"Handles recursive base-2 functions with limit ordinal epsilon-0, limit of the Wainer’s hierarchy."
Rule 1 (with empty curly brackets or with just 0 in it, or if the base is 0 or 1):
A{} = A{0} = 2^A
0{X}{Q} = 1{Q}, where X can be anything inside the curly brackets, and {Q} can be any sequence of the function – First degeneration rule.
1{X}{Q} = 2{Q}, where X can be anything inside the curly brackets, and {Q} can be any sequence of the function – Second degeneration rule.
Rule 2 (with natural numbers inside {} or being appended at the end inside {}; {Q} can be any sequence of the function):
A{X + 1}{Q} = A{X}{X}{X}…{X}{X}{X}{Q} with A copies of {X}, and X can be any sequences nested inside it
Rule 3 (where the inner bracket contains {} or {0}, {S} indicates the rest of the expression inside the operating curly bracket):
A{{S}{}}{Q} = A{{S}{0}}{Q} = A{{S} + 1}{Q}
Rule 4 (using the “dot” multiplication to copy the same brackets; plus signs are optional between two brackets):
A{{S}{X}}{Q} = A{{S}+{X}}{Q}
A{{S}{}.F}{Q} = A{{S}{0}.F}{Q} = A{{S} + F}{Q}
A{{S}{X}.F}{Q} = A{{S}{X}{X}{X}…{X}{X}{X}}{Q} with F copies of {X}
We can also use the “asterisk” (*) operator as an alternative to the “dot” (.) operator for bracket multiplication as well.
Rule 5 (the rightmost part of the sequence is not “added” by a natural number):
A{{S}{1}}{Q} = A{{S} + A}{Q}
A{{S}{Y + 1}}{Q} = A{{S}{Y}.A}{Q}
A{{S}{{…{{{S}{Y + 1}}}…}}}{Q} = A{{S}{{…{{{S}{Y}.A}}…}}}{Q}
In particular, you must consult the rules for the innermost bracket level first until the bracket sequence is added by a natural number at the end before applying the rules for the particular sequence one bracket level lower.
It is important to note that the brackets are operated from left to right, and each variable in the rules must be a non-negative integer.
We can also have:
A{X}^N{Q} = A{X}{X}{X}…{X}{X}{X}{Q} with N copies of {X}.
Before we analyze the growth rate of the function, we have to provide some examples using the binary bisector notation as follows (with Knuth’s up-arrow notation for approximations in addition):
0{} = 0{0} = 2^0 = 1 (trivial solution using the rule 1)
1{} = 1{0} = 2^1 = 2 (another trivial solution using the rule 1)
8{} = 8{0} = 2^8 = 256 (using the rule 1)
4{}{} = 4{0}{0} = (2^4){0} = 16{0} = 2^16 = 65,536 (using the rule 1 again, with two or more consecutive brackets at the outer layer of the expression, operate it from left to right first)
2{1} = 2{0}{0} = (2^2){0} = 4{0} = 2^4 = 16 (using the rule 2 then rule 1)
3{1} = 3{0}{0}{0} = 8{0}{0} = 256{0} = 2^256 ≈ 1.157920892373162 × 10^77 (using the rule 2 first)
4{1} = 4{0}{0}{0}{0} = 16{0}{0}{0} = 65,536{0}{0} = (2^65,536){0} = 2^2^65,536 = 2^^6 (same rules as in the previous example)
5{1} = 5{0}{0}{0}{0}{0} = 32{0}{0}{0}{0} = (2^32){0}{0}{0} = 4,294,967,296{0}{0}{0} = (2^4,294,967,296){0}{0} = (2^2^4,294,967,296){0} = 2^2^2^4,294,967,296 (same rules as in the previous example)
2{2} = 2{1}{1} = 2{0}{0}{1} = 4{0}{1} = 16{1} (using the rule 2 first)
= 16{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0} (with 16 {0}’s)
= 65,536{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0}{0} (with 15 {0}’s)
= 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^65,536 (with 15 2’s)
= 2^^19
3{2} = 3{1}{1}{1} = 3{0}{0}{0}{1}{1} (same rules as in the previous example)
= 8{0}{0}{1}{1} = 256{0}{1}{1} = (2^256){1}{1}
= (2^256){0}{0}{0}{0}…{0}{0}{0}{0}{1} with 2^256 {0}’s
≈ (2^^2^256){1} ≈ 2^^2^^2^256
2{3} = 2{2}{2} = 2{1}{1}{2} = 16{1}{2} (same rules as in the previous example)
= (2^^19){2} ≈ 2^^^2^^19
2{4} = 2{3}{3} = 2{2}{2}{3} = 2{1}{1}{2}{3} = 16{1}{2}{3} (same rules as in the previous example)
= (2^^19){2}{3} ≈ (2^^^2^^19){3} ≈ 2^^^^2^^^2^^19
2{5} = 2{4}{4} = 2{3}{3}{4} = 2{2}{2}{3}{4} = 2{1}{1}{2}{3}{4} (same rules as in the previous example)
= 16{1}{2}{3}{4} = (2^^19){2}{3}{4} ≈ (2^^^2^^19){3}{4}
≈ (2^^^^2^^^2^^19){4} ≈ 2^^^^^2^^^^2^^^2^^19
You can see that A{1} is equal to 2^2^2^…^2^2^2^A with A 2’s, and thus is also equal to E[2]A#A in Saibian’s Hyper-E notation, and A{B} is approximately 2^^^…^^^A with B+1 arrows.
