An Introduction to Polygonal Numbers, Their Patterns, and Relationship to Pascal's Triangle

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An Introduction to Polygonal Numbers, Their Patterns, and Relationship to Pascal’s Triangle

Tyler Albany

April 22, 2015

1 The ancient Greek mathematician Diophantos was one of the first to study polygonal numbers. A polygonal number can be defined as a sum of equidistant dots used to represent a polygon of a certain size. For example, if you have a square number with rank one it is one, rank two is four because you expand the length and width by one dot each and fill in the outer layer, then rank three would be nine and it continues in this fashion. The rank of a polygonal numbers is the number of dots on a side of the outermost layer of the polygonal number. This holds true for all polygonal numbers. (see figure 1.1) A polygonal number is denoted by Pd(n) where d is the number of sides to the corresponding polygon and n is the rank, or order, of the polygonal number. For instance P5(4) would be a pentagonal number with rank four. All polygonal numbers with rank one equal one, and all polygonal numbers of rank two are equal to the number of sides on the corresponding polygon. Furthermore, you can find (d−2)n2+(4−d)n any polygonal number by using the formula Pd(n) = 2 . Note that d, n ∈ N, and d ≥ 3 since less than three sides would not form a polygon. Throughout this paper I will show some of the various properties of polygonal numbers, such as: the closed formula for a triangular number of rank n, the closed formula to find any polygonal number, that all hexagonal numbers are also triangular numbers, a polygonal number can be prime if and only if it has a rank of two and the number of sides for the polygon is prime, and finally, the relationship between Pascal’s triangle and triangular numbers.

Triangular numbers are the most basic polygonal numbers. This stems from how triangular numbers are formed. Suppose you have a triangular number of rank n, to get the triangular number of rank n + 1 simply add n + 1 dots. For example, since the first triangular number is one, to get the second you add two, to get the third you add three and so on until you have found your desired triangular number. While this method will always work and it is quick to use for small triangular numbers it would take a very long time to find the 1000th triangular number. To find the formula for the nth triangular number we will proceed by induction.

Deﬁnition 1.1: tn = 1 + 2 + ··· + n th n(n+1) n(n+1) Theorem 1: The n triangular number is equal to 2 , tn = 2 . Proof.

1(1+1) Base Case: n = 1, t1 = 1 = 2 n(n+1) Inductive Step: Assume tn = 2 By deﬁnition 1.1

tn+1 = 1 + 2 + 3 + ··· + n + (n + 1) n(n + 1) = + (n + 1) 2 n(n + 1) 2n + 2 = + 2 2 n2 + 3n + 2 = 2 (n + 1)(n + 2) = 2

While knowing the formula to find any triangular number is nice to have what if you wanted to find the 4th pentagonal number or even the 1357th octagonal number. There is no feesable way to simply count to find these numbers it would take far too long. In order to find the formula for any polygonal number I will use some properties of finite calculus. Moreover, in order to find the formula

2 for any polygonal number we need to deﬁne the formula for the number of dots added to the (n − 1)st polygonal number in order to receive the nth polygonal number.

Property 2.1: The diﬀerence, similar to a derivative in calculus, of a function f : Z → Z is ∆f : Z → Z and equals f(n + 1) − f(n).

Property 2.11: ∆(cf(n)) = c∆f(n)

Property 2.12: ∆(f(n) + g(n)) = ∆f(n) + ∆g(n)

Property 2.2: The number of dots added to the (n − 1)st polygonal number to get the nth polygonal number is equal to 1 + (d − 2)(n − 1).

Lemma 2.3: ∆tn−1 = n Proof.

∆tn−1 = tn − tn−1 n(n + 1) n(n − 1) = − 2 2 n2 + n n2 − n = − 2 2 2n = 2 = n

Lemma 2.4: ∆n = 1

Proof.

