Defining Katie’s Triangle and Common Properties with Pascal’s Triangle

 

Charles Lundelius, Howard Community College
Jake Hoffman, Howard Community College

Mentored by: Mike Long, Ed.D. & Loretta FitzGerald Tokoly, Ph.D.

Abstract

A triangular array, known as Katie’s Triangle, was derived from Pascal’s Triangle. The talking about it in the abstractarray was initially derived directly using the elements in Pascal’s Triangle, but later a general form that uses an iterative process with the elements of Katie’s Triangle was derived.  For any element in Katie’s Triangle, a term in the n + 1 row takes the larger numerator and denominator of the adjacent upper left and upper right terms of the nth row.  Once a derivation for Katie’s Triangle was found, the structure of it was studied and compared to that of Pascal’s Triangle to determine if there were similar properties.   Some of the properties were found to exist in both:  including the flowers and the Triangular Numbers.  While the flowers property held exactly the same for both triangles, the Triangular Numbers existed albeit in a different arrangement than in Pascal’s Triangle.  Other properties were found to either not exist at all, or were not one-to-one matches with Pascal’s properties.  These include sum of the rows, sum of the first n Figurate Numbers, and the existence of most figurate numbers.  Despite being derived initially from Pascal’s Triangle, Katie’s Triangle stands on its own.

Paper

A colleague, Katie, was analyzing Pascal’s Triangle when she decided to multiply each element to get the next one in a row. The resulting needed factors were then removed and formed into their own triangle. From here, the new triangle was analyzed for properties similar to Pascal’s Triangle.  After generating Katie’s Triangle, the question arose as what to relationships, if any, were translated from Pascal’s Triangle to this new triangle.  A second question was if there were new properties that existed in Katie’s Triangle.  Katie’s Triangle is constructed starting with LaTeX: \frac{1}{1}  at the peak.

Triangle0.PNG

Each successive row has one more entry than the previous row. The numerators start at one and increase by one moving across the row from right to left while the denominators start at one and increase by one moving from left to right. Any term in the triangle can be expressed by the following general formula with the peak defined as row zero and diagonal zero:

LaTeX: K_n^k=\frac{k+1}{n-k+1}

where LaTeX: n represents the cardinality of the row and LaTeX: k represents the diagonal indexed down the left side of the triangle. In addition, LaTeX: n \ge 0LaTeX: k \ge 0, with LaTeX: k, n being Natural Numbers. LaTeX: \frac{1}{1}  is defined as the 0th row and 0th diagonal.  Figure 1 shows rows zero through seven of Katie’s Triangle and also diagonals zero through seven.  A diagonal refers to the string of numbers continuing down and to the right as depicted in Figure 1. The indices for the diagonals start from the top center and then incrementing by one as you travel to each new diagonal that starts on the left hand side of the triangle. Therefore LaTeX: \frac{1}{1} starts the 0th diagonal with LaTeX: \frac{1}{2} being that diagonal’s second term and LaTeX: \frac{2}{1} starts the 1st diagonal with LaTeX: \frac{2}{2} being that diagonal’s second term.

Triangle1.PNG
Figure 1

       It was decided to look at the following structures from Pascal’s Triangle:  flowers, sum of rows, sum of the Triangular Numbers, figurate/ polygonal numbers, and triangular numbers.

In Pascal’s Triangle, choose an interior entry and call that the center of the flower.  There surrounding six entries referred to as petals.  The product of any alternating set of three petals equals the product of the remaining three petals (see Figure 2).

Triangle2.PNG
Figure 2: Pascal Petals [1]

In Katie’s Triangle, this property also exists (see Figure 3). For example, if one takes the flower around the entry LaTeX: \frac{3}{3}, one can see that this property holds true (see Figure 4).

Triangle34.PNG

       While some properties of Pascal’s Triangle translate directly to Katie’s Triangle, some do not.  In Pascal’s Triangle, the 0th diagonal is composed entirely of 1’s; the next diagonal, moving inward, contains the Natural Numbers; the 2nd diagonal contains the Triangular Numbers, and from there each subsequent diagonal is the set of the next higher Figurate Number. In Katie’s Triangle, this pattern does not hold. Each successive diagonal has a constant numerator and an increasing denominator from right-to-left.