Moving on the nested brackets:
4{{0}} = 4{1} = 2^2^65,536 = 2^^6 (using the rule 3, see the detailed step-by-step solution in the previous examples)
4{{1}}
= 4{{0}.4} = 4{4} (using the rules 4 and 5)
= 4{3}{3}{3}{3} = 4{2}{2}{2}{2}{3}{3}{3} = 4{1}{1}{1}{1}{2}{2}{2}{3}{3}{3}
= 4{0}{0}{0}{0}{1}{1}{1}{2}{2}{2}{3}{3}{3} (using the rule 2)
= 16{0}{0}{0}{1}{1}{1}{2}{2}{2}{3}{3}{3} (using the rule 1)
= 65,536{0}{0}{1}{1}{1}{2}{2}{2}{3}{3}{3} (using the rule 1)
= (2^65,536){0}{1}{1}{1}{2}{2}{2}{3}{3}{3} (using the rule 1)
= (2^2^65,536){1}{1}{1}{2}{2}{2}{3}{3}{3} (using the rule 1)
≈ (2^^2^2^65,536){1}{1}{2}{2}{2}{3}{3}{3} (using the rule 2 again)
≈ (2^^2^^2^2^65,536){1}{2}{2}{2}{3}{3}{3} (using the rule 2 again)
≈ (2^^2^^2^^2^2^65,536){2}{2}{2}{3}{3}{3} (using the rule 2 again, etc.)
≈ (2^^^2^^2^^2^^2^2^65,536){2}{2}{3}{3}{3} ≈ (2^^^2^^^2^^2^^2^^2^2^65,536){2}{3}{3}{3}
≈ (2^^^2^^^2^^^2^^2^^2^^2^2^65,536){3}{3}{3} ≈ (2^^^^2^^^2^^^2^^^2^^2^^2^^2^2^65,536){3}{3}
≈ (2^^^^2^^^^2^^^2^^^2^^^2^^2^^2^^2^2^65,536){3}
≈ 2^^^^2^^^^2^^^^2^^^2^^^2^^^2^^2^^2^^2^2^65,536 ≈ 2^^^^^5
2{{1} + 1}
= 2{{1}}{{1}} (using the rule 2)
= 2{2}{{1}} (using the rule 5)
= 2{1}{1}{{1}} = 2{0}{0}{1}{{1}} (using the rule 2)
= 4{0}{1}{{1}} = 16{1}{{1}} (using the rule 1)
= (2^^19){{1}} (using the rule 2, detailed steps is already explained for 16{1} in the previous examples)
= (2^^19){{0}.(2^^19)} = (2^^19){2^^19} (using the rule 5)
≈ f{ω}(f{3}(17)) in the fast-growing hierarchy (using the Wainer’s hierarchy) – we can approximate in the fast-growing hierarchy that N{1} ≈ f3(N − 1) for N ≥ 5.
