自创表示法之GEAF(G式爆炸数阵函数)【葛立恒数二吧】

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自创表示法之GEAF(G式爆炸数阵函数)

g^{n}(n)＝G(n){n}
G(1){0}＝g(1)≈f_ω＋1(1)
G(2){0}＝g(2)≈f_ω＋1(2)
G(G(1){0}){0}＝g(g(1))≈f_ω＋1(f_ω＋1(1))
G(1){1}＝g(1)≈f_ω＋2(1)＝f_ω＋1(1)
G(2){1}＝g(g(2))≈f_ω＋2(2)
G(G(1){1}){1}＝g(…g(g(1))…)≈f_ω＋2(f_ω＋1(1))
G(1){2}＝g(1)≈f_ω＋3(1)＝f_ω＋2(1)＝f_ω＋1(1)
G(2){2}＝f_ω＋3(2)
G(1){3}＝g(1)≈f_ω＋1(1)
G(2){0,1}＝f_ω2(2)
G(2){1,1}＝G(G(2){0,1}){0,1}＝f_ω2＋1(2)
G(2){0,2}＝f_ω3(2)
G(n){0,0,1}＝f_ω^2(n)
G(n){0(1)1}＝f_ω^ω(n)
G(n){0(0/1)1}＝f_ε0(n)＝f_ψ(0)(n)
G(n){0(0[lbk]1[rbk]1)1}＝f_ψ(Ω^ω)(n)
G(n){0(0[lbk]0/1[rbk]1)1}＝f_ψ(Ω^Ω)(n)
G(n){0(0[lbk]0[lbk]1[rbk]1[rbk]1)1}＝f_ψ(ψ_1(0))(n)＝BHO
G(n){0(0/_{1}1)1}＝f_ψ(Ω_2)(n)
G(n){0(0[lbk]1[rbk]_{1}1)1}＝f_ψ(Ω_2^ω)(n)
G(n){0(0/_{2}1)1}＝f_ψ(Ω_3)(n)
G(n){0(0/_{0,1}1)1}＝f_ψ(Ω_ω)(n)
G(n){0(0/_{0/1}1)1}＝f_ψ(Ω_Ω)(n)
G(n){0(0/_{0/_{1}1}1)1}＝f_ψ(Ω_Ω_2)(n)
G(n){0(0/_{0/_{0/1}1}1)1}＝f_ψ(Ω_Ω_Ω)(n)
G(n){0(0/_{0|1}1)1}＝f_ψ(ψ_I(0))(n)＝EBO
楼下继续分析

送TA礼物

IP属地:四川

来自Android客户端1楼2024-02-10 21:14回复

直接从头分析，免得又有什么不良定义
G(n){0(0/_{0|1}1)1}＝f_ψ(ψ_I(0))(n)
G(n){1(0/_{0|1}1)1}＝f_ψ(ψ_I(0))＋1(n)
G(n){0,1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ω(n)
G(n){0,1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ω(n)
G(n){0(1)1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ω^ω(n)
G(n){0(0/1)1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ψ(0)(n)
G(n){0(0/_{1}1)1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ψ(Ω_2)(n)
G(n){0(0/_{0,1}1)1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ψ(Ω_ω)(n)
G(n){0(0/_{0/1}1)1(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))＋ψ(Ω_Ω)(n)
G(n){0(0/_{0|1}1)2}
＝f_ψ(ψ_I(0))2(n)
G(n){0(0/_{0|1}1)0,1}
＝f_ψ(ψ_I(0))ω(n)
G(n){0(0/_{0|1}1)0(0/_{0|1}1)1}
＝f_ψ(ψ_I(0))^2(n)
G(n){0(1/_{0|1}1)1}
＝f_ψ(ψ_I(0))^ω(n)
G(n){0(2/_{0|1}1)1}
＝f_ψ(ψ_I(0))^(ω^2)(n)
G(n){0(0(0/1)1/_{0|1}1)1}
＝f_ψ(ψ_I(0))^ψ(0)(n)
G(n){0(0(0/_{0|1}1)1/_{0|1}1)1}
＝f_ψ(ψ_I(0))^ψ(ψ_I(0))(n)
G(n){0(0/_{0|1}2)1}
＝f_ψ(ψ_I(0)＋1)(n)
G(n){0(0/_{0|1}0/1)1}
＝f_ψ(ψ_I(0)＋Ω)(n)
G(n){0(0/_{0|1}0/_{0|1}1)1}
＝f_ψ(ψ_I(0)2)(n)
G(n){0(0/_{0|1}1/_{0|1}1)1}
＝f_ψ(ψ_I(0)2＋1)(n)
G(n){0(0/_{0|1}0/_{0|1}2)1}
＝f_ψ(ψ_I(0)3)(n)
G(n){0(0/_{0|1}1/_{0|1}0/_{0|1}1)1}
＝f_ψ(ψ_I(0)^2)(n)
G(n){0(0[1]_{0|1}1)1}
＝f_ψ(ψ_I(0)^ω)(n)
G(n){0(0[0/1]_{0|1}1)1}
＝f_ψ(ψ_I(0)^Ω)(n)
G(n){0(0[0/_{0|1}1]_{0|1}1)1}
＝f_ψ(ψ_I(0)^ψ_I(0))(n)
G(n){0(0[0[1]_{0|1}1]_{0|1}1)1}
＝f_ψ(ψ_(Ω_ψ_I(0)＋1)(0))
G(n){0(0[0[{0|1}出现2次]_{0|1}1]_{0|1}1)1}
＝
f_ψ(ψ_(Ω_ψ_I(0)＋1)(ψ_(Ω_ψ_I(0)＋1)))
G(n){0(0/_{1|1}1)1}
＝f_ψ(Ω_ψ_I(0)＋1)

