduction
- 2 Rules
- 3 Extended Rules
- 4 Naming System
- 5 Examples
NooberRevival's Average Notation or NAN for short, is a notation created by NooberRevival. It is one of many notations made by NooberRevival.
NAN uses positive integers as entries, because for hyperoperators above exponentiation, it is ill-defined to have a non-integer value.
- Rule 1: N[a] = a
- Rule 2: N[a1,a2,a3,...,an,1] = N[a1,a2,a3,...,an]
- Rule 2.1: N[a,1] ≠ N[a] but N[a,0] = N[a]
- Rule 3: Let {0} be multiplication and {-1} be addition in BEAF, N[a1,a2,a3,...,an-1,an] = N[a1,a2,a3,...,(an-1){n-3}an,an-1]
In plain English:
- Rule 1: If there is only one entry, then the value is the entry itself.
- Rule 2 and 2.1: If the last entry is 1, it may be removed, except the second entry where it needs to be 0 to…
What is the speed of this function?
α[0]=(0)=1
α[1]=(0)(1)=ω
α[2]=(0)(1)(2)=ω^ω
α[3]=(0)(1)(2)(3)=ω^ω^ω
α[ω]=(0)(1,1)=ψ(Ω)
α[ω+1]=(0)(1,1)(2)=ψ(Ω*ω)
α[ω+2]=(0)(1,1)(2)(3)=ψ(Ω*ω^ω)
α[ω*2]=(0)(1,1)(2)(3,1)=ψ(Ω*ψ(Ω))
α[ω*2+1]=(0)(1,1)(2)(3,1)(4)=ψ(Ω*ψ(Ω*ω))
α[ω*3]=(0)(1,1)(2)(3,1)(4)(5,1)=ψ(Ω*ψ(Ω*ψ(Ω)))
α[ω^2]=(0)(1,1)(2,1)=ψ(Ω^2)
α[ω^2+1]=(0)(1,1)(2,1)(3)=ψ(Ω^ω)
α[ω^2+2]=(0)(1,1)(2,1)(3)(4)=ψ(Ω^ω^ω)
α[ω^2+ω]=(0)(1,1)(2,1)(3)(4,1)=ψ(Ω^ψ(Ω))
α[ω^2+ω*2]=(0)(1,1)(2,1)(3)(4,1)(5)(6,1)=ψ(Ω^ψ(Ω*ψ(Ω)))
α[ω^2*2]=(0)(1,1)(2,1)(3)(4,1)(5,1)=ψ(Ω^ψ(Ω^2))
α[ω^2*2+1]=(0)(1,1)(2,1)(3)(4,1)(5,1)(6)=ψ(Ω^ψ(Ω^ω))
α[ω^2*3+1]=(0)(1,1)(2,1)(3)(4,1)(5,1)(6)(7,1)(8,1)(9)=ψ(Ω^ψ(Ω^ψ(Ω^ω)))
α[ω^3]=(0)(1,1)(2,1)(3,1)=ψ(Ω^Ω)
α[ω^3+1]=(0)(1,1)(2,1)(3,1)(4)=ψ(Ω^Ω^ω)
α[ω^3*2]=(0)(1,1)(2,1)(3,1)(4)(5,1)(6,1)(7,1)=ψ(Ω^Ω^ψ(Ω^Ω))
α[ω…
LSN growth estimates
Special thanks to Sho Phuane
s1(x)=f0(x)
s2(x)=f1(x)
s2(x,x) around f2(x)
s3(x) around f3(x)
s3(x,x) around f4(x)
s3(x,x,x) around fw(x)
s4(x) around fw+1(x)
s4(x,x) around fw+2(x)
s4(x,x,x) around fw2(x)
s4(x,x,x,x) around fw2+1(x)
sn(x,...,x) around fw^w(n)
C-like function with a "shift" for 2
(0,0)=1
(0,(0,0))=2
(1,0)=ω
(1,1)=ω+1
(1,(1,0))=ω*2
(2,0)=ω^2
((1,0),0)=ω^ω
(((1,0),0),0)=ω^ω^ω
•((Ω,0),0)=ψ(Ω)
((Ω,0),1)=ψ(Ω)+1
((Ω,0),((Ω,0),0))=ψ(Ω)+ψ(Ω)
((Ω,1),0)=ψ(Ω+1)
((Ω,2),0)=ψ(Ω+2)
((Ω,((Ω,0),0)),0)=ψ(Ω+ψ(Ω))
((Ω,((Ω,((Ω,0),0)),0)),0)=ψ(Ω+ψ(Ω+ψ(Ω)))
((Ω,(Ω,0)),0)=ψ(Ω*2)
((Ω,(Ω,(Ω,0))),0)=ψ(Ω*3)
((Ω+1,0),0)=ψ(Ω*ω)
((Ω+2,0),0)=ψ(Ω*ω^2)
