StirlingS2[n,m]
给出第二类
的 Stirling 数.
StirlingS2
StirlingS2[n,m]
给出第二类
的 Stirling 数.
更多信息
- 整数型数学函数,同时适合符号和数值运算.
- StirlingS2 定义为从连续积分的 Power 到离散积分
的 FactorialPower 的转换矩阵,其中
.
给出把
个元素放入
个非空子集中的方法数.- StirlingS2 自动线性作用于列表.
范例
打开所有单元 关闭所有单元基本范例 (1)
范围 (2)
StirlingS2 按元素线性作用于列表:
StirlingS2[{2, 4, 6}, 2]TraditionalForm 格式:
StirlingS2[n, m]//TraditionalForm应用 (5)
ListPlot3D[Table[Log[Abs[StirlingS2[n, m]] + 1], {n, 60}, {m, 60}]]ArrayPlot[Mod[Table[StirlingS2[n, m], {n, 0, 100}, {m, 0, 100}], 2]]setPartitions[{x_}] := {{{x}}};
setPartitions[{r__, x_}] := Apply[Join, ReplaceList[#, {b___, {S__}, a___} | {b__} :> {b, {S, x}, a}]& /@ setPartitions[{r}]]n = 4;
parts = setPartitions[Range[n]]KeySort[Counts[Length /@ parts]]StirlingS2[n, Range[n]]D[f[Exp[z]], {z, 6}]Sum[Exp[k z] StirlingS2[6, k]Derivative[k][f][Exp[z]], {k, 6}]一个
-面色子独立投掷
次. 所有的面都至少出现一次的概率以第二类斯特令数的形式给出:
classicalOccupancyPr[n_, r_] := (r!/r^n)StirlingS2[n, r]DiscretePlot[classicalOccupancyPr[n, 6], {n, 5, 24}]With[{n = 12, rep = 10 ^ 5}, Map[If[Complement[Range[6], #] === {}, 1.0, 0.0]&, RandomVariate[DiscreteUniformDistribution[{1, 6}], {rep, n}]]]//MeanclassicalOccupancyPr[12, 6]//N属性和关系 (7)
Series[BellB[12, t], {t, 0, 5}]Table[StirlingS2[12, m], {m, 5}]With[{n = 12},
Table[n! / m!SeriesCoefficient[Series[(E^t - 1)^m, {t, 0, n}], n], {m, 5}]]Table[StirlingS2[12, m] , {m, 5}]第二类的 Stirling 数是第一类 Stirling 数的逆:
Table[Sum[StirlingS2[l, n] StirlingS1[m, l], {l, 0, Max[m, n] + 1}], {n, 0, 6}, {m, 0, 6}]//MatrixFormWith[{n = 120, m = 10, ρ = 10}, Re[-n! / m! NIntegrate[Evaluate[(Exp[x] - 1) ^ m / (x ^ (n + 1)) ρ Exp[I φ] /. x -> ρ Exp[I φ]], {φ, 0, 2Pi}]] / (I 2Pi)]StirlingS2[120, 10]//Ns___⊗a⊗a^†⊗t___ := s⊗a^†⊗a⊗t + s⊗ta^†⊗a⊗a^†⊗a⊗a^†⊗a⊗a^†⊗a⊗a^†⊗a⊗a^†⊗aSum[StirlingS2[6, k] CircleTimes@@Flatten[{Array[a^†&, k], Array[a&, k]}], {k, 6}]Nest[(# /. x -> x + 1) - #&, x ^ 10, 4] / 4!% /. x -> 0StirlingS2[10, 4]Table[BellY[n, m, Table[1, {n}]], {n, 6}, {m, 1, n}]Table[StirlingS2[n, m], {n, 6}, {m, 1, n}]可能存在的问题 (2)
StirlingS2 对适度大小的参数可以有较大的值:
StirlingS2[1000, 500]//ShortN[%]StirlingS2[0, 0]巧妙范例 (2)
技术笔记
-
▪
- 组合函数
历史
1988年引入 (1.0)
文本
Wolfram Research (1988),StirlingS2,Wolfram 语言函数,https://reference.wolfram.com/language/ref/StirlingS2.html.
CMS
Wolfram 语言. 1988. "StirlingS2." Wolfram 语言与系统参考资料中心. Wolfram Research. https://reference.wolfram.com/language/ref/StirlingS2.html.
APA
Wolfram 语言. (1988). StirlingS2. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/StirlingS2.html 年
BibTeX
@misc{reference.wolfram_2026_stirlings2, author="Wolfram Research", title="{StirlingS2}", year="1988", howpublished="\url{https://reference.wolfram.com/language/ref/StirlingS2.html}", note=[Accessed: 03-November-2026]}
BibLaTeX
@online{reference.wolfram_2026_stirlings2, organization={Wolfram Research}, title={StirlingS2}, year={1988}, url={https://reference.wolfram.com/language/ref/StirlingS2.html}, note=[Accessed: 03-November-2026]}