【图片】分析BMS的第一天【葛立恒数二吧】

单行与PrSS没什么差异
(0)=1
(0)(0)=2
(0)(0)(0)=3
(0)(1)=(0)(0)(0)(0)… =ω
(0)(1)(0)=ω+1
(0)(1)(0)(0)=ω+2
(0)(1)(0)(1)=(0)(1)(0)(0)(0)… =ω+ω=ω2
(0)(1)(0)(1)(0)=ω2+1
(0)(1)(0)(1)(0)(1)=(0)(1)(0)(1)(0)(0)… =ω2+ω=ω3
(0)(1)(0)(1)(0)(1)(0)(1)=ω4
(0)(1)(0)(1)(0)(1)(0)(1)(0)(1)=ω5
(0)(1)(1)=(0)(1)(0)(1)(0)(1)…=ω×ω=ω²
(0)(1)(1)(0)=ω²+1
(0)(1)(1)(0)(1)=ω²+ω
(0)(1)(1)(0)(1)(0)(1)=ω²+ω2
(0)(1)(1)(0)(1)(1)=(0)(1)(1)(0)(1)(0)(1)(0)(1)… =ω²2
(0)(1)(1)(0)(1)(1)(0)(1)=ω²2+ω
(0)(1)(1)(0)(1)(1)(0)(1)(0)(1)=ω²2+ω2
(0)(1)(1)(0)(1)(1)(0)(1)(1)=ω²3
(0)(1)(1)(0)(1)(1)(0)(1)(1)(0)(1)=ω²3+ω
(0)(1)(1)(0)(1)(1)(0)(1)(1)(0)(1)(1)=ω²4
(0)(1)(1)(1)=(0)(1)(1)(0)(1)(1)… =ω²×ω=ω³
(0)(1)(1)(1)(0)=ω³+1
(0)(1)(1)(1)(0)(1)=ω³+ω
(0)(1)(1)(1)(0)(1)(1)=ω³+ω²
(0)(1)(1)(1)(0)(1)(1)(1)=ω³2
(0)(1)(1)(1)(0)(1)(1)(1)(0)(1)(1)(1)=ω³3
(0)(1)(1)(1)(1)=(0)(1)(1)(1)(0)(1)(1)(1)…=ω⁴
(0)(1)(1)(1)(1)(0)(1)(1)(1)(1)=ω⁴2
(0)(1)(1)(1)(1)(1)=ω⁵
(0)(1)(1)(1)(1)(1)(1)=ω⁶
(0)(1)(1)(1)(1)(1)(1)(1)=ω⁷
(0)(1)(2)=(0)(1)(1)(1)… =ω^ω
(0)(1)(2)(0)=ω^ω+1
(0)(1)(2)(0)(1)=ω^ω+ω
(0)(1)(2)(0)(1)(1)=ω^ω+ω²
(0)(1)(2)(0)(1)(1)(1)=ω^ω+ω³
(0)(1)(2)(0)(1)(2)=ω^ω2
(0)(1)(2)(0)(1)(2)(0)(1)(2)=ω^ω3
(0)(1)(2)(1)=ω^ω×ω=ω^(ω+1)
(0)(1)(2)(1)(0)(1)(2)(1)=ω^(ω+1)2
(0)(1)(2)(1)(1)=ω^(ω+2)
(0)(1)(2)(1)(1)(1)=ω^(ω+3)
(0)(1)(2)(1)(2)=ω^(ω2)
(0)(1)(2)(1)(2)(1)=ω^(ω2+1)
(0)(1)(2)(1)(2)(1)(1)=ω^(ω2+2)
(0)(1)(2)(1)(2)(1)(2)=ω^(ω3)
(0)(1)(2)(1)(2)(1)(2)(1)=ω^(ω3+1)
(0)(1)(2)(1)(2)(1)(2)(1)(2)=ω^(ω4)
(0)(1)(2)(2)=(0)(1)(2)(1)(2)(1)(2)…=ω^(ω×ω)=ω^ω²
(0)(1)(2)(2)(1)=ω^(ω²+1)
(0)(1)(2)(2)(1)(2)=ω^(ω²+ω)
(0)(1)(2)(2)(1)(2)(1)(2)=ω^(ω²+ω2)
(0)(1)(2)(2)(1)(2)(2)=ω^(ω²2)
写不动了

