我决定独立建站，给博客搬家。感谢佐拉的帮助。新地址:

http://www.liuxiaochuan.org

回顾这几年写博客的历程，我收获很大。我写过乱七八糟的东西当中，少数文章确实有不小影响。其中一篇译作甚至被一个山东大学办的杂志抄袭， 这可能算是体现出有影响的“好现象”吧。我有过很多张狂的言论，也认识了很多人，打开了自己的视野。正如学数学需要对很熟悉的理论反复的思考和理解，不断 的重新“温故知新”，我发现生活中处处有类似的感觉。我个人也享受重新做一件有价值的事情的过程。我打算将之前写过的东西扔在原来的wordpress上 不动。希望大家们到我的新站继续关注。如今的互联网允许每一个人都是自己的出版商，这是表达自己的机会。每个人都应该试试。

有一些老生常谈，我近来有比较深刻的切身体会，我想重复一遍，给自己提个醒儿。

1.不应该试图改变其他人。

2.试图改变自己，要充分意识到困难，而且往往积年累月也只改变一点点，甚至这一点点也需要非凡的毅力。

3.与人相处直接简单，礼貌谦逊，还有就是不卑不亢。

Solutions to last problem: be a prime, and let denote the congruence classes in modulo . also admits multiplication, making a finite field.

Now we fix a and define . Then is a subgroup of , and we can split as coset space .

Since is a surjective homomorphism from to , we have from the the fundamental theorem on homomorphisms that is the number of solutions of the function in .

When , color x with . This gives an coloring of . If , then by theorem 2 there is a monochromatic Schur triple, that is,  there are integers , none of which is divisible by p, such that , since is not divisible by it follows that  , completing the proof.

Theorem 1. Let . Then there exists an integer such that for all primes the congruence has a solution in the integers, such that does not divide .

Theorem 2. (Schur’s theorem)Let . Then there is a natural number , such as if and if the numbers are colored with colors, then there are three of them of the same color satisfying the equation: .

Solutions to last problem: Suppose a abstract simplicial have vertices . We say are geometric independent when the vectors are linearly independent. We first show the fact that we can find in points such that any of them are geometric independent.

We induct on the number and suppose that we have already found points such that any (or less) vertices are geometric independent. Since any (or less) vertices are geometric independent, they have to lie in a space of at most dimension. These dimensional subspaces cannot make full of the whole . We then choose a point which doesn’t lie in any of these dimensional subspaces.Thus we complete our induction by choosing an additional making any (or less) vertices choosing from are geometric independentm, completing the induction step.

Then we can choose points to corespond the abstract vertices . We now verify they are the geomeric realization of the abstract simplicial . We only need to check the the intersection of any two simplices is a face of both and . This is easily seen because there are totally no more than points involved and thus can form a simplicial . This means is either empty set or a common face of and .

Solutions to last problem: Denote the length of a permutation as . On one hand, every transposition can change the number of invrsions by one. So one easily sees that the length of a permutation is no less than the number of “inversions” in .

Now we only need to use exactly transpositions to  get to .

For any , we begin by counting the number of elements which are larger than and lie to the left to in . We denote it as . Then we find that the number of inversions in equals .

conversely, for a given sequence , we build up as the following n steps.

For (which has to be zero), we write down n.

For ,(which should be less than ) , we write down such that there is element on its left.

For , (which should be less than ), we insert such that there are elements on its left-hand side.

Take the ith step from above, what we have done is change to , which is equivalent with multiplying transpositions, namely, . Thus, we get to our conclusion.

Concave compositions of even length is the ordered pair of partitions with distinct parts with the property that they have the same length and that the last part are equal. We also note that zero part is allowed, too.

Recently, I figured out a combinatorial proof of the generating function of concave compositions of even length. I just submitted this work. I include the first draft of this paper here.

Problem 6: Denote the transpositions in the symmetric group as . Define the length of a permutation as the least number of transpositions that formulate the permutation: . Show that the length of a permutation is the number of “inversions”: the number of pairs $latx i<j$ for which .

