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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 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 2 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 num- bers 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 polyg- onal 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]jP (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 2 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 2 N and pk 6= 0 n Suppose = 1; (d − 2)n + 4 − d = p 2 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 jP (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 2 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 2 P rimes 2 j d − 2 + 4 − d ) = p 2 j 2 = 1 6= p 2 j Conversely, let d = pm 2 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.
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