送TA礼物

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来自Android客户端1楼2024-05-03 10:18回复

xyl
TREE
10

当时第一天我可是一口气分析到BO的

IP属地:天津

来自Android客户端2楼2024-05-03 11:48

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区块城市（广州）科技发展

虚拟与现实的结合，中国支持pi币吗，一个去中心化的平行世界，立即拥有你的数字分身区块链应用、中国支持pi币吗，快速开启元宇宙生涯

2024-06-16 16:53广告

立即查看

第一天续:
(0)(1)(2)(2)(1)(2)(2)(1)(2)=ω^(ω²2+ω)
(0)(1)(2)(2)(1)(2)(2)(1)(2)(2)=ω^(ω²3)
(0)(1)(2)(2)(2)=ω^ω³
(0)(1)(2)(2)(2)(1)(2)(2)(2)=ω^(ω³2)
(0)(1)(2)(2)(2)(2)=ω^ω⁴
(0)(1)(2)(2)(2)(2)(2)=ω^ω⁵
(0)(1)(2)(3)=ω^ω^ω
(0)(1)(2)(3)(0)=ω^ω^ω+1
(0)(1)(2)(3)(1)=ω^ω^ω×ω=ω^(ω^ω+1)
(0)(1)(2)(3)(1)(2)=ω^(ω^ω+ω)
(0)(1)(2)(3)(1)(2)(3)=ω^(ω^ω2)
(0)(1)(2)(3)(2)=ω^(ω^ω×ω)=ω^(ω^(ω+1))
(0)(1)(2)(3)(2)(2)=ω^(ω^(ω+2))
(0)(1)(2)(3)(2)(3)=ω^(ω^(ω2))
(0)(1)(2)(3)(2)(3)(2)(3)=ω^(ω^(ω3))
(0)(1)(2)(3)(3)=ω^ω^ω²
(0)(1)(2)(3)(3)(2)=ω^ω^(ω²+1)
(0)(1)(2)(3)(3)(2)(3)=ω^ω^(ω²+ω)
(0)(1)(2)(3)(3)(2)(3)(3)=ω^ω^(ω²2)
(0)(1)(2)(3)(3)(3)=ω^ω^ω³
(0)(1)(2)(3)(3)(3)(2)(3)(3)(3)=ω^ω^(ω³2)
(0)(1)(2)(3)(3)(3)(3)=ω^ω^ω⁴
(0)(1)(2)(3)(4)=ω^ω^ω^ω
(0)(1)(2)(3)(4)(0)=ω^ω^ω^ω+1
(0)(1)(2)(3)(4)(1)=ω^(ω^ω^ω+1)
(0)(1)(2)(3)(4)(1)(2)(3)(4)=ω^(ω^ω^ω2)
(0)(1)(2)(3)(4)(2)=ω^ω^(ω^ω+1)
(0)(1)(2)(3)(4)(2)(3)(4)=ω^ω^(ω^ω2)
(0)(1)(2)(3)(4)(3)=ω^ω^ω^(ω+1)
(0)(1)(2)(3)(4)(3)(4)=ω^ω^ω^(ω2)
(0)(1)(2)(3)(4)(4)=ω^ω^ω^ω²
(0)(1)(2)(3)(4)(4)(4)=ω^ω^ω^ω³
(0)(1)(2)(3)(4)(5)=ω^ω^ω^ω^ω
(0)(1)(2)(3)(4)(5)(0)=ω^ω^ω^ω^ω+1
(0)(1)(2)(3)(4)(5)(1)=ω^(ω^ω^ω^ω+1)
(0)(1)(2)(3)(4)(5)(2)=ω^ω^(ω^ω^ω+1)
(0)(1)(2)(3)(4)(5)(3)=ω^ω^ω^(ω^ω+1)
(0)(1)(2)(3)(4)(5)(4)=ω^ω^ω^ω^(ω+1)
(0)(1)(2)(3)(4)(5)(5)=ω^ω^ω^ω^ω²
(0)(1)(2)(3)(4)(5)(5)(5)=ω^ω^ω^ω^ω³
(0)(1)(2)(3)(4)(5)(6)=ω^ω^ω^ω^ω^ω
(0)(1)(2)(3)(4)(5)(6)(6)=ω^ω^ω^ω^ω^ω²
(0)(1)(2)(3)(4)(5)(6)(7)=ω^ω^ω^ω^ω^ω^ω
(0)(1)(2)(3)(4)(5)(6)(7)(8)=ω^ω^ω^ω^ω^ω^ω^ω
(0)(1)(2)(3)(4)(5)(6)(7)(8)(9)=ω^ω^ω^ω^ω^ω^ω^ω^ω
(0)(1,1)=(0)(1)(2)(3)…=ω^ω^ω^… =ε₀
0 1
0 1