∆n = n + 1 − n = 1

(d−2)n2+(4−d)n Theorem 2: The formula to find any polygonal number with d sides and rank n is 2 . Proof. Let Pd(n) be a polygonal number with d sides and rank n. Therefore ∆Pd(n) = Pd(n + 1) − Pd(n) Pd(n+1)−Pd(n) is simply the polygonal number with d sides and rank n subtracted from the polygonal number with d sides and rank n + 1. The difference between the two will give you the number of dots you add to Pd(n) to get Pd(n + 1). By definition 2.2 we know that to get the (n + 1)st polygonal number we will need to add 1 + (d − 2)n dots. In other words, Pd(n+1)−Pd(n) is equivalent to 1+(d−2)n. Therefore if we can find the fuction that gives us 1 + (d − 2)n when we take its difference we will find the formula for the nth polygonal

3 number. From Lemma 2.3 and 2.4 we know

∆tn−1 = n, ∆n = 1 ∴ ∆[(d − 2)tn−1 + n] = (d − 2)∆tn−1 + ∆n = (d − 2)n + 1

We found the formula we were looking for that gives us 1 + (d − 2)n when you take its diﬀerence. Now if we work backwords we can ﬁnd the formula for any polygonal number.

∆Pd(n) = Pd(n + 1) − Pd(n)

= (d − 2)[∆tn−1] + ∆n

= ∆[(d − 2)tn−1 + n] ∴ Pd(n) = (d − 2)tn−1 + n n(n − 1) = (d − 2) + n 2 n2 − n 2n = (d − 2) + 2 2 (d − 2)n2 − n(d − 2) + 2n = 2 (d − 2)n2 + n(2 + 2 − d) = 2 (d − 2)n2 + (4 − d)n = 2

Now that we know the formula for any polygonal number we can start to see some patterns with these numbers. For instance if we list some of the first triangular numbers: 1, 3, 6, 10, 15, 21, 28 and some of the first hexagonal numbers: 1, 6, 15, 28, 45 you can see that some of these numbers coincide. More importantly you can see that the odd rank triangular numbers listed are also hexagonal numbers. 4k2−2k Lemma 3.1: The formula for a hexagonal number of rank k is 2 Proof. (d−2)n2+(4−d)n We know the formula for any polygonal number is Pd(n) = 2 where d is the number of sides and n is the rank Since we want to find only hexagonal numbers we will set d = 6

(6 − 2)n2 + (4 − 6)n ⇒ P (n) = 6 2 4n2 − 2n = 2

Theorem 3: All triangular numbers with odd rank are also hexagonal numbers.

4 Proof. 4k2−2k Let the hexagonal number with rank k be called hk therefore hk = 2 (n+1)(n+2) Because tn = 2 and we want only odd n and n ≥ 1, let n = 2k − 1

(2k − 1)(2k − 1 + 1) ⇒ t = n 2 (2k − 1)(2k) = 2 4k2 − 2k = 2 = hk

Hexagonal numbers also being triangular numbers is not the only pattern that can be found, or in this case not found. A polygonal number can only be prime if and only if it has rank two and it has a prime number of sides.

Theorem 4: A polygonal number can be prime iﬀ it is of rank two and the number of sides is prime. There are two cases:

Proof. Case 1: (d − 2)n2 + (4 − d)n P (n) = d 2 n = [(d − 2)n + 4 − d] 2 n ⇒ and [(d − 2)n + 4 − d]|P (n) 2 d n n By the deﬁniton of a prime number if 2 is prime then [(d − 2)n + 4 − d] is 1, or 2 is 1 and [(d − 2)n + 4 − d] is prime. n Suppose = p ∈ P rimes, (d − 2)n + 4 − d = 1 2 k n = 2pk

⇒ (d − 2)2pk + 4 − d = 1

d(2pk − 1) = 4pk − 3 4p − 3 d = k 2pk + 1 2(2p − 2) + 1 = k 2pk + 1 1 d 6= 2 + 2pk + 1 ∵ d ∈ N and pk 6= 0 n Suppose = 1, (d − 2)n + 4 − d = p ∈ P rimes 2 i n = 2

