User:Polygon/Better lower bound for BB(4,3)

Machine: 0RB1RZ0RB_1RC1LB2LB_1LB2RD1LC_1RA2RC0LD (bbch)

Better lower bound for BB(4,3)

Definitions

$f(n)=2^{2^{n+1}}$

$g(n)=\frac{5\times 2^{2^{f^n(0)+1}+2}-8}{9}$

$n0=\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9}$

$n1=2^{2^{32}+1}-4$

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7$

Lower bound on f^n(0)

$f(n)=2^{2^{n+1}}$

$f(0)=2^{2^1};f(f(0))=2^{2^{(2^{2^1}+1)}}>(2\uparrow {)}^{4}1;f^3(0)=2^{2^{(2^{2^{(2^{2^1}+1)}}+1)}}>(2\uparrow {)}^{6}1$

$f^k(0)=2^{2^{f^{k-1}(0)+1}}>2^{2^{f^{k-1}(0)}}=>(2\uparrow {)}^2{f}^{k-1}(0)=>(2\uparrow {)}^{2a}{f}^{k-a}(0)$

$=>f^k(0)>(2\uparrow {)}^{2k}1=2\uparrow \uparrow 2k$

$=>f^n(0)>2\uparrow \uparrow 2n$

Upper bound

$f_u(n)=2^{2^{2^n}};f_u(0)=2^{2^2};f_u^0(0)=1$

$f_u(n)>f(n)$ as $2^n>n+1$

$f_u^k(0)=(2\uparrow {)}^3{f}_u^{k-1}(0)$

$f_u^2(0)=f_u(2^{2^2})=2^{2^{2^{2^{2^2}}}}=>(2\uparrow {)}^{6}1$

$f_u^k(0)=(2\uparrow {)}^{3k}1=2\uparrow \uparrow 3k;f_u(n)>f(n)=>2\uparrow \uparrow 3k>f(n)$

$f^n(0)<2\uparrow \uparrow 3n$

Lower bound on g^k(n)

$g(n)=\frac{5\times 2^{2^{f^n(0)+1}+2}-8}{9}>\frac{5\times 2^{2^{2\uparrow \uparrow 2n}+2}-8}{9}>2^{2^{2\uparrow \uparrow 2n}}=2\uparrow \uparrow (2n+2)$

$g^2(n)>\frac{5\times 2^{2^{f^{2\uparrow \uparrow (2n+2)}(0)+1}+2}-8}{9}>\frac{5\times 2^{2^{2\uparrow \uparrow 2\uparrow \uparrow (2n+2)}+2}-8}{9}>2\uparrow \uparrow 2\uparrow \uparrow (2n+2)$

$g^k(n)>2\uparrow \uparrow (2\times g^{k-1}(n)+2)=(2\uparrow \uparrow {)}^1(2\times g^{k-1}(n)+2)=>>(2\uparrow \uparrow {)}^a(2\times g^{k-a}(n)+2)$

$g^k(n)>(2\uparrow \uparrow {)}^k(2n+2)>(2\uparrow \uparrow {)}^{k}2n$

Upper bound

$g(n)=\frac{5\times 2^{2^{f^n(0)+1}+2}-8}{9}<\frac{5\times 2^{2^{(2\uparrow \uparrow 3n)+1}+2}-8}{9}<5\times 2^{2^{(2\uparrow \uparrow 3n)+1}+2}<2\uparrow \uparrow (3n+3)$

$g^2(n)<2\uparrow \uparrow (3\times (2\uparrow \uparrow (3n+3))+3)<2\uparrow \uparrow 2\uparrow \uparrow (3n+4)$

$g^k(n)<2\uparrow \uparrow (3\times (g^{k-1}(n))+3)<2\uparrow \uparrow 2\uparrow g^{k-1}(n)$

$g^k(n)<(2\uparrow \uparrow {)}^k(3n+k+2)$

Lower bound on g^n1(n0)

$g^{n1}(n0)=g^{2^{2^{32}+1}-4}(\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9})>g^{2^{2^{32}+1}-4}(2^{2^{2^{32}}})>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}$ ; Note that $2\uparrow \uparrow 6<2^{2^{2^{32}}}<2\uparrow \uparrow 7$