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来自Android客户端3楼2024-05-03 12:40

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第二天
0 1 0
0 1 0
(0)(1,1)(0)=ε₀+1
0 1 1
0 1 0
(0)(1,1)(1)=ε₀ω
0 1 1 2
0 1 0 0
(0)(1,1)(1)(2)=ε₀ω^ω
0 1 1 2
0 1 0 0
(0)(1,1)(1)(2)(3)=ε₀ω^ω^ω
0 1 1 2
0 1 0 1
(0)(1,1)(1)(2,1)=ε₀²
0 1 1 2 1
0 1 0 1 0
(0)(1,1)(1)(2,1)(1)=ε₀²ω
0 1 1 2 1 2
0 1 0 1 0 0
(0)(1,1)(1)(2,1)(1)(2)=ε₀²ω^ω
0 1 1 2 1 2 3
0 1 0 1 0 0 0
(0)(1,1)(1)(2,1)(1)(2)(3)=ε₀²ω^ω^ω
0 1 1 2 1 2
0 1 0 1 0 1
(0)(1,1)(1)(2,1)(1)(2,1)=ε₀³
0 1 1 2 1 2 1
0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(1)(2,1)(1)=ε₀³ω
0 1 1 2 1 2 1 2
0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(1)(2,1)(1)(2)=ε₀³ω^ω
0 1 1 2 1 2 1 2
0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(1)(2,1)(1)(2,1)=ε₀⁴
0 1 1 2 1 2 1 2 1
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(1)(2,1)(1)(2,1)(1)=ε₀⁴ω
0 1 1 2 1 2 1 2 1 2
0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(1)(2,1)(1)(2,1)(1)(2,1)=ε₀⁵
0 1 1 2 1 2 1 2 1 2 1 2
0 1 0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(1)(2,1)(1)(2,1)(1)(2,1)(1)(2,1)=ε₀⁶
0 1 1 2 2
0 1 0 1 0
(0)(1,1)(1)(2,1)(2)=ε₀^ω
0 1 1 2 2 1 2
0 1 0 1 0 0 1
(0)(1,1)(1)(2,1)(2)(1)(2,1)=ε₀^(ω+1)
0 1 1 2 2 1 2 1 2
0 1 0 1 0 0 1 0 1
(0)(1,1)(1)(2,1)(2)=ε₀^(ω+2)
0 1 1 2 2 1 2 2
0 1 0 1 0 0 1 0
(0)(1,1)(1)(2,1)(2)(1)(2,1)(2)=ε₀^(ω2)
0 1 1 2 2 1 2 2 1 2
0 1 0 1 0 0 1 0 0 1
(0)(1,1)(1)(2,1)(2)(1)(2,1)(2)(1)(2,1)=ε₀^(ω2+1)
0 1 1 2 2 1 2 2 1 2 2
0 1 0 1 0 0 1 0 0 1 0
(0)(1,1)(1)(2,1)(2)(1)(2,1)(2)(1)(2,1)(2)=ε₀^(ω3)
0 1 1 2 2 2
0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(2)=ε₀^ω²
0 1 1 2 2 2 1 2 2 2
0 1 0 1 0 0 0 1 0 0
(0)(1,1)(1)(2,1)(2)(2)(1)(2,1)(2)(2)=ε₀^(ω²2)
0 1 1 2 2 2 2
0 1 0 1 0 0 0
(0)(1,1)(1)(2,1)(2)(2)(2)=ε₀^ω³
0 1 1 2 2 2 2 2
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(2)(2)(2)=ε₀^ω⁴
0 1 1 2 2 3
0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3)=ε₀^ω^ω
0 1 1 2 2 3 2
0 1 0 1 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(2)=ε₀^ω^(ω+1)
0 1 1 2 2 3 2 3
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(2)(3)=ε₀^ω^(ω2)
0 1 1 2 2 3 3
0 1 0 1 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(3)=ε₀^ω^ω²
0 1 1 2 2 3 3 3
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(3)(3)=ε₀^ω^ω³
0 1 1 2 2 3 4
0 1 0 1 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)=ε₀^ω^ω^ω
字数竟然满了