⇒ (d − 2)2 + 4 − d = pi

5 2d − 4 + 4 − d = pi

d = pi

Case 2: (d − 2)n2 + (4 − d)n P (n) = d 2 (d − 2)n + 4 − d = n 2 (d − 2)n + 4 − d ⇒ n and |P (n) 2 d

(d−2)n+4−d (d−2)n+4−d By the deﬁniton of a prime number if n is prime then 2 is 1, or n is 1 and 2 is prime. (d − 2)n + 4 − d Suppose n = p ∈ P rimes, = 1 n 2 (d − 2)p + 4 − d ⇒ n = 1 2 (d − 2)pn + 4 − d = 2

d(pn − 1) = 2pn − 2 2p − 2 d = n pn − 1 d 6= 2 ∵ d ≥ 3 (d − 2)n + (4 − d)n Suppose n = 1, = p ∈ P rimes 2 j d − 2 + 4 − d ⇒ = p 2 j 2 = 1 6= p 2 j

Conversely, let d = pm ∈ Primes and n = 2 (p − 2)22 + (4 − p )2 P (2) = m m pm 2 = 2pm − 4 + 4 − pm

= pm

Perhaps the most interesting property of polygonal numbers comes from a seemingly unrelated object. All triangular numbers are included in Pascal’s triangle. However, not only are they included they are all in the third diagonal (see ﬁgure 5.1). In order to prove that the third diagonal of Pascal’s triangle contains all triangular numbers it is important to recall an important property of Pascal’s triangle.

n n! Property 5.1: Any number in Pascal’s triangle can be found using k = k!(n−k)! where n is the row and k is the position in the row starting at the leftmost position, and n ≥ 0, 0 ≤ k ≤ n. In other

6 words, the ﬁrst row in Pascal’s triangle is row zero and the leftmost position in a given row is also zero.

Theorem 5: The third diagonal of Pascal’s triangle includes all triangular numbers.

Proof. Notice that the third diagonal of Pascal’s triangle starts in row two and position zero, the next number in the diagonal is in the third row and position one. k+2 Therefore by property 5.1 we can deﬁne each number in the third diagonal by k .

Let tk be a triangular number of rank k. k + 2 (k + 2)! = k k!(2 + k − k)! (k + 2)! = 2!k! 1 (k + 2)(k + 1)(k) ··· (2)(1) = · 2! (k)(k − 1) ··· (2)(1) (k + 2)(k + 1) = 2 = tk+1

Here even though we have tk+1 we still have every triangular number because k begins at zero and n begins at one.

Furthermore, to ﬁnd a triangular number we now have three methods: the standard formula for n(n+1) (d−2)n2+(4−d)n triangular numbers, tn = 2 , the formula to ﬁnd any polygonal number Pd(n) = 2 , k+2 and the newly found method using Pascal’s triangle, k where k = n − 1.

In conclusion, we can also find a formula for tetrahedral numbers using Pascal’s triangle. To visualize a tetrahedral number think of triangular number in dots, then stack each triangular number as levels, so the first level is one, t1, the next level is three, t2, and so on (see figure 6.1). For each triangular number you add the rank goes up by one. If you look at Pascal’s triangle you will see that the fourth diagonal includes all tetrahedral numbers. In fact, evry diagonal of Pascal’s triangle represents a unique dimension. The first diagonal is all 1’s so it is 0 dimensional, a point, the next is 1 dimensional, a line, and as we saw the third and fourth diagonals are two and three dimensional, respectively. This pattern continues throughout Pascals’s triangle.

7 References

[1] Elliot Forhan, Polygonal Numbers and Finite Calculus, 2007, 13/11/2013

[2] ”Polygonal Number.” Wikipedia. Wikimedia Foundation, 13/11/2013.

[3] ”Pascals Triangle.” All You Ever wanted to Know and More. 15/11/2013