Jon J.

# What is the rule for the sequence/pattern (X) that produces the following numbers (Y): (1,3), (2,9), (3,30), (4,108), (5,408), etc.....?

In other words:

Pattern 1= 3 line segments (1 triangle)
Pattern 2= 9 line segments (4 triangles)
Pattern 3= 30 line segments (16 triangles)
etc.......

Andre W.

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01/29/14

Kenneth G.

Andre,  I don't agree that the problem, as stated, has a "unique" answer.   However, I am very interested to see your solution!
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01/29/14

Andre W.

Kenneth, the problem is poorly worded. I believe there is an entire paragraph missing before "in other words". It should read like this:
Find the number of line segments in the n-th term pattern of a sequence of triangles defined iteratively as follows:
Pattern 1: 1 equilateral triangle with 3 line segments
Pattern n: 4 triangles of pattern (n-1) arranged as an equilateral triangle.
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01/29/14

Kenneth G.

Andre,   Thanks.  I like your re-statement!
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01/29/14

Kenneth G.

Andre,

Still waiting for your answer.  I'm going to hypothesize that it's something like this based on your statement of the problem, but I can't prove it yet.

If Sn is the number of sides in the nth step,  So S1 = 3, S2 = 9, etc.   Then I hypothesize a recursive formula that looks something like this:  Sn = 3Sn-1 +  ∑k=0 to n; n-k ≥ 3 (3kSn-k-2).

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01/29/14

Andre W.

Kenneth,
I'm not sure I understand your summation, k=0 to n, but n-k ≥ 3 (which means k ≤ n-3 ?).

Rather than adding segments to 3Sn-1, I subtracted segments from 4Sn-1 (namely, the outer edges of the inner triangle). The final formula I got when I solved the recursion relation for Sn was Sn = 3*(2*4n-1 + 2n-1).
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01/30/14

Kenneth G.

Here is another potential solution:

# triangles = 4n-1

# sides = # of unshared sides - (number of shared sides)/2

So the non-recursive formula would be:

Sn = 3*2n-1 + 3(4n-1 - 2n-1)/2

Proof:  The number of unshared sides doubles each step and the number of triangles increases by a factor of 4.  So if you subtract the number of unshared sides (3*2n-1) from the total number of sides if the triangles were disjoint (3*4n-1) and the divide by 2 you get the total number of sides that are shared.  Then add back the number of unshared sides to get the total number of sides.
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01/30/14

Kenneth G.

Sorry, I didn't see your latest comment.  Yes, now we both have the same answer.  I do think my summation works also.

Let me clarify the summation.  The summation is over all k where k is between 0 and n and also k ≤ n-3. To be in the summation, k must satisfy both conditions.  So for n = 1 or 2, no k satisfies that condition, so the summation term goes away.  Starting with n = 3 the summation comes into play.

I got the summation by watching the behavior of the triangles in the construction.  You can make a structure leaving out the center triangle where no sides are shared.  When you add the center triangle the outer sides aren't counted a second time, so you basically only have to add the sides of the inner central triangle when n = 3.  But for higher values of n you need to add not only the sides of the inner center triangle but also the sides of lower level triangles that fit in the outer triangles of that inner triangle.   I hope that's clear.

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01/30/14

Andre W.

I understand now. The recursion relation is apparently not unique, but the explicit formula for Sn is. Our formulas only look differently because mine starts with n=0 and your starts with n=1.
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01/30/14

Kenneth G.

Andre,  I think you misstated your formula in your comment.  I think it should have been 3*(4n-1 + 2n-1)/2.  The formula you posted above gets the wrong value for n=4 which should be 108.

I'm not sure I understand what you did on your recursion where you said "Rather than adding segments to 3Sn-1, I subtracted segments from 4Sn-1 (namely, the outer edges of the inner triangle)."

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01/30/14

Kenneth G.

I'd like to see your recursion formula and how you resolved it to a non-recursive formula.  I'm not sure how to do that.
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01/30/14

## 3 Answers By Expert Tutors

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Andre W.

I solved the recursion relation by first finding its generating function, f(x) = ∑ anxn, from the recursion relation. Then I used the geometric series (twice) for f(x) to get the formula for an.
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01/30/14

Kenneth G.

Thanks.
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01/30/14

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