((Ω+((Ω,0),0),0),0)=ψ(Ω*ψ(Ω))
((Ω+((Ω+((Ω,0),0),0),0),0),0)=ψ(Ω*ψ(Ω*ψ(Ω)))
((Ω+(Ω,0)),0)=ψ(Ω^2)
((Ω+(Ω+(Ω,0)),0)),0)=ψ(Ω^3)
((Ω*2,0),0)=ψ(Ω^ω)
((Ω*2,1),0)=ψ((Ω^ω)+1)
((Ω*2,((Ω*2,0),0)):0)=ψ((Ω^ω)+ψ(Ω^ω))
((Ω*2,(Ω,0)),0)=ψ((Ω^ω)+Ω)
((Ω*2,(Ω*2,0)),0)=ψ(Ω^ω)*2)
((Ω*2+1,0),0)=ψ((Ω^ω)*ω)
((Ω*2+(Ω,0),0),0)=ψ(Ω^ω+1)
((Ω*2+(Ω,(Ω,0)),0),0)=ψ(Ω^ω+2)
((Ω*2+(Ω*2,0),0),0)=ψ(Ω^ω*2)
((Ω*3,0),0)=ψ(Ω^ω^2)
((Ω*4,0),0)=ψ(Ω^ω^3)
((Ω*ω,0),0)=ψ(Ω^ω^ω)
((Ω*…
List of Numbers
- 1 Introduction
- 2 Counting Era (0 - 7.5 × 108)
- 3 Exponential Era (109 - 109×109+3 )
- 4 Tetrational Era (101010 - 10^^900,000):
- 5 Knuth Arrow Era (10^^1,000,000 - 10^^10^^1,000,000,000):
This is a list of numbers from 0 to unimaginably large numbers.
Please note that this list only includes natural numbers and transfinite numbers. This list also include some numbers/googolisms that isn't in the "List of googolisms" pages but made by my (limited) knowledge.
Credits: douglasshamlinjr.392 on YouTube and various wiki pages.
Subitizing Range (0 - 6):
- Zero | 0
- One | 1
- Two | 2
- Three | 3
- Four | 4
- Five | 5
- Six | 6
Palpable Range (7 - 9):
- Seven | 7
- Eight | 8
- Nine | 9
Tens Range (10 - 99):
- Ten | 10
- Eleven | 11
- Twelve | 12
- Thirteen | 13
- Fourteen | 14
- Fifteen | 15
- Sixteen | 16
- Seventeen | 1…
fandomical monkey unicodic number
imagine a monkey sitting on a very large keyboard, which contains all unicode characters version 16.0 with a total of 155,063 characters with code points, covering 168 modern and historical scripts, as well as multiple symbol sets. then the monkey walks randomly and every monkey touches 1 unicode key then the output will appear on a large printer that prints every character stepped on by the monkey
what is the possibility that the monkey can be able to write all the pages in the googologywiki at this 27 september 2024 00:00 UTC based on its source code, this includes templates, image information, warnings, user talk, user blog comments, GeoJSON, modules, policies from googology wiki, user pages, and most importantly blogposts, that means th…
Help with making a website to source my stuff
So I've came up with a new notation system, and wanted to make a page on here, however I have to make first an external source, a.k.a. my own website or something.
