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In space there are given $p_n=\lfloor{en!}\rfloor+1$ points. Each pair of points are connected by a line, and each line is coloured by one of the $n$ colours. Prove that there is at least one triangle of the same coloured sides.

I am not getting how to counter the $e$ here. Please help.

I however made one guess though.. Is $p_n = R(\underbrace{3,3,\cdots,3}_{n \ times})$ ?. I guessed so just because It holds true for $n=2,3$.

NB: $R$ is the Ramsey number.

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    The $e$ is there related to the fact that $\lfloor en! \rfloor= n! + \frac{n!}{1!}+\cdots +\frac{n!}{n!}$. – 2017-02-02
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    The question implies $p_n \ge R(\underbrace{3,3,\cdots,3}_{n \ times})$ – 2017-02-02
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    @HenningMakholm $\lfloor en!\rfloor=\lfloor e\times n!\rfloor$ – 2017-02-02
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    You are making the problem infinitely harder than the already hard one that was posed. The problem asks to prove that $R(\underbrace{3,3,\cdots,3}_{n \ times}) \leq \lfloor n!\, e\rfloor $. If your guess that this Ramsey number is $p_n$ were right, it would mean be that you had to prove that $p_n$ is the exact value of the Ramsey numer. That is assuredly well beyond the capability of current human mathematics! $ \lfloor n!\, e\rfloor $ is for $n>3$ probably much larger than the minimum number of vertices needed to assure a monochrome triangle; you "just" have to show that it does assure one. – 2017-02-02
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    Isn't this in essence the same question as [How do I get this upper bound for Ramsey numbers: $R_k \le \left \lfloor k!e \right \rfloor + 1$?](http://math.stackexchange.com/q/1718493) – 2017-02-02

1 Answers 1

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Let $f(n)$ be the maximum size of a complete graph that can have its edges colored with $n$ colors without triangles.

Notice that $f(n+1)\leq (n+1)(f(n))+1$. To see this pick an arbitrary vertex $v$ and consider the connections to the remaining $f(n+1)-1$ vertices. If we had $f(n+1)-1>(n+1)f(n)$ then by the pigeon-hole principle there would be a set with $f(n)+1$ vertices that are all connected to $v$ by the same color, this is a contradiction, because those $f(n)+1$ vertices would have to have all of their connections with only the remaining $n$ colors, and a triangle would be formed.

We have $f(2)=5\leq \lfloor e2!\rfloor $.

We now prove by induction that $f(n)\leq\lfloor en!\rfloor$

We just have to prove that $(n+1)\lfloor en!\rfloor+1\leq \lfloor e(n+1)!\rfloor$.

this is equivalent to proving that $(n+1)\lfloor en!\rfloor<\lfloor e(n+1)!\rfloor=\lfloor en!(n+1)\rfloor$

All we have to do is show that the fractional part of $en!$ is at least $\frac{1}{n+1}$ but the fractional part of $en!$ is $\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}\dots$

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    Your first "notice that..." statement is far from obvious to me. How do you show that? I tied starting by separating the vertices on the right into $n+1$ groups with one left over, but got nowhere. – 2017-02-02
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    Can you please explain your `Notice that` ?.. I am a beginner in Graph theory, so... It went over my head. – 2017-02-02
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    I added a paragraph explaining it. – 2017-02-02
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    @Qwerty The argument is similar to upper bound on Ramsey numbers – 2017-02-02
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    @Axolotl do you study math? – 2017-02-02
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    Your added explanation works for me. – 2017-02-02
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    @JorgeFernándezHidalgo Yeah. Why are you asking? – 2017-02-02
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    @Axolotl maybe we study at the same college and you we can meet or something. – 2017-02-02
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    I loved to be so, But It isn't Unfortunately. – 2017-02-02
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    oh, which one is yours? – 2017-02-02
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    I'm not in mexico. – 2017-02-02
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    O.o did you participate in the Olimpiada? – 2017-02-02
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    You can probably start with $f(1)=2$ rather than $f(2)=5$? – 2017-02-02