勾股数新公式

蔡家雄

求：\(\sqrt{23}\) 的渐近分数的 分子 的通解公式，

求：\(\sqrt{23}\) 的渐近分数的 分母 的通解公式，

Treenewbee

蔡家雄 发表于 2023-12-4 04:27
由 72+9=9^2，72 又能被 9 整除，还会有 无穷多组 解吗？

求解方程：\(y^2=72*x^2+9\) 的正整数解，

\[(3y')^2=9(8x^2+1)\rightarrow y'^2=8x^2+1\]

王守恩

Treenewbee 发表于 2023-12-3 14:25
暂未找到通项公式。

{1, 2, 9, 40, 77, 342, 1519, 2924, 12987, 57682, 111035, 493164, 2190397, 4216406, 18727245,
83177404, 160112393, 711142146, 3158550955, 6080054528, 27004674303, 119941758886}

\(x_{n}=\lfloor\frac{(Mod[n-1,3]+1)^{2-Mod[n^2,3]} }{12\sqrt{10}}(19+6\sqrt{10})^{\lceil n/3\rceil}\rfloor\)

Table[Floor[(Mod[n-1,3]+1)^(2-Mod[n^2,3])/(12Sqrt[10])(19+6Sqrt[10])^Ceiling[n/3]],{n,24}]

Treenewbee

王守恩 发表于 2023-12-4 11:00
{1, 2, 9, 40, 77, 342, 1519, 2924, 12987, 57682, 111035, 493164, 2190397, 4216406, 18727245,
831 ...

{1,2,9,38,76,342,1443,2886,12987,54796,109592,493164,2080805,4161610,18727245,79015794,158031588,711142146,3000519367,6001038734,27004674303,113940720152,227881440304,1025466481368}

这个出入有点大

Treenewbee

蔡家雄 发表于 2023-12-4 04:35
求：\(\sqrt{23}\) 的渐近分数的 分子 的通解公式，

求：\(\sqrt{23}\) 的渐近分数的 分母 的通解公式 ...

1. Convergents[Sqrt@23, 30]

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\[\left\{4,5,\frac{19}{4},\frac{24}{5},\frac{211}{44},\frac{235}{49},\frac{916}{191},\frac{1151}{240},\frac{10124}{2111},\frac{11275}{2351},\frac{43949}{9164},\frac{55224}{11515},\frac{485741}{101284},\frac{540965}{112799},\frac{2108636}{439681},\frac{2649601}{552480},\frac{23305444}{4859521},\frac{25955045}{5412001},\frac{101170579}{21095524},\frac{127125624}{26507525},\frac{1118175571}{233155724},\frac{1245301195}{259663249},\frac{4854079156}{1012145471},\frac{6099380351}{1271808720},\frac{53649121964}{11186615231},\frac{59748502315}{12458423951},\frac{232894628909}{48561887084},\frac{292643131224}{61020311035},\frac{2574039678701}{536724375364},\frac{2866682809925}{597744686399}\right\}\]

Treenewbee

蔡家雄 发表于 2023-12-4 04:35
求：\(\sqrt{23}\) 的渐近分数的 分子 的通解公式，

求：\(\sqrt{23}\) 的渐近分数的 分母 的通解公式 ...

1. Table[{m=Mod[n,4],r=Ceiling[n/4],a=1909 m-1725 m^2+368 m^3,b=6-406 m+363 m^2-77 m^3};{n,Round[Sqrt[23]/276 ((a +b Sqrt[23] ) (24+5 Sqrt[23])^r)],Round[1/276 ((a+b Sqrt[23]) (24+5 Sqrt[23])^r)]},{n,30}]//FullSimplify

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{{1,4,1},{2,5,1},{3,19,4},{4,24,5},{5,211,44},{6,235,49},{7,916,191},{8,1151,240},{9,10124,2111},{10,11275,2351},{11,43949,9164},{12,55224,11515},{13,485741,101284},{14,540965,112799},{15,2108636,439681},{16,2649601,552480},{17,23305444,4859521},{18,25955045,5412001},{19,101170579,21095524},{20,127125624,26507525},{21,1118175571,233155724},{22,1245301195,259663249},{23,4854079156,1012145471},{24,6099380351,1271808720},{25,53649121964,11186615231},{26,59748502315,12458423951},{27,232894628909,48561887084},{28,292643131224,61020311035},{29,2574039678701,536724375364},{30,2866682809925,597744686399}}

Treenewbee

令\[m=(n \bmod 4)\]
\[a=368 m^3-1725 m^2+1909 m\]
\[b=-77 m^3+363 m^2-406 m+6\]
\[\sqrt{23}=\frac{y_n}{x_n}\]
则有：
\[y_n=\text{Round}\left[\frac{\sqrt{23}}{276}  \left(\left(5 \sqrt{23}+24\right)^\left\lceil \frac{n}{4}\right\rceil\left(a+\sqrt{23} b\right)\right)\right]\]
\[x_n=\text{Round}\left[\frac{1}{276} \left(5 \sqrt{23}+24\right)^\left\lceil \frac{n}{4}\right\rceil\left(a+\sqrt{23} b\right)\right]\]

王守恩

Treenewbee 发表于 2023-12-4 11:17
{1,2,9,38,76,342,1443,2886,12987,54796,109592,493164,2080805,4161610,18727245,79015794,158031588,7 ...

\(x_{n}=\lfloor(\frac{2^{\lfloor5 (n + 1)/3\rfloor -\lfloor5n/3\rfloor]}+3^{Mod[-n,3]}}{4 \sqrt{10}}-\frac{Mod[-n, 3]}{2})(19+6\sqrt{10})^{\lceil n/3\rceil}\rfloor\)
Table[Floor[((2^(Floor[5 (n + 1)/3] - Floor[5 n/3])+3^(Mod[-n, 3]))
/(4 Sqrt[10]) -  Mod[-n, 3]/2) (19 + 6 Sqrt[10])^Ceiling[n/3]], {n,24}]

Treenewbee

\[x_n=\left\lfloor\left(\frac{\frac{105}{32}-2 \left(n \% 3-\frac{9}{8}\right)^2}{\sqrt{10}}-\frac{(2n)\% 3}{2}\right) \left(3+\sqrt{10}\right)^{2 \left\lceil \frac{n}{3}\right\rceil } \right\rfloor\]

Treenewbee

或者
\[x_n=\left\lfloor \left(\sqrt{10}+3\right)^{2 \left\lceil \frac{n}{3}\right\rceil } \left(\frac{(n \% 3) (18-8 (n \% 3))+3}{4 \sqrt{10}}-\frac{1}{2} ((2 n) \% 3)\right)\right\rfloor\]