Solution to last problem:

Define the rank of a partition to be , which would be the difference of the length of the first part of and the number of parts of . Then the following finishes the proof.

Lemma 1: (Euler’s pentagonal number theorem)

We use to denote the number of partitions of with distinct parts and the largest part is odd. Similarly, is the number of partitions of with distinct parts and the largest part is even.

lemma 2:

Solution to last problem: First we prove that this is equivalent with .

Given the MacMahon diagram. The involution is defined as follows. Denote the smallest part of as and then the smallest part of as .

When , remove the last column of and attach it under the last row; mark the last square if the vertical line does not contain a marked square, or vice versa.

If , we have several situations.

Case one is that the last column has the last square marked. If so, we make it unmarked, remove it and put it in the among the last a few rows of the same length. (If there is already a marked square there, then we must insert this new row above the marked row.)

Case two is when the last column has the last square unmarked. In this situation, there are two subcases. The first is when there is one row with the smallest length which is not marked. We marked this one and take it to the right hand side of the partition. Another subcase is when there is only one row with the smallest length and it is marked, too. We unmark it and put it to the right hand side of the partition .

When , we remove the last row and attach it to the right hand side of the graph; mark the last square if the vertical line does not contain a marked square, or vice versa.

In this definition, the only staying-put situations are the diagrams, with or without marked squares.

在整个初中阶段，我不得不说，我的记忆之中留下的百分之九十九的人和事都跟你有关系。你是我妈妈从小到大学毕业她唯一能叫出名字的同学。是跟你在一起让我这辈子第一次知道什么叫做哥们儿。那年张兮维和你我三人吃饭到今天，我就再也没有见到你们俩。我过去的几年活得一直很激烈，可是矛盾的是，我这样生活却是为了将来能活的平和，能不激烈。忽然接到你的电话，那一个小时的牛吹了下来，挂了电话之后我居然有一种失落感。我知道这感觉不可能停留太久，我还有那么多事情堆成了堆。但是，我想为你我的友谊写一点东西，并让它留在这儿，给我们两人的未来祈一个福。

回想曾经什么时候，我们也无忧无虑过。那是每天放学天也不黑，我们跑到互相家里玩仙剑奇侠传的日子。那时候谁知道这世界到底是怎么回事啊？在我感到艰难的时候，我会忽然想到那段日子，然后我告慰自己，好歹我还有过童年，至少算是童年的一个尾巴，不是么？初一刚开学我是插班晚进的。当时教导主任领我到班里的大门口，对陈老师说：“这个真的是个好学生。”哈，我的虚荣让我对这句话记得这么清楚。陈老师似乎不那么相信。她把我领进屋，让我坐在了宁悦含的身边。是最左前方的位置。那时候你对我会有印象么？

我对你的第一次印象却是另外一番光景。有一天课间休息，你突然跑过来嘴里含着水，喷了我一身。我感到愤怒，上中学以后还没有人欺侮过我呢。我也跑去水房含了一口的水，找你报仇。多亏下课时间只有10分钟。不然我们俩估计没法完整的进教室了。我们的教室出门就是一个常常的回廊，很是狭小。陈老师对学生的管理堪称军队式的，她让我们写那个说明书，还记得不记得，动不动就是几千字上万字，而且还让刘浏记下来每个人写说明书的“光辉历史”。有那么一次，你骑车带我赶下午的课，冲进教室被老陈一顿奚落，最后说，“二人共乘啊！”我好像被罚写说明书最狠的一次就是那次了。

暑假呢？我们整天耗在一起。我常常跑到你家里一直到很晚，至今还记得你妈妈做的鱼。那味道，就像夏天，像初恋，像世界上一切美好的东西，像友谊。哦，不，那就是友谊。

记得李萧么？我还很清晰的记得她。那时候我跟她应该算做有朦胧的爱情吧。我后来很奇怪，你那时候真的是心理一点没发育吗？怎么会完全没有暧昧就走过了整个初中？难道是地下的？不可能啊，难道连我都不知道！

