角谷猜想的证明

水流成林

[size=15.0000pt]角谷猜想的证明

[size=12.0000pt]刘 承 宁 [size=12.0000pt]13533050293

[size=12.0000pt]Email:lcn620422@163.com

[size=12.0000pt]
[size=15.0000pt]摘要[size=14.0000pt]：[size=10.5000pt]本文甪数学归纳法证明了角谷猜想，即假设[size=10.5000pt]3，5，7一直到2m+1猜想成立，用角谷猜想变换法则证明2m+3中m无论是奇数还是偶数猜想都成立，彻底证明了角谷猜想。
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[size=10.5000pt]关键词：整数的奇偶性，缩放法，数学归纳法。[size=10.5000pt]
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[size=15.0000pt]引言[size=15.0000pt]：[size=10.5000pt]本文用数学归纳法解决了数学中至今公开渠道还未彻底解决的角谷猜想，该猜想

[size=10.5000pt]通过计算机验证到了 [size=10.5000pt]7*10^11[size=10.5000pt]，于是人们有充份理由猜想它是成立的，但缺少一个令人信服[size=10.5000pt]的证明，我经过几十年的努力先是于 [size=10.5000pt]2010 [size=10.5000pt]年部份证明了它并得到数学专业人士的肯定，针对他们的建议十年以后终于完全证明了它，使角谷猜想成为角谷定理。
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[size=10.5000pt]角谷猜想指的是当 [size=10.5000pt]N[size=10.5000pt] 是偶数时做 [size=10.5000pt]N/2,N[size=10.5000pt] 是奇数时做 [size=10.5000pt]3N+1[size=10.5000pt]，如此反复运算其结果必然等于 [size=10.5000pt]1。验证：
[size=10.5000pt]N=3,有 10-5-16-8-4-2-1, N=5,有 16-8-4-2-1，
[size=10.5000pt]N=7,有 22-11-34-17-52-26-13-40-20-10-5-16-8-4-2-1，
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[size=10.5000pt]明显只要证明 [size=10.5000pt]N[size=10.5000pt] 是奇数时角谷猜想成立即可，设 [size=10.5000pt]N[size=10.5000pt] 大于 [size=10.5000pt]7，N=2m+1,m[size=10.5000pt] 只有奇数和偶数两种情[size=10.5000pt]形，先证明 [size=10.5000pt]m[size=10.5000pt] 是奇数时的情形，因 [size=10.5000pt]3，5，7[size=10.5000pt] 猜想验证成立，故可假设 [size=10.5000pt]m[size=10.5000pt] 直到奇数时,[size=10.5000pt]3，5， 7...2m+1[size=10.5000pt] 猜想成立，做 [size=10.5000pt]N+2=2m+3[size=10.5000pt].如能证明 [size=10.5000pt]2m+3[size=10.5000pt] 角谷猜想成立，则 [size=10.5000pt]m [size=10.5000pt]是奇数时该猜想成立。[size=10.5000pt]([size=10.5000pt]数学归纳法[size=10.5000pt])
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[size=10.5000pt]因 [size=10.5000pt]2m+3[size=10.5000pt] 是奇数，做 [size=10.5000pt]3*（2m+3）+1=6m+10[size=10.5000pt] 是偶数，除以 [size=10.5000pt]2[size=10.5000pt] 等于 [size=10.5000pt]3m+5[size=10.5000pt] 是偶数，设 [size=10.5000pt]3m+5=2[size=10.5000pt]k[size=10.5000pt]n[size=10.5000pt]，
[size=10.5000pt]（k 是指数，n 是奇数）
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[size=10.5000pt]当[size=10.5000pt] [size=10.5000pt]m[size=10.5000pt] [size=10.5000pt]大于[size=10.5000pt] [size=10.5000pt]3[size=10.5000pt] [size=10.5000pt]时，[size=10.5000pt]2m+[size=10.5000pt]1[size=10.5000pt]>[size=10.5000pt]（[size=10.5000pt]3m+[size=10.5000pt]5[size=10.5000pt]）[size=10.5000pt]/2>[size=10.5000pt]([size=10.5000pt]3m+[size=10.5000pt]5[size=10.5000pt])/2[size=10.5000pt]2[size=10.5000pt]>……[size=10.5000pt]>(3[size=10.5000pt]m[size=10.5000pt]+5)[size=10.5000pt]/[size=10.5000pt]2[size=10.5000pt]k[size=10.5000pt]，即[size=10.5000pt] [size=10.5000pt]N+[size=10.5000pt]2[size=10.5000pt] [size=10.5000pt]落于[size=10.5000pt] [size=10.5000pt]3[size=10.5000pt]，[size=10.5000pt]5[size=10.5000pt]，[size=10.5000pt]7[size=10.5000pt]……[size=10.5000pt]2m[size=10.5000pt]-[size=10.5000pt]1[size=10.5000pt]中，此前假设成立，故 [size=10.5000pt]m [size=10.5000pt]是奇数时角谷猜想成立。
