Creating a Tromino with n=8 and Missing Square at (4,6)

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I'm trying to create a tromino where n=8 and the missing square is four from the left and six from the top.

Here's what I have so far:

Input: n, a power of 8; and the location L of the missing square
Output: A tiling of an n x n deficient board
1. tile(n,L) {
2. if (n==8) {


I'm unsure of how to denote where the missing square is. Can anyone help?
 
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What's a tromino?
 
StatusX said:
What's a tromino?

That doesn't bode well for me - but a tromino is an object made up of three squares. It can also be called a polyomino.
 
Well, that doesn't really narrow it down. But I can guess what you mean is one of the two 2D shapes consisting of 3 squares, where each shares a side with one of the others, ie, they're in a straight line or they make an L.

But then what does n refer to? Or do you mean for there to be n squares, where n isn't necessarily 3? (I'm betting this is actually what "polyomino" refers to)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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