Exponentiating by squaring is an algorithm used for the fast computation of large integer powers of a number. It is also known as the square-and-multiply algorithm or binary exponentiation. In additive groups the appropriate name is double-and-add algorithm. It implicitly uses the binary expansion of the exponent. It is of quite general use, for example in modular arithmetic.

    一个从上而下的n层格子，m_i 为第i层的格子数，当m_i>=m_i+1(i=1,2,^...,n-1) ，即上层的格子数不少于下层的格子数时，称之为Ferrers图像，如图(2-6-2)示。 

                                 图   (2-6-2)

    Ferrers图像具有如下性质： 

    1.每一层至少有一个格子。 

    2.第一行与第一列互换，第二行于第二列互换，…，即图(2-6-3)绕虚线轴旋转所得的图仍然是Ferrers图像。两个Ferrers 图像称为一对共轭的Ferrers图像。 

    利用Ferrers图像可得关于整数拆分的十分有趣的结果。 

    (a)整数n拆分成k个数的和的拆分数，和数n拆分成个数的和的拆分数相等。 

    因整数n拆分成k个数的和的拆分可用一k行的图像表示。所得的Ferrers图像的共轭图像最上面一行有k个格子。例如： 

图   (2-6-3)      

    (b)整数n拆分成最多不超过m个数的和的拆分数，和n拆分成最大不超过m的拆分数相等。 理由和(a)相类似。 

    因此，拆分成最多不超过m个数的和的拆分数的母函数是 

    拆分成最多不超过m-1个数的和的拆分数的母函数是 

    所以正好拆分成m个数的和的拆分数的母函数为 

    (c)整数n拆分成互不相同的若干奇数的和的的拆分数,和n拆分成自共轭的Ferrers图像的拆分数相等. 设 

其中n₁>n₂>^...>n_k 

    构造一个Ferrers图像，其第一行，第一列都是n₁+1格，对应于2n₁+1，第二行，第二列各n₂+1格，对应于2n₂+1。以此类推。由此得到的Ferres图像是共轭的。反过来也一样。 

    例如 17=9+5+3 对应为Ferrers图像为

图   (2-6-4)

假定n拆分为n=n1+n2+n3+……+nk，且n1>=n2>=n3>=……>=nk

我们将它排列成阶梯形，左边看齐，我们可以得到一个类似倒阶梯图像，这种图像我们称之为Ferrers图像，如对于20=10+5+4+1，我们有图像：

对于Ferrers图像，我们很容易知道以下两条性质：

（1）      每层至少一个格子

（2）      行列互换，所对应的图像仍为Ferrers图像，他应该为该图像的共轭图像

   任意的Ferrers图像对应一个整数的拆分，而可用Ferrers图像方便地证明：

（1）      n拆分为k个整数的拆分数，与n拆分成最大数为k的拆分数相等

（2）      n拆分为最多不超过k个数的拆分数，与n拆分成最大数不超过k的拆分数相等

（3）      n拆分为互不相同的若干奇数的拆分数，与n拆分成图像自共轭的拆分的拆分数相等

Wikipedia上的Catalan numbers:

In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named for the Belgian mathematician Eugène Charles Catalan (1814–1894).

The n^th Catalan number is given directly in terms of binomial coefficients by

The first Catalan numbers (sequence A000108 in OEIS) for n = 0, 1, 2, 3, … are

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, …

Properties

An alternative expression for C_n is

This shows that C_n is a natural number, which is not a priori obvious from the first formula given. This expression forms the basis for André's proof of the correctness of the formula (see below under second proof).

The Catalan numbers satisfy the recurrence relation

They also satisfy:

which can be a more efficient way to calculate them.

Asymptotically, the Catalan numbers grow as

in the sense that the quotient of the n^th Catalan number and the expression on the right tends towards 1 for n → ∞. (This can be proved by using Stirling's approximation for n!.)

Applications in combinatorics

There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the case C₃ = 5.

• C_n is the number of Dyck words of length 2n. A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's (see also Dyck language). For example, the following are the Dyck words of length 6:

• Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, C_n counts the number of expressions containing n pairs of parentheses which are correctly matched:

• C_n is the number of different ways n + 1 factors can be completely parenthesized (or the number of ways of associating n applications of a binary operator). For n = 3 for example, we have the following five different parenthesizations of four factors:

• Successive applications of a binary operator can be represented in terms of a binary tree. It follows that C_n is the number of rooted ordered binary trees with n + 1 leaves:

If the leaves are labelled, we have the quadruple factorial numbers.

