53 lines
1.2 KiB
Markdown
53 lines
1.2 KiB
Markdown
根据137题有
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$$
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A_G(x)=\frac{x+3x^2}{1-x-x^2}
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$$
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所以
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$$
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(n+3)x^2+(n+1)x-n=0
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$$
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解此方程，得
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$$
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x=\frac{\sqrt{5n^2+14n+1}-(n+1)}{2(n+3)}
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$$
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只有 $5n^2+14n+1$ 是平方数时， $x$ 才会是有理数。
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尝试分析 $N=5n^2+14n+1$ 是平方数时 $n$ 的变化规律
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| N | increasing rate of N |
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| -------- | -------------------- |
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| 2 | |
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| 5 | 2.5 |
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| 21 | 4.2 |
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| 42 | 2.0 |
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| 152 | 3.619047619047619 |
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| 296 | 1.9473684210526316 |
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| 1050 | 3.5472972972972974 |
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| 2037 | 1.94 |
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| 7205 | 3.5370643102601864 |
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| 13970 | 1.9389312977099236 |
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| 49392 | 3.535576234788833 |
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| 95760 | 1.9387755102040816 |
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| 338546 | 3.535359231411863 |
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| 656357 | 1.9387527839643652 |
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| 2320437 | 3.535327573256627 |
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| 4498746 | 1.9387494683113569 |
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| 15904520 | 3.535322954441082 |
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观察发现 $1.9387\ldots*3.5353\ldots=\frac{\sqrt{5}+1}{2}$.
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进一步尝试
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$$
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5-2=3 \\
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42-21=21 \\
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296-152=144
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$$
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$$
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21-5=16=2*8 \\
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152-42=110=2*55 \\
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1050-296=754=2*377
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$$
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$3,21,144,\ldots,8,55,377\ldots$ 它们都是 Fibonacci 数。 |