IP属地:四川

来自Android客户端2楼2024-02-10 21:32

你这个g函数是葛立恒函数吗

IP属地:江苏

来自iPhone客户端3楼2024-02-10 21:46

收起回复

G(n){0(0/_{0/1|1}1)1}
＝f_ψ(Ω_ψ_I(0)＋Ω)(n)
G(n){0(0/_{0/_{0|1}1|1}1)1}
＝f_ψ(Ω_ψ_I(0)2)(n)
G(n){0(0/_{0/_{1|1}1|1}1)1}
＝f_ψ(Ω_Ω_ψ_I(0)＋1)(n)
G(n){0(0/_{0|2}1)1}
＝f_ψ(ψ_I(1))(n)
G(n){0(0/_{0|0/1}1)1}
＝f_ψ(ψ_I(Ω))(n)
G(n){0(0/_{0|0/_{0|1}1}1)1}
＝f_ψ(ψ_I(ψ_I(0)))(n)
G(n){0(0/_{0|0/_{0|2}1}1)1}
＝f_ψ(ψ_I(ψ_I(1)))(n)
G(n){0(0/_{0|0/_{0|0/_{0|1}1}1}1)1}
＝f_ψ(ψ_I(ψ_I(ψ_I(0))))(n)
G(n){0(0/_{0|0|1}1)1}
＝f_ψ(I)(n)＝f_ψ(I^2)(n)
如果以上分析正确那么
G(n){0(0/_{0{1}1}1)1}
＝f_ψ(I^ω)(n)
G(n){0(0/_{0{0/_{0|1}1}1}1)1}
＝f_ψ(I^ψ_I(0))(n)
G(n){0(0/_{0{0/_{0|2}1}1}1)1}
＝f_ψ(I^ψ_I(1))(n)
G(n){0(0/_{0{0/_{0{1}1}1}1}1)1}
＝f_ψ(I^ψ_I(I^ω))(n)
G(n){0(0/_{0|_{1}1}1)1}
＝f_ψ(ζ_(I＋1))(n)
G(n){0(0/_{0|_{2}1}1)1}
＝f_ψ(ζ_(I＋2))(n)
G(n){0(0/_{0|_{0/1}1}1)1}
＝f_ψ(ζ_(I＋Ω))(n)
G(n){0(0/_{0|_{0/_{0|1}1}1}1)1}
＝f_ψ(ζ_(I＋ψ_I(0)))(n)
G(n){0(0/_{0|_{0/_{0|_{1}1}1}1}1)1}
＝f_ψ(ζ_(ζ_(I＋1)))(n)
G(n){0(0/_{0|_{0|1}1}1)1}
＝f_ψ(η_I＋1)(n)＝f_ψ(Ω_I＋1^3)
后面不会了

IP属地:四川

来自Android客户端4楼2024-02-10 21:47