• C_n is the number of non-isomorphic full binary trees with n vertices that have children, usually called internal vertices or branches. (A rooted binary tree is full if every vertex has either two children or no children.)

• C_n is the number of monotonic paths along the edges of a grid with n × n square cells, which do not cross the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up". The following diagrams show the case n = 4:

• C_n is the number of different ways a convex polygon with n + 2 sides can be cut into triangles by connecting vertices with straight lines. The following hexagons illustrate the case n = 4:

• C_n is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences.

• C_n is the number of noncrossing partitions of the set { 1, ..., n }. A fortiori, C_n never exceeds the nth Bell number. C_n is also the number of noncrossing partitions of the set { 1, ..., 2n } in which every block is of size 2. The conjunction of these two facts may be used in a proof by mathematical induction that all of the free cumulants of degree more than 2 of the Wigner semicircle law are zero. This law is important in free probability theory and the theory of random matrices.

Proof of the formula

There are several ways of explaining why the formula

solves the combinatorial problems listed above. The first proof below uses a generating function. The second and third proofs are examples of bijective proofs; they involve literally counting a collection of some kind of object to arrive at the correct formula.

First proof

We start with the observation that several of the combinatorial problems listed above can easily be seen to satisfy the recurrence relation

For example, every Dyck word w of length ≥ 2 can be written in a unique way in the form

w = Xw₁Yw₂

with (possibly empty) Dyck words w₁ and w₂.

The generating function for the Catalan numbers is defined by

As explained in the article titled Cauchy product, the sum on the right side of the above recurrence relation is the coefficient of xⁿ in the product

Therefore

Multiplying both sides by x, we get

So we have

and hence

The square root term can be expanded as a power series using the identity

which can be proved, for example, by the binomial theorem, (or else directly by considering repeated derivatives of ) together with judicious juggling of factorials. Substituting this into the above expression for c(x) produces, after further manipulation,

Equating coefficients yields the desired formula for C_n.

Second proof

This proof depends on a trick due to D. André, which is now more generally known as the reflection principle (not to be confused with the Schwarz reflection theorem in complex analysis). It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see above).

Figure 1. The green portion of the path is being flipped.

Suppose we are given a monotonic path in an n × n grid that does cross the diagonal. Find the first edge in the path that lies above the diagonal, and flip the portion of the path occurring after that edge, along a line parallel to the diagonal. (In terms of Dyck words, we are starting with a sequence of n X's and n Y's which is not a Dyck word, and exchanging all X's with Y's after the first Y that violates the Dyck condition.) The resulting path is a monotonic path in an (n − 1) × (n + 1) grid. Figure 1 illustrates this procedure; the green portion of the path is the portion being flipped.

Since every monotonic path in the (n − 1) × (n + 1) grid must cross the diagonal at some point, every such path can be obtained in this fashion in precisely one way. The number of these paths is equal to

.

Therefore, to calculate the number of monotonic n × n paths which do not cross the diagonal, we need to subtract this from the total number of monotonic n × n paths, so we finally obtain

which is the nth Catalan number C_n.

Third proof

The following bijective proof, while being more involved than the previous one, provides a more natural explanation for the term n + 1 appearing in the denominator of the formula for C_n.

Figure 2. A path with exceedance 5.

Suppose we are given a monotonic path, which may happen to cross the diagonal. The exceedance of the path is defined to be the number of pairs of edges which lie above the diagonal. For example, in Figure 2, the edges lying above the diagonal are marked in red, so the exceedance of the path is 5.

Now, if we are given a monotonic path whose exceedance is not zero, then we may apply the following algorithm to construct a new path whose exceedance is one less than the one we started with.

• Starting from the bottom left, follow the path until it first travels above the diagonal.
• Continue to follow the path until it touches the diagonal again. Denote by X the first such edge that is reached.
• Swap the portion of the path occurring before X with the portion occurring after X.

The following example should make this clearer. In Figure 3, the black circle indicates the point where the path first crosses the diagonal. The black edge is X, and we swap the red portion with the green portion to make a new path, shown in the second diagram.

