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June 20th, 2014, 11:21 AM   #11
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I was looking forward for your question.

This can be done by counting endpoints of diagonals, I think, which avoids counting diagonals and their meetpoints altogether.

The number of triangles in the interior of diagonals with three endpoints is $\binom{6}{3}$. The number of triangles in the interior of diagonals with four endpoints is $4\binom{6}{4}$. Similarly, the number of triangles in the interior of diagonals with five endpoints is $5 \binom{6}{5}$ (note that the diagonals with 5 endpoints must meet themselves in a hexagon)

Thus, the number of triangles is

$$\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} = 110$$
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June 20th, 2014, 02:33 PM   #12
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This is also A006600 in the OEIS.
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June 20th, 2014, 03:38 PM   #13
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Quote:
Originally Posted by CRGreathouse View Post
For those interested in the nonmodular answer in its 5782-digit glory, here's a GP script:
Code:
val(n,p)=my(t);while(n\=p,t+=n);t
t=1;forprime(p=2,2014,t*=p^val(2014,p)+1);t
Here is another, hope you'll like it:

Code:
prod(i=1,primepi(2014),g=2014;p=prime(i);(p^sum(j=0,log(g)\log(p),g\=p)+1))
The inaccuracy in log doesn't matter here. Since we start from j = 0, most of the time, 0 is added once too much.
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June 20th, 2014, 04:16 PM   #14
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I'm amused by how similar our programs are and yet how different they appear. The prod syntax is nicer (no need for a temporary variable to hold the sum) but generating the n-th prime at each step is inefficient. Not that it matters when you're only looping up to 2014, of course!

I think we're ready for a new question?
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June 21st, 2014, 03:31 AM   #15
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Here are the question, the one who gets one of the hard questions below gets the control.

Easy : Let $f$ be a univariate complex function satisfying $f(x) = f(x + \omega_1) = f(x+\omega_2) = f(x + \omega_3)$ with $\omega_1, \omega_2, \omega_3$ being $\Bbb R$-linearly independent. Can a function like that exist? If so, give example otherwise show proof. [Bonus point 2]

Hard : Let $f$ be defined as above except $\omega_1, \omega_2, \omega_3$ being $\Bbb Q$-linearly independent. Can such a function exist? If so, give example otherwise show proof.

Hard (this one is for real analysis lovers) : Let $f : \Bbb R \to \Bbb R$ be a function such that for all $x, y \in \Bbb R$ one has $f(x + y^2) - f(x) \geq y$. Can such a function exist? If so, give example otherwise show proof.

Balarka
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Last edited by mathbalarka; June 21st, 2014 at 03:36 AM.
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June 21st, 2014, 04:39 AM   #16
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What does it mean to say that the $\omega_i$ are $\mathbb{R}$ linearly independant? Aren't they constants?
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Last edited by v8archie; June 21st, 2014 at 04:41 AM.
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June 21st, 2014, 06:22 AM   #17
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\begin{align*}
f(x + y^2) - f(x) &\ge y \implies f(x + y) - f(x) \ge \sqrt{y} \qquad \text{(the positive root since $\sqrt{y} \ge -\sqrt{y}$)} \\[12pt]
\text{For } a \in \mathbb{R}^+, n \in \mathbb{Z}^+ \qquad f(x + \frac{a}{n}) - f(x) &\ge \sqrt{\frac{a}{n}} \\
f((x + \frac{a}{n}) + \frac{a}{n}) - f(x + \frac{a}{n}) &\ge \sqrt{\frac{a}{n}} \\
f((x + \frac{2a}{n}) + \frac{a}{n}) - f(x + \frac{2a}{n}) &\ge \sqrt{\frac{a}{n}} \\
&\cdots \\
f((x + \frac{(n-1)a}{n}) + \frac{a}{n}) - f(x + \frac{(n-1)a}{n}) &\ge \sqrt{\frac{a}{n}} \\[6pt]
\text{Summing the above equations gives} \qquad f(x+a) - f(x) &\ge \sqrt{an} \\
\end{align*}
Then, letting $n \to \infty$ we see that $f(x+a) - f(x) \to \infty$. This is true for all $x$ and all $a \gt 0$ so no such function exists.
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Last edited by v8archie; June 21st, 2014 at 06:25 AM.
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June 21st, 2014, 06:47 AM   #18
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Quote:
Originally Posted by v8archie View Post
What does it mean to say that the $\omega_i$ are $\mathbb{R}$ linearly independant? Aren't they constants?
Yes. For example, $1$ and $\pi$ are $\Bbb Q$-linearly independent.
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June 21st, 2014, 06:51 AM   #19
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Then, letting $n \to \infty$ we see that $f(x+a) - f(x) \to \infty$. This is true for all $x$ and all $a \gt 0$ so no such function exists.
Nice work. You get to post the next question.

PS : There is actually a swifter argument, IMHO, using the Basel problem.
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June 21st, 2014, 07:31 AM   #20
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I'm not familiar with the Basel problem. I first proved it to myself using l'Hopital's Rule on$$\lim_{y \to 0} \frac{f(x + y^2) - f(x)}{y} \ge 1$$
but I wasn't completely happy with the validity of the argument.
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