The thing is, I would need some help on how to form the website so that it doesn't look like a random dump of numbers and symbols.
poly-multi-adic factorial type 6
For type 6 here, the definition is actually very simple and perhaps the slowest growing, this number combines factorial with triangular numbers such as 1+2+3+4+5+..., this series can be simplified to
poly-multi-adic factorial type
poly-multi-adic factorial is a type of function that uses the results of the product or combination of the product and sum that comes from factorial or different, some functions produce properties of functions that are not complicated enough, there are also functions that produce factorial at the end, these functions all use the letter gamma, there are several types here
for type 1 inspired by nth order polynomials, meaning each term is added, examples of polynomial equations of each order see below
order 0, \(a\)
order 1, \(ax+b\) This is the equation for a line and will output the shape of the line
order 2, \(ax^2+bx+c\) This is quadratic equation and will make a parabola
order 3, \(ax^3+bx^2+cx+d\)
order 4, \(ax^4+bx^3+cx^2+dx+e\)
This means t…
Xsan part lV
expected limit-ψ(Μ_2)
(1(1~2)2)=ψ(Ω(1,0))
(1(1~2)(1)2)=ψ(Ω(1,0)+1)
(1(1~2)(1(1~2)(1)2)2)=ψ(Ω(1,0)+ψ(Ω(1,0)+1))
(1(1~2)(1-2)2)=ψ(Ω(1,0)+Ω)
(1(1~2)(1(1-2-)2)2)=ψ(Ω(1,0)+Ω_Ω)
(1(1~2)(1(1(1-2-)2-)2)2)=ψ(Ω(1,0)+Ω_Ω_Ω)
(1(1~2)(1(1~2-)2)2)=ψ(Ω(1,0)+Ω(Ω(1,0)))
(1(1~2)(2(1~2-)2)2)=ψ(Ω(1,0)+Ω(Ω(1,0))+1))
(1(1~2)(1(1~2-)3)2)=ψ(Ω(1,0)+Ω(Ω(1,0))+Ω(Ω(1,0)))
(1(1~2)(1(1~2-)(1)2)2)=ψ(Ω(1,0)+Ω(Ω(1,0)+1))
(1(1~2)(1(1~2-)(1-2)2)2)=ψ(Ω(1,0)+Ω(Ω(1,0)+Ω))
(1(1~2)(1(1~2-)(1(1~2-)2)2)2)=ψ(Ω(1,0)+Ω(Ω(1,0)+Ω(Ω(1,0))))
(1(2~2)2)=ψ(Ω(1,0)*2)
(1(3~2)2)=ψ(Ω(1,0)*3)
(1(1,2~2)2)=ψ(Ω(1,0)*ω)
(1(1-2~2)2)=ψ(Ω(1,0)*Ω)
(1(1(1-2-)2~2)2)=ψ(Ω(1,0)*Ω_Ω)
(1(1(1~2-)2~2)2)=ψ(Ω(1,0)*Ω(Ω(1,0)))
(1(1(1(1~2-)2~2-)2~2)2)=ψ(Ω(1,0)*Ω(Ω(1,0)*Ω(Ω(1,0))))
(1(1~3)2)=ψ(Ω(1,0)^2)
(1(1~4)2)=ψ(Ω(1,0)^3)
(1(1~(1)2)2)=ψ(…
2. اگر مساوی کو مساوی میں شامل کیا جائے تو پورے برابر ہوتے ہیں۔2
(|4) =psi(Ω^4)
(|5) =psi(Ω^5)
(|6) =psi(Ω^6)
(|7) =psi(Ω^7)
(|8) =psi(Ω^8)
(|(|0)) =psi(Ω^w)
(|((|0)|0)) =psi(Ω^w^w)
(|(((|0)|0)|0)) =psi(Ω^w^w^w)
(|((((|0)|0)|0)|0)) =psi(Ω^w^w^w^w)
(|(|1)) =psi(Ω^psi(Ω))
(|((|0)|1)) =psi(Ω^psi(Ω+w))
(|(((|0)|0)|1)) =psi(Ω^psi(Ω+w^w))
(|((((|0)|0)|0)|1)) =psi(Ω^psi(Ω+w^w^w))
(|(((((|0)|0)|0)|0)|1)) =psi(Ω^psi(Ω+w^w^w^w))
(|((|1)|1)) =psi(Ω^psi(Ω+psi(Ω)))
(|(((|1)|1)|1)) =psi(Ω^psi(Ω+psi(Ω+psi(Ω))))
(|((((|1)|1)|1)|1)) =psi(Ω^psi(Ω+psi(Ω+psi(Ω+psi(Ω)))))
(|{1}1) = psi(Ω^Ω) (Feferman–Schütte ordinal)
