查看“佩尔数”的源代码

'''佩尔数'''是一个自古以来就知道的整数数列，由[[递推关系]]定义，与[[斐波那契数]]类似。佩尔数呈指数增长，增长速率与[[白银比]]的幂成正比。它出现在[[2的算術平方根]]的近似值以及[[三角平方数]]的定义中，也出现在一些组合数学的问题中。

==定义==

佩尔数由以下的[[递推关系]]定义：
:<math>P_n=\begin{cases}0&\mbox{if }n=0;\\1&\mbox{if }n=1;\\2P_{n-1}+P_{n-2}&\mbox{otherwise.}\end{cases}</math>

也就是说，佩尔数的数列从0和1开始，以后每一个佩尔数都是前面的数的两倍加上再前面的数。最初几个佩尔数是：
:[[0]], [[1]], [[2]], [[5]], [[12]], [[29]], [[70]], [[169]], 408, 985, 2378…… {{OEIS|id=A000129}}。

佩尔数也可以用通项公式来定义：
:<math>P_n=\frac{(1+\sqrt2)^n-(1-\sqrt2)^n}{2\sqrt2}.</math>
对于较大的''n''，<math>\scriptstyle (1+\sqrt 2)^n</math>的项起主要作用，而<math>\scriptstyle (1-\sqrt 2)^n</math>的项则变得微乎其微。因此佩尔数大约与[[白银比]]<math>\scriptstyle (1+\sqrt 2)</math>的幂成正比。

第三种定义是以下的[[矩阵]]公式：
:<math>\begin{pmatrix} P_{n+1} & P_n \\ P_n & P_{n-1} \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}^n.</math>

从这些定义中，可以推出或证明许多恒等式；例如以下的恒等式，与斐波那契数的[[卡西尼恒等式]]类似：
:<math>P_{n+1}P_{n-1}-P_n^2 = (-1)^n,</math>
这个恒等式是以上矩阵公式的直接结果（考虑矩阵的[[行列式]]）。

==2的算術平方根的近似值==

佩尔数出现在[[2的算術平方根]]的[[丢番图逼近|有理数近似值]]中。如果两个大的整数''x''和''y'' 是[[佩尔方程]]的解：
:<math>\displaystyle x^2-2y^2=\pm 1,</math>
那么它们的比<math>\tfrac{x}{y}</math>就是<math>\scriptstyle\sqrt 2</math>的一个较精确的近似值。这种形式的近似值的数列是：
:<math>1, \frac32, \frac75, \frac{17}{12}, \frac{41}{29}, \frac{99}{70}, \dots</math>
其中每一个分数的分母是佩尔数，分子则是这个数与前一个佩尔数的和。也就是说，佩尔方程的解具有<math>\tfrac{P_{n-1}+P_n}{P_n}</math>的形式。<math>\sqrt 2\approx\frac{577}{408}</math>是这些近似值中的第八个，在公元前3或4世纪就已经为印度数学家所知。公元前5世纪的古希腊数学家也知道这个近似值的数列；他们把这个数列的分母和分子称为“边长和直径数”，分子也称为“有理对角线”或“有理直径”。

这些近似值可以从<math>\scriptstyle\sqrt 2</math>的[[连分数]]展开式推出：
:<math>\sqrt 2 = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{\ddots\,}}}}}.</math>
取这个展开式的有限个项，便可以产生<math>\scriptstyle\sqrt 2</math>的一个近似值，例如：
:<math>\frac{577}{408} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2}}}}}}}.</math>

==素数和平方数==

'''佩尔素数'''是既是佩尔数又是[[素数]]的数。最初几个佩尔素数是：
:2, 5, 29, 5741, …… {{OEIS|id=A086383}}。
与斐波那契素数相似，仅当''n''本身是素数时<math>P_n</math>才有可能是素数。

唯一的既是佩尔数又是平方数、立方数或任意整数次方的数是0, 1, 以及169 = 13<sup>2</sup>。

然而，佩尔数与[[三角平方数]]有密切的关系。它们出现在以下佩尔数的恒等式中：
:<math>\bigl((P_{k-1}+P_k)\cdot P_k\bigr)^2 = \frac{(P_{k-1}+P_k)^2\cdot\left((P_{k-1}+P_k)^2-(-1)^k\right)}{2}.</math>
等式的左面是[[平方数]]，等式的右面是[[三角形数]]，因此是三角平方数。

Santana和Diaz-Barrero在2006年证明了佩尔数与平方数之间的另外一个恒等式，并证明了从<math>P_1</math>到<math>P_{4n+1}</math>的所有佩尔数的和总是平方数：
:<math>\sum_{i=0}^{4n+1} P_i = \left(\sum_{r=0}^n 2^r{2n+1\choose 2r}\right)^2 = (P_{2n}+P_{2n+1})^2.</math>
例如，从<math>P_1</math>到<math>P_5</math>的和是<math>0+1+2+5+12+29=49</math>，是<math>P_2+P_3=2+5=7</math>的平方。<math>P_{2n}+P_{2n+1}</math>就是这个和的平方根：
:1, 7, 41, 239, 1393, 8119, 47321, …… {{OEIS|id=A002315}}。

==勾股数==

[[File:Pell right triangles.svg|thumb|300px|边长为整数的直角三角形，其直角边几乎相等，由佩尔数引出。]]
如果一个直角三角形的边长为''a''、''b''和''c''（必须满足[[勾股定理]]''a''<sup>2</sup>+''b''<sup>2</sup>=''c''<sup>2</sup>），那么(''a'',''b'',''c'')称为[[勾股数]]。Martin在1875年描述，佩尔数可以用来产生勾股数，其中''a''和''b''相差一个单位。这个勾股数具有以下形式：
:<math>(2P_{n}P_{n+1}, P_{n+1}^2 - P_{n}^2, P_{n+1}^2 + P_{n}^2=P_{2n+1}).</math>
用这种方法产生的勾股数的序列是：
:(4,3,5), (20,21,29), (120,119,169), (696,697,985), ……

==佩尔-卢卡斯数==

'''佩尔-卢卡斯数'''由以下的递推关系定义：
:<math>Q_n=\begin{cases}2&\mbox{if }n=0;\\2&\mbox{if }n=1;\\2Q_{n-1}+Q_{n-2}&\mbox{otherwise.}\end{cases}</math>

也就是说，数列中的最初两个数都是2，后面每一个数都是前一个数的两倍加上再前面的一个数。这个数列的最初几个项是{{OEIS|id=A002203}}：[[2]], [[2]], [[6]], [[14]], [[34]], [[82]], [[198]], [[478]]……

佩尔-卢卡斯数的通项公式为：
:<math>Q_n=(1+\sqrt 2)^n+(1-\sqrt 2)^n.</math>

这些数都是偶数，每一个数都是以上<math>\scriptstyle\sqrt 2</math>的近似值中的分子的两倍。

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</div>

==外部链接==

*{{mathworld | title = Pell Number | urlname = PellNumber}}

[[Category:整数数列|P]]