CatalanNumber
给出了第 n
个 卡塔兰数
.
更多信息
- CatalanNumber[n] 的一般定义为
. - 对于整数参数,卡塔兰数为整数且出现在各种树状列举问题中.
- CatalanNumber 可与 Interval 和 CenteredInterval 对象一起使用: »
范例
打开所有单元 关闭所有单元范围 (9)
CatalanNumber[100000]//ShortCatalanNumber[5 / 2]CatalanNumber[2.3]CatalanNumber[1.2 + I]Plot[CatalanNumber[n], {n, -3, 3}]计算涉及 CatalanNumber 的和:
Sum[1 / CatalanNumber[n], {n, 1, Infinity}]Sum[(1/4^n) CatalanNumber[n], {n, m - 1}]CatalanNumber 按元素线性作用于列表:
CatalanNumber[{1, 2, 3, 4}]CatalanNumber 可与 Interval 和 CenteredInterval 对象一起使用:
CatalanNumber[Interval[{0.5, 0.6}]]CatalanNumber[CenteredInterval[1, 1 / 100]]TraditionalForm 格式:
CatalanNumber[n]//TraditionalForm应用 (3)
SetAttributes[f, {Flat, OneIdentity}]分配到 CirclePlus 中的列表上:
e : CirclePlus[___, _List, ___] := Distribute[Unevaluated[e], List]parenthesizedlist = f[a, b, c, d] //. {f[x__] :> ReplaceList[f[x], f[u_, v_] :> CirclePlus[u, v]]}//FlattenLength[parenthesizedlist]CatalanNumber[3]卡塔兰数 CatalanNumber[n] 可描述为唯一一组使得两个汉克尔行列式都等于一的数字. 对前几种情况进行验证:
Table[Det[HankelMatrix[CatalanNumber[Range[0, n]], CatalanNumber[Range[n, 2n]]]] == Det[HankelMatrix[CatalanNumber[Range[n + 1]], CatalanNumber[Range[n + 1, 2n + 1]]]] == 1, {n, 9}]Assuming[n∈Integers && n >= 0, FullSimplify[CatalanNumber[n] == (2^2n + 1(2n - 1)!!/(2n + 2)!!)]]属性和关系 (6)
GeneratingFunction[CatalanNumber[n], n, x]Series[%, {x, 0, 10}]Table[CatalanNumber[n], {n, 0, 10}]CatalanNumber[n] == Binomial[2n, n] - Binomial[2n, n + 1]Table[%, {n, -6, 6}]FunctionExpand[%%]//FullSimplifyTable[BellY[Transpose[{1 / Range[n, 1, -1]!, Range[n]!}]], {n, 10}]Table[CatalanNumber[n], {n, 10}]可以用 DifferenceRoot 来表示 CatalanNumber:
DifferenceRootReduce[CatalanNumber[k], k]FindSequenceFunction 可以识别 CatalanNumber 序列:
Table[CatalanNumber[n], {n, 10}]FindSequenceFunction[%, n]CatalanNumber 的指数母函数:
ExponentialGeneratingFunction[CatalanNumber[n], n, x]可能存在的问题 (1)
巧妙范例 (2)
只有形式为 CatalanNumber[2k-1] 的 Catalan 数为奇:
Table[Mod[CatalanNumber[2^n - 1], 10], {n, 20}]Table[Det[HankelMatrix[CatalanNumber[Range[0, n]] + CatalanNumber[Range[n + 1]], CatalanNumber[Range[n, 2n]] + CatalanNumber[Range[n + 1, 2n + 1]]]], {n, 0, 9}]Table[Fibonacci[2n + 3], {n, 0, 9}]参见
技术笔记
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- 组合函数
文本
Wolfram Research (2007),CatalanNumber,Wolfram 语言函数,https://reference.wolfram.com/language/ref/CatalanNumber.html (更新于 2014 年).
CMS
Wolfram 语言. 2007. "CatalanNumber." Wolfram 语言与系统参考资料中心. Wolfram Research. 最新版本 2014. https://reference.wolfram.com/language/ref/CatalanNumber.html.
APA
Wolfram 语言. (2007). CatalanNumber. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/CatalanNumber.html 年
BibTeX
@misc{reference.wolfram_2026_catalannumber, author="Wolfram Research", title="{CatalanNumber}", year="2014", howpublished="\url{https://reference.wolfram.com/language/ref/CatalanNumber.html}", note=[Accessed: 07-August-2026]}
BibLaTeX
@online{reference.wolfram_2026_catalannumber, organization={Wolfram Research}, title={CatalanNumber}, year={2014}, url={https://reference.wolfram.com/language/ref/CatalanNumber.html}, note=[Accessed: 07-August-2026]}