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[size=10.5000pt]如 [size=10.5000pt]m [size=10.5000pt]是偶数，可设 [size=10.5000pt]m= [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]n[size=10.5000pt]，[size=10.5000pt](n [size=10.5000pt]是奇数[size=10.5000pt]) [size=10.5000pt]此时可假设 [size=10.5000pt]3[size=10.5000pt]，[size=10.5000pt]5[size=10.5000pt]，[size=10.5000pt]7...2m+1 [size=10.5000pt]猜想成立，做 [size=10.5000pt]N+2=2m+3,
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[size=10.5000pt]即 [size=10.5000pt]N+2= 2[size=10.5000pt]k [size=10.5000pt]+[size=10.5000pt]1 [size=10.5000pt]n+3 [size=10.5000pt]是奇数，做 [size=10.5000pt]3(N+2)+1=3 2[size=10.5000pt]k [size=10.5000pt]+[size=10.5000pt]1 [size=10.5000pt]n+10 [size=10.5000pt]是偶数，除以 [size=10.5000pt]2 [size=10.5000pt]得 [size=10.5000pt]3 2[size=10.5000pt]k [size=10.5000pt]n+5 [size=10.5000pt]是奇数，做
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[size=10.5000pt]3(3 2[size=10.5000pt]k [size=10.5000pt]n+5)+1= 3[size=10.5000pt]2 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]n+16 [size=10.5000pt]是偶数，除以 [size=10.5000pt]16 [size=10.5000pt]即 [size=10.5000pt]2[size=10.5000pt]4 [size=10.5000pt]得[size=10.5000pt]3[size=10.5000pt]2 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]4 [size=10.5000pt]n+1 [size=10.5000pt]是奇数，做 [size=10.5000pt]3( 3[size=10.5000pt]2 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]4 [size=10.5000pt]n+1)+1[size=10.5000pt]，
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[size=10.5000pt]即[size=10.5000pt]3[size=10.5000pt]3 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]4 [size=10.5000pt]n+4 [size=10.5000pt]是偶数，除以 [size=10.5000pt]4 [size=10.5000pt]得[size=10.5000pt]3[size=10.5000pt]3 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]6 [size=10.5000pt]n+1 [size=10.5000pt]是奇数，做 [size=10.5000pt]3( 3[size=10.5000pt]3 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]6 [size=10.5000pt]n+1)+1= 3[size=10.5000pt]4 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]6 [size=10.5000pt]n+4 [size=10.5000pt]是偶数，除以 [size=10.5000pt]4 [size=10.5000pt]得[size=10.5000pt]3[size=10.5000pt]4 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]8 [size=10.5000pt]n+1 [size=10.5000pt]是奇数，做 [size=10.5000pt]3( 3[size=10.5000pt]4 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]8 [size=10.5000pt]n+1)+1 [size=10.5000pt]即[size=10.5000pt]3[size=10.5000pt]5 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]8 [size=10.5000pt]n+4 [size=10.5000pt]是偶数，除以 [size=10.5000pt]4 [size=10.5000pt]得[size=10.5000pt]3[size=10.5000pt]5 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]10 [size=10.5000pt]n+1[size=10.5000pt]，