       While not in the same way as Pascal’s Triangle, the Triangular Numbers do exist in Katie’s Triangle.  Starting in row 1, and then each subsequent row, if the last two elements on the left side of each row are multiplied together, they produce the Triangular Numbers. Table 1 below show this for five rows of Pascal’s Triangle.

n 2nd to last element from the left Last element on the left Product
1 LaTeX: \frac{1}{2} LaTeX: \frac{2}{1} 1
2 LaTeX: \frac{2}{2} LaTeX: \frac{3}{1} 3
3 LaTeX: \frac{3}{2} LaTeX: \frac{4}{1} 6
4 LaTeX: \frac{4}{2} LaTeX: \frac{5}{1} 10
5 LaTeX: \frac{5}{2} LaTeX: \frac{6}{1} 15

Table 1

       This can be proved using the last two elements on the left side in any row (beginning with row LaTeX: n=1) . Since the two elements in any row will always be from the 0th and 1st diagonal, we know that for the next to last element from the right, LaTeX: k=1, and the for the last element from the left, LaTeX: k=0. Let any element in Katie’s Triangle be:

LaTeX: K_n^k=\frac{n-k+1}{k+1}

The product of two numbers of this form will be:

LaTeX: \left(\frac{n-k+1}{k+1}\right)\left(\frac{n-k+1}{k+1}\right)

Since the first element taken has LaTeX: k=1 and the next has LaTeX: k=0, the product now takes the form:

LaTeX: \left(\frac{n-(1)+1}{(1)+1}\right)\left(\frac{n-(0)+1}{(0)+1}\right)=
LaTeX: \left(\frac{n}{2}\right)\left(\frac{n+1}{1}\right)=
LaTeX: \frac{n(n+1)}{2}

The resulting form is the general form for all Triangular Numbers. On the right side of the Triangle, a similar result holds true. If instead of selecting the last two elements of a row, one selects the first two elements on the right side of the row (beginning with row LaTeX: n=1) then the resulting product of those two numbers will be the reciprocals of the Triangular Numbers. This can be proven generally as well. If we again call the product of the elements as before:

LaTeX: \left(\frac{n-k+1}{k+1}\right)\left(\frac{n-k+1}{k+1}\right)

The first element in every row will have LaTeX: k=n. Therefore, second element will have LaTeX: k=n-1. The product now takes the form.

LaTeX: \left(\frac{n-(n)+1}{(n)+1}\right)\left(\frac{n-(n-1)+1}{(n-1)+1}\right)=
LaTeX: \left(\frac{1}{n+1}\right)\left(\frac{n-n+1+1}{n}\right)=
LaTeX: \left(\frac{1}{n+1}\right)\left(\frac{2}{n}\right)=
LaTeX: \frac{2}{n(n+1)}

       This is the reciprocal of the Triangular Numbers since the generating formula for the Triangular Numbers is LaTeX: \frac{n(n+1)}{2}.

       Some properties of Pascal’s Triangle do not translate at all to Katie’s Triangle.  In Pascal’s Triangle, each subsequent row sums to the next highest power of two, where the sum of the LaTeX: n^{th} row is LaTeX: 2^n, starting with the 0th row.  An attempt was made to find a similar pattern in Katie’s Triangle, but no discernible pattern was identified. Table 2 compares the sum of rows 0 to 4 in Pascal’s Triangle and Katie’s Triangle.

n Pascal’s Katie’s
0 1 1
1 2 LaTeX: 2\frac{1}{2}
2 4 LaTeX: 4\frac{1}{3}
3 8 LaTeX: 6\frac{5}{12}
4 16 LaTeX: 8\frac{7}{10}

Table 2

       A pattern may still be present in the sum of the elements in each row of Katie’s Triangle’s, but that is for further study. What is known is that the product of the elements of each row of Katie’s Triangle is 1.

The pattern that the sum of the first n Figurate Numbers is in the adjacent diagonals is certainly in Pascal’s Triangle and can be seen in Figure 5.

Triangle5.PNG
Figure 5: Sum of diagonals property [2]

Using Katie’s Triange (written as [row, diagonal]) we can replicate this form.

          LaTeX: [n,k]

                    LaTeX: [n+1, k]

                              LaTeX: [n+2, k]

                                        LaTeX: [...]

                                                  LaTeX: [n+j, k]

                                                            LaTeX: [n+j+1, k+1]

Looking at just the 3rd (LaTeX: k = 2) diagonal of Katie’s Triangle, a counter example shows LaTeX: \frac{3}{1}+\frac{3}{2}+\frac{3}{2} does not equal LaTeX: \frac{4}{3}. Therefore this property does not hold true for Katie’s Triangle.

Finally, the Figurate Numbers and other patterns found on the diagonals of Pascal’s Triangle are not found in Katie’s Triangle. The very structure of Katie’s Triangle makes finding these impossible, since the diagonal would have to be made of a subset of the Natural Numbers, and only the outside left Diagonal, which is the set of Natural Numbers, has this property.

From the results of the research it can be concluded that Katie’s triangle is a distinct entity to Pascal’s Triangle as it maintains some similar properties and yet others do not translate directly or do not work entirely.

Contact: charles.lunkelius@howardcc.edu, jacob.hoffman@howardcc.edu

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References

[1] The Math Forum. (n.d.) Pascal Petals. Retrieved from http://mathforum.org/workshops/usi/pascal/petals_pascal.html

[2] The Math Forum. (n.d.) Sum of Diagonals Property. Retrieved from http://mathforum.org/mathimages/index.php/Image:Pascalhockey.jpg

 

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