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来自Android客户端4楼2024-05-04 08:11

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xyl
TREE
10

这辈子也分析不到ζ0（）

IP属地:天津

来自Android客户端6楼2024-05-04 15:29

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xyl
TREE
10

(0)(1,1)(1)(2,1)(2)(3,1)=？
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)=?
(0)(1,1)(1,1)=?
(0)(1,1)(1,1)(1)(2,1)=?
(0)(1,1)(1,1)(1)(2,1)(2,1)=?
(0)(1,1)(1,1)(1,1)=?
(0)(1,1)(2)=?
(0)(1,1)(2)(1,1)=?
(0)(1,1)(2)(2)=?
(0)(1,1)(2)(3)=?
(0)(1,1)(2)(3,1)=?
(0)(1,1)(2)(3,1)(4)(5,1)=?
(0)(1,1)(2,1)=?

IP属地:天津

来自Android客户端7楼2024-05-04 20:50

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xyl
TREE
10

(0)(1,1)(2,1)(1)=?
(0)(1,1)(2,1)(1)(2,1)(3,1)=?
(0)(1,1)(2,1)(1)(2,1)(3,1)(2)(3,1)(4,1)=?
(0)(1,1)(2,1)(1,1)=?
(0)(1,1)(2,1)(1,1)(1,1)=?
(0)(1,1)(2,1)(1,1)(2)=?
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)=?
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(2)=?
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(2)(3,1)(4,1)=?
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(3)=?
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(3)(4,1)(5,1)=?
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(3,1)=?
(0)(1,1)(2,1)(1,1)(2,1)=?

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来自Android客户端8楼2024-05-04 21:14

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第二天续
0 1 1 2 2 3 4 1
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(1)=ε₀^(ω^ω^ω+1)
0 1 1 2 2 3 4 2
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(2)=ε₀^ω^(ω^ω+1)
0 1 1 2 2 3 4 3
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(3)=ε₀^ω^ω^(ω+1)
0 1 1 2 2 3 4 4
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(4)=ε₀^ω^ω^ω²
0 1 1 2 2 3 4 4 4
0 1 0 1 0 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(4)(4)=ε₀^ω^ω^ω³
0 1 1 2 2 3 4 5
0 1 0 1 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(5)=ε₀^ω^ω^ω^ω
0 1 1 2 2 3 4 5 5
0 1 0 1 0 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(5)(5)=ε₀^ω^ω^ω^ω²
0 1 1 2 2 3 4 5 6
0 1 0 1 0 0 0 0 0
(0)(1,1)(1)(2,1)(2)(3)(4)(5)(6)=ε₀^ω^ω^ω^ω^ω
0 1 1 2 2 3
0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)=ε₀^ε₀
0 1 1 2 2 3 1
0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(1)=ε₀^ε₀ω
0 1 1 2 2 3 1 2
0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2)=ε₀^ε₀ω^ω
0 1 1 2 2 3 1 2
0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2,1)=ε₀^(ε₀+1)
0 1 1 2 2 3 1 2 2
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2,1)(2)=ε₀^(ε₀+ω)
0 1 1 2 2 3 1 2 2 3
0 1 0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2,1)(2)(3)=ε₀^(ε₀+ω^ω)
0 1 1 2 2 3 1 2 2 3
0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2,1)(2)(3,1)=ε₀^(ε₀2)
0 1 1 2 2 3 1 2 2 3 1 2
0 1 0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2,1)(2)(3,1)(1)(2,1)=ε₀^(ε₀2+1)
0 1 1 2 2 3 1 2 2 3 1 2 2 3
0 1 0 1 0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(1)(2,1)(2)(3,1)(1)(2,1)(2)(3,1)=ε₀^(ε₀3)
0 1 1 2 2 3 2 1 2 2 3
0 1 0 1 0 1 0 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(2)(1)(2,1)(2)(3,1)=ε₀^(ε₀ω+ε₀)
0 1 1 2 2 3 2 1 2 2 3 2
0 1 0 1 0 1 0 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(2)(1)(2,1)(2)(3,1)(2)=ε₀^(ε₀ω2)
再分析下去我就要释怀的死了