(|{1}2) =psi(Ω^Ω+1)
(|{1}3) =psi(Ω^Ω+2)
(|{1}4) =psi(Ω^Ω+3)
(|{1}5) =psi(Ω^Ω+5)
(|{1}(|0)) =psi(Ω^Ω+w)
(|{1}((|0)|0)) =psi(Ω^Ω+w^w)
(|{1}(((|0)|0)|0)) =psi(Ω^Ω+w^w^w)
(|{1}((((|0)|0)|0)|0)) =psi(Ω^Ω+w^w^w^w)
(|{1}(|1)) =psi(Ω^Ω+psi(Ω))
(|…
ٹیٹریشنل چوتھا کثیر الثانی بطور ایک نئی فیکٹریل شکل
Tetronomial is a higher level of polynomial, polynomial uses exponents, coefficients and variables, exponents use exponentiation, coefficients use multiplication and each term is added using addition
but tetronomial is higher again, each term is separated by multiplication and coefficients use exponentiation, the variables are tetramed with certain numbers, an example of tetronomial is
Full decimal expansion of super-leviathan
Full decimal expansion of super-leviathan = \(9^{9^{9}}\) was calculated with . It has 369,693,100 digits.
- 999.zip (171MB, )
First 100 digits
Last 100 digits
a hard coercion for nested hyperfactorial (small gamma) functions, PART IIIb
now I want to revise all of this, remember that the power tower \(\varepsilon_0\) is already represented by nesting brackets so we don't need to align it with the buchholz ocf, this nesting brackets represents the summation in the buchholz ocf, but then there is also an inner multiplication in the buchholz ocf like \(\psi(\Omega \alpha)\) where \(\alpha\) is an ordinal, this multiplication corresponds to the subscript in \(\varepsilon_{\alpha}\), this is the gateway to \(\zeta_0\), if we know the concept we will be able to continue
to solve this problem, curly brackets will be made in the left space signs such as {}|, the brackets should be curly brackets, meaning that it requires special care when writing in mathjax, the {}| sign is filled…
cosecant quectoradian
cosecant quectoradian is a number that is hypothesized as an irrational number although I have not proven it, this number is a fairly large number that is comparable to nonillion but even so this number is still very small when compared to other googolisms and is still in the lowest googology class 2,
because radians are derived international units that have dimensions \(L^0\) or 0, then this unit also has an SI prefix just like other units in SI, although radians cannot be greater than \(2\pi\) because it will return to the beginning but prefixes such as deciradian, centiradian and milliradian are possible, so that means we can make cosecant quectoradian or \(\mathsf{csc(10^{-30} \ rad)}\)
in the madhava series, this number can be written a…
my extended operation names
WIP
Bolded entries are not created by me.
