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

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

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

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:

$g^{n1}(n0)>(2\uparrow \uparrow )^{2^{2^{32}+1}-3}6$

$f^{g^{n1}(n0)}(0)>f^{(2\uparrow \uparrow )^{2^{2^{32}+1}-3}6}(0)>2\uparrow \uparrow (2\times (2\uparrow \uparrow )^{2^{2^{32}+1}-3}6)>2\uparrow \uparrow (2\uparrow \uparrow )^{2^{2^{32}+1}-3}6=>(2\uparrow \uparrow )^{2^{2^{32}+1}-2}6$

$f^{g^{n1}(n0)}(0)>(2\uparrow \uparrow )^{2^{2^{32}+1}-2}6$

Upper bound

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

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

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

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, more precisely:

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7>2^{2^{(2\uparrow \uparrow )^{2^{2^{32}+1}-2}6}}>(2\uparrow \uparrow )^{2^{2^{32}+1}-2}6$

Σ > $(2\uparrow \uparrow )^{2^{2^{32}+1}-2}6$

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

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 < $(2\uparrow \uparrow )^{2^{2^{32}+1}-2}9$