Jon J.

asked • 01/29/14# 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.......

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## 3 Answers By Expert Tutors

Kenneth G. answered • 01/29/14

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jon,

Five points do not make a pattern, but only evidence of a pattern. Questions like this have multiple answers, and are asking one to guess what's in the mind of the question's author.. Unfortunately, these types of problems masquerade as Math problems when they really belong, in my opinion, only in parlour games.

Here is one solution to the pattern: fit some polynomial of degree 4 or higher, for example f(x) = Ax

^{4}+ Bx^{3}+ Cx^{2}+D^{x}+ E, to the five points, so that f(1) = 3, f(2) = 9, f(3) = 30, f(4) - 108, f(5) = 408, and then use this polynomial to generate the rest of the "pattern".3 = f(1) = A + B + C + D + E

9 = f(2) = 16A + 8B + 4C + 2D + E

30

_{ }= f(3) = 81A + 27B + 9C + 3D + E108 = f(4) = 256A + 64B + 16C + 4D + E

408 = f(5) = 625A + 125B + 25C + 5D + E

So the solution is: A = 41/8, B = -177/4, C = 1159/8, D = -783/4, E = 93

And the next entry in the sequence is:

f(6) = 1296A + 216B + 36C + 6D + E = 1218, so we have (6,1218)

f(7) = 2401A + 343B + 49C + 7D + E = 2949, so we have (7,2949)

Using this formulation, we can even continue the "pattern" in the other direction:

f(0) = 93, giving (0,93)

f(-1) = 483, giving (-1,483)

f(-2) = 1500, giving (-2,1500)

If you object that the numbers must be divisible by 3 because of the mention of triangles, note that all of these new results are indeed divisible by 3.

Andre W. answered • 01/29/14

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This pattern formation is an iterative (recursive) process and leads to a so-called

*fractal*. In step 0 you start with one 3-segment triangle. Then in each following step you take four of the triangles of the previous step and put them together to form a new triangle (with one of the four triangles "upside down"). The number of segments in each step, however, will not just be 4 times the number of segments of the previous step, because for one of the added triangles (the "upside down", or interior, one) you have to subtract those segments that are already part of the other three triangles, lest you count them twice. In the n-th step, each of the other three triangle from the (n-1)th step contributes 2^{n-1}segments to the interior triangle, for a total of 3*2^{n-1}segments that need to be subtracted. Therefore, the recursion relation for the number of segments in the n-th step isa

_{n }= 4*a_{n-1}- 3*2^{n-1}, with a_{0 }= 3.You get

a

_{1}= 4*a_{0}- 3*2^{0}= 12-3 = 9a

_{2}= 4*a_{1}- 3*2^{1}= 36-6 = 30a

_{3}= 4*a_{2}- 3*2^{2}= 120-12 = 108a

_{4}= 4*a_{3}- 3*2^{3}= 432-24 = 408a

_{5}= 4*a_{4}- 3*2^{4}= 1632-48 = 1584a

_{6}= 4*a_{5}- 3*2^{5}= 6336-96 = 6240etc.

When you solve the recursion relation for a

_{n}you geta

_{n}= 3*(2*4^{n-1}+ 2^{n-1}),e.g., a

_{6}= 3*(2*4^{5}- 2^{5}) = 6240.In the limit as n→∞, the pattern becomes a fractal: a self-similar pattern whose dimension lies between 1 and 2.

Looking at the problem differently, Jon writes Pattern 1, Pattern 2, Pattern 3, and so on.

Looking at these separate patterns individually will yield different rules for each Pattern.

To begin with you have one triangle with the base at the bottom and a vertex at the top(3 segments and

1 triangle). Notice I'm looking at ordered pairs involving segments and triangles; that is (s,t) where

s=# of segments and t=# of triangles.

/\(this is the first triangle)

Now make a second triangle under the first; 5 segments, 2 triangles.

Make a third triangle to the right of the first; 7 segments, 3 triangles

Make a fourth triangle under the third; 9 segments, 4 triangles

Make a fifth triangle at the top to the right of the third one you made; 11 segments, 5 triangles

This is the first Pattern(not including the first triangle because these patterns occur in groups of four.)

(5,2), (7,3), (9,4), (11,5)

The sixth triangle in the pattern only needs one more segment giving us 12 segments, 6 triangles.

The seventh triangle in the pattern yields 14 segments and 7 triangles.

This is the second Pattern of four ordered pairs:

(12,6), (14,7), (16,8), (18,9)

The third Pattern yields the ordered pairs (19,10), (21,11), (23,12), (25,13)

The fourth Pattern yields (26,14), (28,15), (30,16), (32,17)

The fifth Pattern yields (33,18), (35,19), (37,20), (39,21)

You can continue this pattern of triangles for as long as you like.

The sequence for the ordered pairs (s,t) in Pattern 1 would be t=(s-1)/2

The sequence for the ordered pairs (s,t) in Pattern 2 would be t=s/2

Pattern 3 you would have t=(s+1)/2

Pattern 4 you would have t=(s+2)/2

Pattern 5 you would have t=(s+3)/2

These rules themselves have a pattern to them(s is being increased by 1 every four newly created triangles.)

Now if we could take all these rules and put them together into one rule we would have another solution

to the problem that would not be changed or interrupted every four triangles.

I discovered these patterns looking at the problem in terms of segments and triangles.

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Andre W.

01/29/14