初二开物理课了，我物理一直学的很好跟吴老师有很大关系。吴红员老师你还记得不？她是我这么多年来一直联系的唯一初中老师。今年我最艰难的3，4月份，她还打电话告诉我不要为别人，为面子上的事，非得去完成什么，要好好的生活。我不知道自己是不是幸运，似乎每一个阶段我都能碰到一个真心对我的长者和一个能持续一生的朋友。虽然基本上也就只有一个。初二我感觉有些变化了，我当了班干部，成了一个政治人物，我和张博之间在陈老师那里争风吃醋。我现在常常看不起当时的自己。我觉得你才是高手。当我意识到你身上那自然和那灵气，意识到唯有保持这些才能真正活得幸福的时候，已经是新世纪了。

后来我们又搬到新校舍，又中考，分分合合，聚聚散散。高中更加单调无聊，每次聚会仅仅能吃个饭。互相报喜不报忧。再后来，我们到了不同的城市，各自为生活打拼。有一天我跟李萧她聊了几句QQ，她说她完全不能理解博士每天在做什么，我试图解释两句，但是再聊几句发现真的完全不投机。看来有些回忆注定要留在过去的。各自都有自己的生活，我们距离对方越来越遥远，而是你一直坚定的让我俩这故事不散。我已经开始畅想什么时候我们领着自己的儿子，一起玩点什么，然后告诉他们他们俩的爹，当初就这么玩到了现在！虽然这是谎话，但是我希望撒这个谎的一天可以到来。

初中的我们谈到21世纪是多么遥远，如今却已经走过十年。谈过了动真情的恋爱，感受了生活的无奈。我们已经成人，没有一个“家”做靠山，没有同学间那自然的亲近。而相反，我们是走到哪里哪里就是家，遇见谁谁就是“朋友”。可是真的都对得起这两个字么？

你在电话中跟我诉说你的困难，我知道，我理解。我经历过，虽然不会完全相同，但是类似的。人啊，经历的事情十有八九都差不离，可都是到了自己身上难以面对。你可以度过这段艰难的日子的。不要后悔什么，老年人才常常去后悔去怀旧。多去想想下面该怎样，如果你居然敢保证从今天开始不再犯巨大错误，不再走错路，那便了不得了。所以，即使不能够保证，也要小心谨慎的去争取！不要忘了，我们还有“年轻”作为资本。我很喜欢爱默生的一句话，“为了一个特定目标而不断前进的人，全世界都给他让路。”我要把这个送给你。

今年我的申请与十一期间正式开始。我不知道结果会怎样，但是我知道即使我出去到了一个陌生的地方，艰难的压力仍然很大很多。我不时的会停下来，告诉自己奋斗并不是我的全部。而且我常常回忆过去那段日子都会面带微笑。那段时光太美好了，美好的我简直不敢去相信。美好本身原来是力量的源泉。当我每次跨步向前的时候，我是如此的坚定和果敢。

你也一样。

2010-9-30 零点十三分

Hint of last problem: Both sides can be interpreted as a fixed squares and a partition without any restriction.

When we already know some combinatorial results, more than often we are far from satisfaction. We always want to seek bijective proofs in order to convince ourselves that we “actually” understand this result. Why bother? It is always the case that we are not seeking bijections. What we are seeking are “easy bijections”, which, can make more people understand it more easily.

Albert Einstein once said, “Make everything as simple as possible, but not simpler.”I guess the second half of this quote means if things are just that simple, then making it even simpler is actually making it wrong. Bijections and all combinatorial proofs are something people are seriously considering about simplest way to make things.

Recently I am working on something related with partition bijections. I feel that this maybe the best thing for high school students to do if they want to start working on serious research problems before they attend college.

Concave compositions of even length is the ordered pair of partitions with distinct parts with the property that they have the same length and that the last part are equal. We also note that zero part is allowed, too.

Strictly speaking, they are of the form , with .

Interestingly enough, we have the generating function of concave compositions as following:

I call it interesting because of the fact that

And is just the rihgt hand side with variations of sign.

I hope to give a bijective proof of this generating function.