Figure 3. The green and red portions are being exchanged.

Notice that the exceedance has dropped from three to two. In fact, the algorithm will cause the exceedance to decrease by one, for any path that we feed it.

Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedance-decreasing algorithm.

It is also not difficult to see that this process is reversible: given any path P whose exceedance is less than n, there is exactly one path which yields P when the algorithm is applied to it.

This implies that the number of paths of exceedance n is equal to the number of paths of exceedance n − 1, which is equal to the number of paths of exceedance n − 2, and so on, down to zero. In other words, we have split up the set of all monotonic paths into n + 1 equally sized classes, corresponding to the possible exceedances between 0 and n. Since there are

monotonic paths, we obtain the desired formula

Figure 4 illustrates the situation for n = 3. Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. Since there are five rows, C₃ = 5.

Hankel matrix

The n×n Hankel matrix whose (i, j) entry is the Catalan number C_i+j has determinant 1, regardless of the value of n. For example, for n = 4 we have

Note that if the entries are ``shifted", namely the Catalan numbers C_i+j+1, the determinant is still 1, regardless of the size of n. For example, for n = 4 we have

.

The Catalan numbers is the unique sequence with this property.

Quadruple factorial

The quadruple factorial is given by , or . This is the solution to labelled variants of the above combinatorics problems. It is entirely distinct from the multifactorials.

History

The Catalan sequence was first described in the 18th century by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named after Eugène Charles Catalan, who discovered the connection to parenthesized expressions. The counting trick for Dyck words was found by D. André in 1887.

2. 设A, B, C为任意的集合，则
(1) A≈A
(2) 若A≈B，则B≈A
(3) 若A≈B且B≈C，则A≈C

3. Cantor定理
(1) N不与R等势
(2) 设A为任意的集合，则A不与P(A)等势

4. 若一个集合A与某个自然数n等势，则称A是有穷集合，否则称A为无穷集合。

2. 设A为一个集合，如果A中任何元素的元素也是A的元素，则称A为传递集。每个自然数都是传递集。
以下命题等价：
(1) A是传递集
(2) ∪A 包含于 A
(3) 对于任意的y∈A，y包含于A
(4) A包含于P(A)
(5) P(A)为传递集

3. 设A为一个集合，称从A*A到A的函数为A上的二元运算。
另+: N*N -> N，且对于任意的m, n ∈N，+(<m, n>) = A_m(n), 记作m + n，称+为N上的加法运算
另·: N*N -> N，且对于任意的m, n ∈N，·(<m, n>) = M_m(n)，记作m·n，称·为N上的乘法运算。

Peano系统是满足以下公设的有序三元组<M, F, e>，其中M为一个集合，F为M到M的函数，e为首元素。5条公设为：

(1) e ∈M
(2) M在F下是封闭的
(3) e ￠ ranF （暂时只找到这个符号表示“不属于” 囧）
(4) F是单射的
(5) 如果M的子集A满足 (e属于A) 且 (A在F下是封闭的)，则A=M
(5)称为极小性公设

2. 设A为一个集合，称 A∪{A} 为A的后继，记作A⁺, 并称求集合的后继为后继运算。

3. 设A为一个集合，若A满足：
(1) Ø  ∈A，
(2) 若对于一切 a ∈A，则 a⁺ ∈A，
则称A是归纳集。

4. 从归纳集的定义可以看出，Ø，Ø⁺, Ø⁺⁺,... 是所有归纳集的元素，于是可以将它们定义成自然数。
自然数是属于每个归纳集的集合，将Ø，Ø⁺, Ø⁺⁺,...分别记为0, 1, 2, ...
设D={v | v是归纳集|，称∩D为全体自然数集合，记为N.
设N为自然数集合，σ: N -> N，且σ(n) = n⁺, 则<N, σ, Ø>是Peano系统。
这个Peano系统的第(5)条公设提出了证明自然数性质的一种方法，即数学归纳法，称此公设为数学归纳法原理。

2. Forced Moves 排除所有其他可能性后唯一的答案

3. Pinned Squares
Intersection 的加强版，根据更多的情况确定某一个数字在该区域的唯一可能位置。
 

4. Locked Sets

如图一R5C1和R6C1只能填1或8，由此可排除与他们有关的域中的其他格填1和8的可能性，从而R6C2只能填5。