3{{1}.2}
= 3{{1}{1}} = 3{{1} + {1}} = 3{{1} + {0}.3} = 3{{1} + 3} (using the rules 3, 4, and 5)
= 3{{1} + 2}{{1} + 2}{{1} + 2} = 3{{1} + 1}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 2)
= 3{{1}}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 2)
= 3{3}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rules 3, 4, and 5)
= 3{2}{2}{2}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 2)
= 3{1}{1}{1}{2}{2}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 2)
= 3{0}{0}{0}{1}{1}{2}{2}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 2)
= 8{0}{0}{1}{1}{2}{2}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 1)
= 256{0}{1}{1}{2}{2}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 1)
= (2^256){1}{1}{2}{2}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2} (using the rule 1)
= …
≈ f{ω+2}^2(f{ω+1}^2(f{ω}^2(f{4}^2(f{3}^2(f{2}(248)))))) in the fast-growing hierarchy (using the Wainer’s hierarchy)
3{{2}}
= 3{{1}.3} = 3{{1}{1}{1}} = 3{{1}{1} + 3} (using the rules 3, 4, and 5)
= 3{{1}{1} + 2}{{1}{1} + 2}{{1}{1} + 2} (using the rule 2)
= 3{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rule 2)
= 3{{1}{1}}{{1}{1}}{{1}{1}}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rule 2)
= 3{{1} + 3}{{1}{1}}{{1}{1}}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rules 3, 4, and 5)
= 3{{1} + 2}{{1} + 2}{{1} + 2}{{1}{1}}{{1}{1}}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rule 2)
= 3{{1} + 1}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2}{{1}{1}}{{1}{1}}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rule 2)
= 3{{1}}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2}{{1}{1}}{{1}{1}}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rule 2)
= 3{3}{{1}}{{1}}{{1} + 1}{{1} + 1}{{1} + 2}{{1} + 2}{{1}{1}}{{1}{1}}{{1}{1} + 1}{{1}{1} + 1}{{1}{1} + 2}{{1}{1} + 2} (using the rules 3, 4, and 5)
= …
≈ f{ω2+2}^2(f{ω2+1}^2 (f{ω2}^2(f{ω+2}^2(f{ω+1}^2(f{ω}^2(f{5}(3))))))) ≈ f{ω3}(3) = f{ω^2}(3) in the fast-growing hierarchy (using the Wainer’s hierarchy)
4{{{1}}}
= 4{{4}} = 4{{3}.4} = 4{{3}{3}{3}{3}} (using the rules 3, 4, and 5)
= 4{{3}{3}{3}{2}{2}{2}{2}} = 4{{3}{3}{3}{2}{2}{2}{1}{1}{1}{1}} (using the rules 4 and 5)
= 4{{3}{3}{3}{2}{2}{2}{1}{1}{1} + 4} (using the rules 3, 4, and 5)
= 4{{3}{3}{3}{2}{2}{2}{1}{1}{1} + 3}{{3}{3}{3}{2}{2}{2}{1}{1}{1} + 3}{{3}{3}{3}{2}{2}{2}{1}{1}{1} + 3}{{3}{3}{3}{2}{2}{2}{1}{1}{1} + 3} (using the rule 2)
= …
≈ f{ω^ω}(4) = f{ω^4}(4) = f{ω^3*4}(4) = f{ω^3*3+ω^2*4}(4) = f{ω^3*3+ω^2*3+ω4}(4) = f{ω^3*3+ω^2*3+ω3+4}(4) = f{ω^3*3+ω^2*3+ω3+3}^4(4) in the fast-growing hierarchy (using the Wainer’s hierarchy)
4{{{{1}}}} = 4{{{4}}} = 4{{{3}.4}} = 4{{{3}.3 + {3}}} (using the rules 3, 4, and 5)
= 4{{{3}.3 + {2}.4}} = 4{{{3}.3 + {2}.3 + {2}}}} (using the rules 4 and 5)
= 4{{{3}.3 + {2}.3 + {1}.4}}} = 4{{{3}.3 + {2}.3 + {1}.3 + 4}}} (using the rules 3, 4, and 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.4}} (using the rules 4 and 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 3}}} (using the rules 4 and 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.4}} (using the rule 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}}} (using the rules 4 and 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.3 + {{3}.3 + {2}.3 + {1}.3 + 1}.4}} (using the rule 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.3 + {{3}.3 + {2}.3 + {1}.3 + 1}.3 + {{3}.3 + {2}.3 + {1}.3 + 1}}} (using the rules 4 and 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.3 + {{3}.3 + {2}.3 + {1}.3 + 1}.3 + {{3}.3 + {2}.3 + {1}.3}.4}} (using the rule 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.3 + {{3}.3 + {2}.3 + {1}.3 + 1}.3 + {{3}.3 + {2}.3 + {1}.3}.3 + {{3}.3 + {2}.3 + {1}.3}}} (using the rules 4 and 5)
= 4{{{3}.3 + {2}.3 + {1}.3 + 3}.3 + {{3}.3 + {2}.3 + {1}.3 + 2}.3 + {{3}.3 + {2}.3 + {1}.3 + 1}.3 + {{3}.3 + {2}.3 + {1}.3}.3 + {{3}.3 + {2}.3 + {1}.2}.4}} (using the rules 4 and 5)
= …
≈ f{ε0}(4) = f{ω^ω^ω}(4) = f{ω^ω^4}(4) = f{ω^(ω^3*4)}(4) in the fast-growing hierarchy (using the Wainer’s hierarchy)
Using the Wainer’s hierarchy, we can note that the natural number inside the brackets represents the exponent, and the nested curly brackets represents the omega-exponentiation. For instance, {1} is analogous to ω in the fast-growing hierarchy ordinal, {1} + 1 is analogous to ω + 1, {1}.2 is analogous to ω2, {2} is analogous to ω^2, {3} is analogous to ω^3, {{1}} is analogous to ω^ω, {{{1}}} is analogous to ω^ω^ω, and so on.