Every negative hyperoperation is the same but with In- added to the start. Here are a few examples though.
-6th hyperoperation (\(\sqrt[S,y]{x}\)): Inhexation aka. Yth super-root
-5th hyperoperation (\(\sqrt[y]{x}\)): Inpentation aka. Yth root
-4th hyperoperation (x/y): Intetration aka. Division
-3rd hyperoperation (x-y): Intrination aka. Subtraction
-2nd hyperoperation (x-1): Indunation aka. Predecessor function aka. Predecessation
-1st hyperoperation (x=x): Inmonation aka. Nullation
0th hyperoperation (x+1): Nilation aka. Successor function aka. Successation
1st hyperoperation (x+y): Monation aka. Addition
2nd hyperoperation (x*y): Dunation aka. Multiplication
3rd hyperoperation (x^y): Trination aka. Exponentiation
4th …
List of hierarchy’s
- Chronolegend's Shrinking Hierarchy
- Inching Hierarchy
- Slower Growing Hierarchy
- SO-Growing-Hierarchy
- Slow-growing hierarchy
- Eszett Hierarchy
- X Hierarchy
- Hardy hierarchy
- Reptend hierarchy
- Binomial coefficient hierarchy
- User blog:WaxPlanck/Hyper Operator Hierarchy
- Factorial-growing hierarchy (Guess)
- Notation hierarchy
- Bowers’ Hierarchy
- N-growing hierarchy
- Middle-growing hierarchy
- Moderate-growing hierarchy
- Extreme-Growing Hierarchy
- Ultra-Growing Hierarchy
- LimeLemon192/Hyper-Hierarchy (IDK)
- Super-Growing Hierarchy
- Mahlo hierarchy (idk)
- User blog:Undeadlift/Very Fast Growing Hierarchy (guess)
- User blog:Polymations/Arrow Hierarchy (idk)
- Fast-growing hierarchy
- Continous Fast Growing Hierarchy
- Quick-growing hierarchy
- Hyper Hierarchy
- Array hierarchy
- Extended array hierarchy
- Hype…
Mario’s number
Mario’s number is equal to 22^333^4444^55555^666666^7777777^88888888^999999999 = Mario
{Mario,Mario,Mario,Mario,(Mario,72)Mario} = Luigi number
RayoLuigi number(Luigi number (Mario) = Mario number
TRI function or Pyramid function
TRIANGLE/PYRAMID function
TRI(1) = 1 TRI(2) = 2²2 = ⁴2 = 65536
3
3 3
TRI(3) = 3 3 = 7.6trillion penetrated to 2 ..... TRI(n) = n pentrated to n-1 tetrated to n tetrated to n
how large is this number? (2VP6/VP(VP6)
VP number VP(VP6)
vp⁶ = vp^6 = 6[6]5[6]4[6]3[6]2[6]1
⁶vp = vp↑↑6 = vp^(vp^(vp^(vp^(vp^(vp^(vp^6))))))
₆vp = vp↑↑↑6 = vp↑↑(vp↑↑(vp↑↑(vp↑↑(vp↑↑(vp↑↑vp)))))
vp₆ = vp↑↑↑↑6 = vp↑↑↑(vp↑↑↑(vp↑↑↑(vp↑↑↑(vp↑↑↑(vp↑↑↑vp))))) = 4(vp)6
VP6 = vp₆(vp)6
vpⱽᴾ⁶ = VP6[VP6]VP6 -1.....[VP6]1
ⱽᴾ⁶vp = vp^vp^vp^vp.....^VP6
(VP6 times. )
ᵥₚ₆vp = vp↑↑vp↑↑vp↑↑vp.......↑↑VP6
(VP6 times. )
VPᵥₚ₆ = vp↑↑↑vp↑↑↑vp.....↑↑↑6
(VP6 times. ) = 4(vp)VP6
VP(VP6) = VPᵥₚ₆(vp)VP6