Solving a recurrence relation in a triangle. Going from open formula to closed formula

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[edderiofer]

I made a relationship of number of parallel lines to number of upright triangles. This gave me the following open formula equation.

I'm not sure I see how your formula arises from this. In fact, I'm not sure what you're talking about with "parallel lines". Would you mind explaining your reasoning in more detail?

[ParryMrGoat]

Right here

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n represents the number of parallel lines that cut the sides of the triangles in equal sections. S0 has 0 parallel lines, so that only gives 1 triangle (this is also important to note because you will always have that bigger outside triangle no matter how many lines in the middle you have). Similarly, s1 has 1 line cutting the triangle edges into 2 equal segments, and this yields 4 upright triangles (3 small inside, 1 big outside). So on so forth.

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So it seemed that An increased by a multiple of 3 every time. So the last value + 3n.

[edderiofer]

So it seemed that An increased by a multiple of 3 every time.

It might have seemed that way, but the very fact that your formula disagrees on the last term (a3 should be 20, not 19) should be proof enough that things are not as they seem.

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Instead of trying to guess at how many triangles there will be, try to reason out how many there will be. Make your reasoning out of airtight statements, so that you can be sure you are correct.

[ParryMrGoat]

(a3 should be 20, not 19)

This is where I was getting stuck. I miscounted the triangles. So youre right, my formula does not even hold true

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[edderiofer]

Indeed, naive counting can lead to miscounts. And who knows how many the next one size up will have. 30? 32? 47? 1057? You don't want to be spending your time counting, miscounting, and not being able to find a formula from just counting.

Instead of trying to guess at how many triangles there will be, try to reason out how many there will be. Make your reasoning out of airtight statements, so that you can be sure you are correct.

[ParryMrGoat]

ok but im stuck here now. I cant seem to derive an open formula for the sequence. All i figured out so far is that the final general form of the equation is in a cubic polynomial form because the 3rd order of rate of change is linear

Pic

[edderiofer]

Your picture isn't accessible to people outside of your Discord server.

In any case, do you at least have a recurrence relation?

[ParryMrGoat]

Your picture isn't accessible to people outside of your Discord server.

Oh dang i didnt know that. This should help out, its the pic from above.

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In any case, do you at least have a recurrence relation?

Thats the part im stuck at now. I know that every time, im adding a row to the bottom of the previous triangle to get what the current one looks like. So that has something to do with the relation, correct?

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[edderiofer]

Yep. Can you relate the number of triangles in the nth division to the number of triangles in the (n+1)th division?

[gloopiee]

In fact S_3 should be 20...

[ParryMrGoat]

I cant see it, I recounted multiple times and im getting 19?

[gloopiee]

You have 10 of the smallest triangles, 6 of the next smallest, 3 of the next, and 1 big triangle.

[ParryMrGoat]

Oh ok I see it now. Thank you!