Keep in mind that in the basic level, A{N} is approximately 2^^^…^^^(A + 1) with N + 1 arrows in Knuth’s up-arrow notation, so that it is comparable to f{N + 2}(A) in the fast-growing hierarchy (FGH). But, N{1} ≈ f{3}(N − 3) in the FGH.
The limit of the primitive binary bisector notation is N{{{…{{{1}}}…}}} with N pairs of curly brackets is approximately f{ε0}(N) in the fast-growing hierarchy, the limit of the Wainer’s hierarchy. The limit is also eventually outgrows any function provable recursive in Peano arithmetic (PA) and arithmetical comprehension (ACA0). Before defining the two-entry “pair sequence” level of the binary bisector notation, first, we define N{{0, 1}} and N{{0, 1}, 0} to be equal to N{{{…{{{0}}}…}}} with N pairs of curly brackets, which eventually diagonalizes the limit of the primitive binary bisector notation.
Since A{{1} + 1} is equal to A{{1}}{{1}}{{1}}…{{1}}{{1}}{{1}} with A copies of {{1}}, as 3{{1} + 1} = 3{{1}}{{1}}{{1}} = 3{3}{{1}}{{1}} = 3{2}{2}{2}{{1}}{{1}} = … (keep in mind that 3{1} = 3{0}{0}{0} = 2^256, which is significantly larger than 3↑↑3 (3 tetrated to 3) = 7,625,597,484,987, so 3{2} = 3{1}{1}{1} = 2^256{1}{1} > 3^^^3, and 3{3} > 3^^^^3 = G1), so 3{{1} + 1} is slightly greater than G3, using the existing G function used to define Graham’s number:
G0 = 4
G1 = 3^^^^3
G(n+1) = 3^^^^...^^^^3 with G(n) arrows
How large are G3 and 3{{1} + 1}? Let’s show why.
G3 = 3{3{3^^^^3}3}3 ≈ 3{1}{1}{2}{2}{{1}}{{1}}
2{{1} + 1} = 2{{1}}{{1}} = 2{2}{{1}} = 2{1}{1}{{1}} = 2{0}{0}{1}{{1}} = 4{0}{1}{{1}} = 16{1}{{1}} = (2^^19){{1}}
3{{1} + 1} = 3{{1}}{{1}}{{1}} = 3{3}{{1}}{{1}} = 3{2}{2}{2}{{1}}{{1}} = 3{1}{1}{1}{2}{2}{{1}}{{1}} = 3{0}{0}{0}{1}{1}{2}{2}{{1}}{{1}} = 256{0}{1}{1}{2}{2}{{1}}{{1}} = 2^256{1}{1}{2}{2}{{1}}{{1}}
We know that 3{{1} + 1} > G3, 4{{1} + 1} > G4, 5{{1} + 1} > G5, etc. So that k{{1} + 1} > G(k) for k ≥ 3.
Therefore, 64{{1} + 1} is slightly larger than Graham’s number (G64).
Furthermore, since the binary bisector notation is based entirely on the binary numeral system, while the G function is based on the base-3 numeral system, Graham’s number cannot be expressed in the binary bisector notation at all.