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[size=10.5000pt]如法演算即有： [size=10.5000pt]3[size=10.5000pt]6 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]12 [size=10.5000pt]n+1[size=10.5000pt]， [size=10.5000pt]3[size=10.5000pt]7 [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]-[size=10.5000pt]14 [size=10.5000pt]n+1......
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[size=12.0000pt]发现运算有规律，即[size=12.0000pt] [size=12.0000pt]3[size=12.0000pt] [size=12.0000pt]的指数每增加[size=12.0000pt] [size=12.0000pt]1[size=12.0000pt]，[size=12.0000pt]2[size=12.0000pt] [size=12.0000pt]的指数则减少[size=12.0000pt] [size=12.0000pt]2[size=12.0000pt]，设[size=12.0000pt] [size=12.0000pt]3[size=12.0000pt] [size=12.0000pt]的指数是[size=12.0000pt] [size=12.0000pt]r[size=12.0000pt]，则有
[size=10.5000pt]3[size=10.5000pt]r[size=10.5000pt]   [size=10.5000pt]2[size=10.5000pt]k[size=10.5000pt] [size=10.5000pt]-[size=10.5000pt]2[size=10.5000pt]r[size=10.5000pt]  [size=10.5000pt]n[size=10.5000pt]+[size=10.5000pt]1[size=10.5000pt]，[size=10.5000pt]因为每乘以 [size=10.5000pt]3 [size=10.5000pt]则 [size=10.5000pt]3 [size=10.5000pt]的指数增加 [size=10.5000pt]1[size=10.5000pt]，除以[size=10.5000pt]2[size=10.5000pt]2 [size=10.5000pt]则 [size=10.5000pt]2 [size=10.5000pt]的指数减少 [size=10.5000pt]2[size=10.5000pt]，过程可以有限次重复。
[size=12.0000pt]k[size=12.0000pt]-[size=12.0000pt]2[size=12.0000pt]r[size=12.0000pt] [size=12.0000pt]只能是[size=12.0000pt] [size=12.0000pt]0[size=12.0000pt]，[size=12.0000pt]1[size=12.0000pt]，[size=12.0000pt]k[size=12.0000pt] [size=12.0000pt]是偶数可为[size=12.0000pt] [size=12.0000pt]0[size=12.0000pt]，[size=12.0000pt]k[size=12.0000pt] [size=12.0000pt]是奇数可为[size=12.0000pt] [size=12.0000pt]1[size=12.0000pt]，如[size=12.0000pt] [size=12.0000pt]k[size=12.0000pt]=[size=12.0000pt]2[size=12.0000pt]r[size=12.0000pt]，则有[size=12.0000pt]3[size=12.0000pt]r[size=12.0000pt]  [size=12.0000pt]n[size=12.0000pt]+[size=12.0000pt]1[size=12.0000pt] [size=12.0000pt]是偶数[size=10.5000pt]，
[size=12.0000pt]设[size=12.0000pt]3[size=12.0000pt]r[size=12.0000pt] [size=12.0000pt]n[size=12.0000pt]+[size=12.0000pt]1[size=12.0000pt]=[size=12.0000pt] [size=12.0000pt]2[size=12.0000pt]i[size=12.0000pt] [size=12.0000pt]t[size=10.5000pt]，
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[size=10.5000pt]2m+1[size=10.5000pt]= [size=10.5000pt]2[size=10.5000pt]2[size=10.5000pt]r[size=10.5000pt] [size=10.5000pt]+[size=10.5000pt]1[size=10.5000pt] [size=10.5000pt]n+1[size=10.5000pt]>( [size=10.5000pt]3[size=10.5000pt]r[size=10.5000pt] [size=10.5000pt]n+1)/2[size=10.5000pt]>( [size=10.5000pt]3[size=10.5000pt]r[size=10.5000pt] [size=10.5000pt]n+1)/[size=10.5000pt] [size=10.5000pt]2[size=10.5000pt]2[size=10.5000pt]        [size=10.5000pt]>( 3[size=10.5000pt]r [size=10.5000pt]n+1)/[size=10.5000pt] [size=10.5000pt]2[size=10.5000pt]i