IP属地:四川

来自Android客户端9楼2024-05-04 22:01

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xyl
TREE
10

(0)(1,1)(2,1)(1,1)(2,1)(1,1)=?
(0)(1,1)(2,1)(1,1)(2,1)(1,1)(2)=?
(0)(1,1)(2,1)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(3,1)(4,1)(3,1)=?
(0)(1,1)(2,1)(1,1)(2,1)(1,1)(2,1)=?
(0)(1,1)(2,1)(2)=?
(0)(1,1)(2,1)(2)(1,1)(2,1)=?
(0)(1,1)(2,1)(2)(1,1)(2,1)(2)=?
(0)(1,1)(2,1)(2)(2)=?
(0)(1,1)(2,1)(2)(3,1)=?
(0)(1,1)(2,1)(2)(3,1)(4,1)=?
(0)(1,1)(2,1)(2,1)=?

IP属地:天津

来自Android客户端10楼2024-05-04 22:30

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BMS我还不会

只会单行BMS

IP属地:四川

来自Android客户端11楼2024-05-05 02:10

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第三天(绷不住了，我要提速

)
0 1 1 2 2 3 2 2
0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3,1)(2)(2)=ε₀^(ε₀ω²)
0 1 1 2 2 3 2 3
0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3,1)(2)(3)=ε₀^(ε₀ω^ω)
0 1 1 2 2 3 2 3
0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(2)(3,1)=ε₀^ε₀²
0 1 1 2 2 3 2 3 1
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(2)(3,1)(1)=ε₀^ε₀²ω
0 1 1 2 2 3 2 3 2
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(2)(3,1)(2)=ε₀^(ε₀²ω)
0 1 1 2 2 3 2 3 2 3
0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(2)(3,1)(2)(3,1)=ε₀^ε₀³
0 1 1 2 2 3 3
0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)=ε₀^ε₀^ω
0 1 1 2 2 3 3 3
0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(3)=ε₀^ε₀^ω²
0 1 1 2 2 3 3 4
0 1 0 1 0 1 0 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4)=ε₀^ε₀^ω^ω
0 1 1 2 2 3 3 4
0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)=ε₀^ε₀^ε₀
0 1 1 2 2 3 3 4 1
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(1)=ε₀^ε₀^ε₀ω
0 1 1 2 2 3 3 4 2
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(2)=ε₀^(ε₀^ε₀ω)
0 1 1 2 2 3 3 4 3
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(3)=ε₀^ε₀^(ε₀ω)
0 1 1 2 2 3 3 4 3 4
0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(3)(4,1)=ε₀^ε₀^ε₀²
0 1 1 2 2 3 3 4 4
0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)=ε₀^ε₀^ε₀^ω
0 1 1 2 2 3 3 4 4 5
0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)=ε₀^ε₀^ε₀^ε₀
0 1 1 2 2 3 3 4 4 5 1
0 1 0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)(1)=ε₀^ε₀^ε₀^ε₀ω
0 1 1 2 2 3 3 4 4 5 2
0 1 0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)(2)=ε₀^(ε₀^ε₀^ε₀ω)
0 1 1 2 2 3 3 4 4 5 3
0 1 0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)(3)=ε₀^ε₀^(ε₀^ε₀ω)
0 1 1 2 2 3 3 4 4 5 4
0 1 0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)(4)=ε₀^ε₀^ε₀^(ε₀ω)
0 1 1 2 2 3 3 4 4 5 5
0 1 0 1 0 1 0 1 0 1 0
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)(5)=ε₀^ε₀^ε₀^ε₀^ω
0 1 1 2 2 3 3 4 4 5 5 6
0 1 0 1 0 1 0 1 0 1 0 1
(0)(1,1)(1)(2,1)(2)(3,1)(3)(4,1)(4)(5,1)(5)(6,1)=ε₀^ε₀^ε₀^ε₀^ε₀
又满了