I has an idea to define the extension of N{1} to positive real numbers (specifically non-integers), as I already defined the value of N{1} in the section above.
Before making a formal definition, I will make some ideal and fundamental halves and quarters for each N{1}. Note: While N{0} is generally defined specifically for non-negative integers, in this particular context, N{0} is defined as 2^N for any real numbers.
BP0 = 0{1} = 0
BP0.5 = 0{0} = 1
BP1 = 1{1} = 1{0} = 2 = 0{0}{0}
BP1.25 = (4^(1/4)){0} = (√2){0}= 2^(√2) ≈2.665144142690
BP1.5 = (√(1×4)){0} = (√4){0} = 2{0} = 4 = 1{0}{0}
BP1.75 = (4^(3/4)){0} = (2√2){0} = 2^(2√2) ≈7.102993301316
BP2 = 2{1} = 2{0}{0} = 4{0} = 16 = 1{0}{0}{0}
BP2.25 = (2^(3/4)×8^(1/4)){0}{0} = (2√2){0}{0} = 2^2^(2√2) ≈ 137.47193406
BP2.5 = (√(2×8)){0}{0} = (√16){0}{0} = 4{0}{0} = 16{0} = 65536 = 2{0}{0}{0}
BP2.75 = (2^(1/4)×8^(3/4)){0}{0} = (4√2){0}{0} = 2^2^(4√2) ≈ 5.65685425{0}{0} ≈ 50.45251384{0} ≈ 1540706880518839
BP3 = 3{1} = 3{0}{0}{0} = 8{0}{0} = 256{0} = 2^256 ≈ 1.15792089 × 10^77
BP3.5 = (√(3×16)){0}{0}{0} = (√48){0}{0}{0} ≈ 6.92820323{0}{0}{0} ≈ 121.785894{0}{0} ≈ 10^(4.583604 × 10^36)
BP4 = 4{1} = 4{0}{0}{0}{0} = 16{0}{0}{0} = 65536{0}{0} = 2^2^65536 ≈ 10^(6.031226 × 10^19727)
BP4.5 = (√(4×32)){0}{0}{0}{0} = (√128){0}{0}{0}{0} ≈ 11.31370850{0}{0}{0}{0} ≈ 2545.456153{0}{0}{0} ≈ 10^10^(5.460902 × 10^765)
BP5 = 5{1} = 5{0}{0}{0}{0}{0} = 32{0}{0}{0}{0} = 4294967296{0}{0}{0} = 2^2^2^4294967296 ≈ 10^10^10^1292913986
BP5.5 = (√(5×64)){0}{0}{0}{0}{0} = (√320){0}{0}{0}{0}{0} ≈ 17.88854382{0}{0}{0}{0}{0} ≈ 242654.4547{0}{0}{0}{0} ≈ 10^10^10^(5.598249 × 10^73045)
BP6 = 6{1} = 6{0}{0}{0}{0}{0}{0} = 64{0}{0}{0}{0}{0} = 2^2^2^2^2^64 ≈ 10^10^10^10^5553023288523357132
And here is the formal definition:
Let I to be an integer part, and f to be a fraction part, so x = I + f.
The format: BPx where x can be any positive real numbers (non-integers included).
Rule 1 (terminal rule): BP0 = 0, BP1 = 2
Rule 2 (expansion rule): BPI (for any positive integer I) = I{1} = I{0}{0}{0}…{0}{0}{0} with I copies of {0}
Rule 3 (sub-one rule): BPx = x/2 for 0 < x < 1
Rule 4 (exponential epicenter rule): Otherwise, BPx = (I^(1 - f) × 2^((I + 1) × f)){0}{0}{0}…{0}{0}{0} with I copies of {0}
In the basic level, A{B} ≈ 2^^^…^^^(A + 1) with B + 1 arrows in Knuth’s up-arrow notation.
A{1} is equal to E[2]A#A in Saibian’s Hyper-E notation.
333{0} is the smallest number in the form of N{0} that is larger than a googol, and 334{0}{0} is the smallest number in the form of N{0}{0} that is larger than a googolplex, by the specific properties of the powers of two. In the scientific notation, 333{0} is approximately 1.74980058 × 10^100, and 334{0}{0} is approximately 101.05348492 × 10^100.
2 is the smallest base A in A{X} (X can be any sequences) that do not degenerate the expression.