[size=10.5000pt]即 [size=10.5000pt]N+2[size=10.5000pt] 落于 [size=10.5000pt]3，5，7……2m-1[size=10.5000pt] 中，此前假设成立，故 [size=10.5000pt]m= [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]n[size=10.5000pt]，[size=10.5000pt]k=2r [size=10.5000pt]时角谷猜想成立。[size=10.5000pt]如 [size=10.5000pt]k=2r+1[size=10.5000pt]， [size=10.5000pt]3[size=10.5000pt]r [size=10.5000pt]2n+1 [size=10.5000pt]是奇数，做 [size=10.5000pt]3[size=10.5000pt]( [size=10.5000pt]3[size=10.5000pt]r [size=10.5000pt]2n+1)+1[size=10.5000pt]= [size=10.5000pt]3[size=10.5000pt]r [size=10.5000pt]+[size=10.5000pt]1 [size=10.5000pt]2n+4[size=10.5000pt]，除以 [size=10.5000pt]2 [size=10.5000pt]有[size=10.5000pt]3[size=10.5000pt]r [size=10.5000pt]+[size=10.5000pt]1 [size=10.5000pt]n+2 [size=10.5000pt]是奇数，
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[size=10.5000pt]3( 3[size=10.5000pt]r [size=10.5000pt]+[size=10.5000pt]1 [size=10.5000pt]n+2)+1[size=10.5000pt]，即[size=10.5000pt]3[size=10.5000pt]r [size=10.5000pt]+[size=10.5000pt]2 [size=10.5000pt]n+7 [size=10.5000pt]是偶数，设[size=10.5000pt]3[size=10.5000pt]r [size=10.5000pt]+[size=10.5000pt]2 [size=10.5000pt]n+7= 2[size=10.5000pt]s [size=10.5000pt]w
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[size=10.5000pt]k=2r+1[size=10.5000pt]，[size=10.5000pt]m= [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]n[size=10.5000pt] [size=10.5000pt]时，[size=10.5000pt]        [size=10.5000pt]2m+1=[size=10.5000pt] [size=10.5000pt]2[size=10.5000pt]2r[size=10.5000pt]+[size=10.5000pt]2[size=10.5000pt] [size=10.5000pt]n+1>([size=10.5000pt] [size=10.5000pt]3[size=10.5000pt]r[size=10.5000pt] [size=10.5000pt]+[size=10.5000pt]2[size=10.5000pt] [size=10.5000pt]n+7)/2>([size=10.5000pt] [size=10.5000pt]3[size=10.5000pt]r[size=10.5000pt] [size=10.5000pt]+[size=10.5000pt]2[size=10.5000pt] [size=10.5000pt]n+7)/[size=10.5000pt] [size=10.5000pt]2[size=10.5000pt]2[size=10.5000pt]        [size=10.5000pt]>([size=10.5000pt] [size=10.5000pt]3[size=10.5000pt]r[size=10.5000pt] [size=10.5000pt]+[size=10.5000pt]2[size=10.5000pt] [size=10.5000pt]n+7)/[size=10.5000pt] [size=10.5000pt]2[size=10.5000pt]s
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[size=10.5000pt]即 [size=10.5000pt]N+2[size=10.5000pt] 落于 [size=10.5000pt]3，5，7……2m-1[size=10.5000pt] 中，此前假设成立，故 [size=10.5000pt]m= [size=10.5000pt]2[size=10.5000pt]k [size=10.5000pt]n[size=10.5000pt]，[size=10.5000pt]k=2r+1 [size=10.5000pt]时角谷猜想成立。[size=10.5000pt]综上所述 [size=10.5000pt]m [size=10.5000pt]是奇数或偶数角谷猜想都成立，此猜想获证。
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水流成林

为何用word编辑后提交后成乱码，有人知道如何操作？