IP属地:四川

来自Android客户端12楼2024-05-05 10:46

xyl
TREE
10

(0)(1,1)(2,1)(2,1)(1,1)=?
(0)(1,1)(2,1)(2,1)(1,1)(2,1)=?
(0)(1,1)(2,1)(2,1)(1,1)(2,1)(2,1)=?
(0)(1,1)(2,1)(2,1)(2)=?
(0)(1,1)(2,1)(2,1)(2,1)=?
(0)(1,1)(2,1)(3)=?
(0)(1,1)(2,1)(3)(2,1)=?
(0)(1,1)(2,1)(3)(2,1)(3)=?
(0)(1,1)(2,1)(3)(3)=?
(0)(1,1)(2,1)(3)(4)=?
(0)(1,1)(2,1)(3)(4,1)=?
(0)(1,1)(2,1)(3)(4,1)(5,1)(6)=?
(0)(1,1)(2,1)(3)(4,1)(5,1)(6)(7,1)(8,1)(9)=?
(0)(1,1)(2,1)(3,1)=?

IP属地:天津

来自Android客户端13楼2024-05-05 11:05

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xyl
TREE
10

Γ0=ψ(Ω^Ω)
Γ0*ω=ψ(Ω^Ω+1)
Γ0^2=ψ(Ω^Ω+ψ(Ω^Ω))
Γ0^Γ0=ψ(Ω^Ω+ψ(Ω^Ω+ψ(Ω^Ω)))
ε(Γ0+1)=ψ(Ω^Ω+Ω)
ε(Γ0+2)=ψ(Ω^Ω+Ω2)
ε(Γ0*2)=ψ(Ω^Ω+Ωψ(Ω^Ω))
εε(Γ0+1)=ψ(Ω^Ω+Ωψ(Ω^Ω+Ω))
ζ(Γ0+1)=ψ(Ω^Ω+Ω^2)
η(Γ0+1)=ψ(Ω^Ω+Ω^3)
φ(ω,Γ0+1)=ψ(Ω^Ω+Ω^ω)
φ(ε0,Γ0+1)=ψ(Ω^Ω+Ω^ψ(Ω))
φ(φ(φ(ω,0),0),Γ0+1)=ψ(Ω^Ω+Ω^ψ(Ω^ψ(Ω^ω)))
φ(Γ0,1)=ψ(Ω^Ω+Ω^ψ(Ω^Ω))
φ(Γ0,2)=ψ(Ω^Ω+Ω^ψ(Ω^Ω)*2)
φ(Γ0+1,0)=ψ(Ω^Ω+Ω^(ψ(Ω^Ω)+1))
φ(Γ0*2,0)=ψ(Ω^Ω+Ω^(ψ(Ω^Ω)*2))
φ(ε(Γ0+1),0)=ψ(Ω^Ω+Ω^ψ(Ω^Ω+Ω))
φ(φ(Γ0,1),0)=ψ(Ω^Ω+Ω^ψ(Ω^Ω+Ω^ψ(Ω^Ω)))
φ(φ(φ(Γ0,1),0),0)=ψ(Ω^Ω+Ω^ψ(Ω^Ω+Ω^ψ(Ω^Ω+Ω^ψ(Ω^Ω))))
Γ1=ψ(Ω^Ω*2)

IP属地:天津

来自Android客户端14楼2024-05-05 11:38

xyl
TREE
10

(0)(1,1)(2,1)(3,1)(1,1)=?
(0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)=?
(0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6,1)=?
(0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6,1)(4,1)(5,1)(6)(7,1)(8,1)(9,1)=?
(0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,1)=?
(0)(1,1)(2,1)(3,1)(2)=?
(0)(1,1)(2,1)(3,1)(2)(3,1)(4,1)(5,1)=?
(0)(1,1)(2,1)(3,1)(2,1)=?

IP属地:天津

来自Android客户端15楼2024-05-05 16:07

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续15楼
(0)(1,1)(2,1)(3,1)(2,1)(2,1)=φ(1,2,0)
(0)(1,1)(2,1)(3,1)(2,1)(3)=φ(1,ω,0)
(0)(1,1)(2,1)(3,1)(2,1)(3,1)=φ(1,φ(1,0,0),0)?
(0)(1,1)(2,1)(3,1)(3)=φ(2,0,0)？
再分析下去估计错一大堆

IP属地:四川

来自Android客户端16楼2024-05-05 17:11

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