$(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}2\uparrow \uparrow 6=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}6>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}4=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}2\uparrow \uparrow 2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2$

$g^{n1}(n0)>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2$

Upper bound

$g^{n1}(n0)=g^{2^{2^{32}+1}-4}(\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9})<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(3\times (\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9})+2^{2^{32}+1}-4+2)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}2\uparrow \uparrow 7=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}7$

$g^{n1}(n0)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}7$

or, for a more precise upper bound:

$g^{n1}(n0)=g^{2^{2^{32}+1}-4}(\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9})<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(3\times (\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9})+2^{2^{32}+1}-4+2)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(2^{2^{2^{33}}})$

$g^{n1}(n0)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(2^{2^{2^{33}}})$

Lower bound on f^g^n1(n0)(0)

$g^{n1}(n0)>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2$ and $f^n(0)>2\uparrow \uparrow 2n$

$f^{g^{n1}(n0)}(0)>f^{(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2}(0)>2\uparrow \uparrow (2\times (2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2)>2\uparrow \uparrow (2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2$

$(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2\uparrow \uparrow 1=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}}1=>2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

$f^{g^{n1}(n0)}(0)>2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

or, for a more precise lower bound:

$g^{n1}(n0)>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}$

$f^{g^{n1}(n0)}(0)>f^{(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}}(0)>2\uparrow \uparrow (2\times (2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}})>2\uparrow \uparrow (2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}{2}^{2^{2^{32}}}$

$f^{g^{n1}(n0)}(0)>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}{2}^{2^{2^{32}}}$

Upper bound

$g^{n1}(n0)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}7$ and $f^n(0)<2\uparrow \uparrow 3n$

$f^{g^{n1}(n0)}(0)<2\uparrow \uparrow (3\times (2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}7)<2\uparrow \uparrow (2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}8=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}8$

$g^{n1}(n0)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}8$

or, for a more precise upper bound:

$g^{n1}(n0)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(2^{2^{2^{33}}})$ and $f^n(0)<2\uparrow \uparrow 3n$

$f^{g^{n1}(n0)}(0)<2\uparrow \uparrow (3\times (2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(2^{2^{2^{33}}}))<2\uparrow \uparrow (2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}(2^{2^{2^{33}}}+1)=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}(2^{2^{2^{33}}}+1)$

$f^{g^{n1}(n0)}(0)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}(2^{2^{2^{34}}})$

Lower bound on Σ

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7>5\times 2^{2^{(2\uparrow \uparrow \uparrow 2^{2^{32}+1})+1}+2}+7>2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

Σ > $2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

or, for a more precise lower bound:

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7>2^{2^{(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}{2}^{2^{2^{32}}}}}>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}{2}^{2^{2^{32}}}$

Σ > $(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}{2}^{2^{2^{32}}}$

Upper bound

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7<5\times 2^{2^{((2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}8)+1}+2}+7<2^{2^{2^{(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}8}}}<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}9<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}65536=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2\uparrow \uparrow 4=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}4$

$(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}4=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2\uparrow \uparrow 2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}}2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}}2\uparrow \uparrow 1=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}+1}1=>2\uparrow \uparrow \uparrow (2^{2^{32}+1}+1)$

Σ < $2\uparrow \uparrow \uparrow (2^{2^{32}+1}+1)$

or, for a more precise upper bound:

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7<2^{2^{2^{(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}(2^{2^{2^{33}}}+1)}}}<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}(2^{2^{2^{33}}}+2)$

Σ < $(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}(2^{2^{2^{33}}}+2)$

General bound on Σ

$2\uparrow \uparrow \uparrow 2^{2^{32}+1}$ < Σ < $2\uparrow \uparrow \uparrow (2^{2^{32}+1}+1)$

More precisely: $(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}6$ < Σ < $(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}9$

Even more precisely: $(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}6<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}{2}^{2^{2^{32}}}$ < Σ < $(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}(2^{2^{2^{33}}}+2)<(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}7$