* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download A Generalization of Pascalâ•Žs Triangle - Via Sapientiae
Survey
Document related concepts
Mathematics and architecture wikipedia , lookup
Foundations of mathematics wikipedia , lookup
List of important publications in mathematics wikipedia , lookup
History of mathematics wikipedia , lookup
Ethnomathematics wikipedia , lookup
Location arithmetic wikipedia , lookup
Vincent's theorem wikipedia , lookup
History of trigonometry wikipedia , lookup
System of polynomial equations wikipedia , lookup
Pythagorean theorem wikipedia , lookup
Recurrence relation wikipedia , lookup
Weber problem wikipedia , lookup
Elementary mathematics wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
Transcript
DePaul Discoveries Volume 1 | Issue 1 Article 13 2012 A Generalization of Pascal’s Triangle Eliya Gwetta DePaul University, [email protected] Adrian Pacurar DePaul University, [email protected] Elizabeth Mai Smith DePaul University, [email protected] Follow this and additional works at: http://via.library.depaul.edu/depaul-disc Part of the Physical Sciences and Mathematics Commons Acknowledgements Faculty Advisors: T. Kyle Petersen, Bridget Tenner Recommended Citation Gwetta, Eliya; Pacurar, Adrian; and Smith, Elizabeth Mai (2012) "A Generalization of Pascal’s Triangle," DePaul Discoveries: Vol. 1: Iss. 1, Article 13. Available at: http://via.library.depaul.edu/depaul-disc/vol1/iss1/13 This Article is brought to you for free and open access by the College of Science and Health at Via Sapientiae. It has been accepted for inclusion in DePaul Discoveries by an authorized administrator of Via Sapientiae. For more information, please contact [email protected], [email protected]. Gwetta et al.: A Generalization of Pascal’s Triangle D E PA U L D I S C O V E R I E S ( 2 O 1 2 ) A Generalization of Pascal’s Triangle Eliya Gwetta, Adrian Pacurar and Elizabeth Mai Smith* Department of Mathematical Sciences 1. Introduction The branch of mathematics particularly interested in the study of discr ABSTRACT called Combinatorics. Enumerative Combinatorics is an area that dea Combinatorics is a branch of mathematics interested in the study of finite, or countable, sets. In particular, Enumerative Combinatorics is an area interested inpatterns counting many ways patterns are created, such for finite set canhow be formed, and developing counting techniques as counting permutations and combinations. Brenti and Welker, authors of “The Veronese Construction for Formal an interesting relationship between Eulerian Power Series and Graded Algebras,” seek an explanation for a illustrates combinatorial identity posed in their research. Using numbers and b techniques practiced in this area of mathematics, we have discovered that certain numbers appearing in their very important combinatorial quantities. Brentiidentity and Welker, in [2], hold properties similar to properties of the well-known binomial coefficients. analogous to Worpitzky's identity. When this new identity is expanded, possess properties similar to the properties of the binomial coefficient coefficients as geonomial coefficients; they represent the bulk of our res INTRODUCTION PA S C A L’ S T R I A N G L E A N D B I N O M I A L 2. Pascal's Triangle C O E F F I C I Eand N TS Binomial Coefficients The branch of mathematics particularly interested in the study of discrete and finite structures is called Combinatorics. Enumerative Combinatorics is an A well-known construction in mathematics, called A well-known construction in mathematics, called Pascal's Triangle Pascal’s Triangle, contains many beautiful properties. sets. Worpitzky’s identity [1] illustrates an interesting The numbersThe constructed here originally arose from properties. numbers constructed here originally arose from Hindu Hindu studies of combinatorics, and will be defined later. and will be defined Before the we reader identifyshould any properties, the reader Before we identify anylater. properties, take relationship between Eulerian numbers and binomial some time them to discover them in the following table. discover in the following table. area that deals with how many ways patterns can be formed, and developing counting techniques for finite coefficients, two very important combinatorial quantities. Brenti and Welker, in [2], introduce a new identity analogous to Worpitzky’s identity. When this TA B L E 1 : PA S C A L’ S T R I A N G L E new identity is expanded, the coefficients that arise possess properties similar to the properties of the binomial coefficients. We refer to these new coefficients as geonomial coefficients; they represent the bulk of our research. * Author names listed alphabetically n 0 1 2 3 4 5 6 7 Faculty Advisors: T. Kyle Petersen, Bridget Tenner Table 1: Pascal's Triangle k = 0,, n 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 Department of Mathematical Sciences, Summer 2011 Author contact: [email protected] The reader has most likely seen these numbers before, whether or not t — 168 — Published by Via Sapientiae, 2012 1 DePaul Discoveries, Vol. 1 [2012], Iss. 1, Art. 13 A G E N E R A L I Z AT I O N O F PA S C A L ' S T R I A N G L E Recall from high school algebra that students were SY M M E T RY asked to expand (multiply) the following: (a+b)(a+b). The most direct observation would be that symmetry lies A helpful technique usually learned at the time, called in each row. By cutting the triangle directly through the “FOIL” (acronym for first, outside, inside, last), made center, it is clear that the numbers mirror one another. this computation more straightforward. This method, The following observation should provide some intuition however, quickly becomes tedious as the number of for why this happens: by choosing 0 elements from a set binomials to multiply increases. To confirm this claim, of n , there are exactly n elements remaining, so therefore try to compute the following using the “FOIL” method: the number of ways to choose 0 is the same as the number (a+b)(a+b) (a+b)(a+b). Certainly this method will give of ways to choose n ; there is only one way. Similarly, a correct solution, but it is an inefficient approach. A the number of ways to choose 1 element from this set helpful theorem, named the binomial theorem, provides is exactly the number of ways to choose n — 1 elements. a more canny procedure. The following list is easily The number of ways to choose 2 elements is exactly the computed by the binomial theorem. number of ways to choose n — 2 elements from the same Recall from high school algebra that students were asked to expand (multiply) the following: set, and so on. (a b)(a b) . A helpful technique usually learned at the time, called ``FOIL" (acronym for first, (a+b) = 1a+1b outside, inside, last), made this computation more straightforward. This method, however, quickly (a+b)(a+b) = 1a2 + 2ab + 1b2 Recall from school thattostudents asked To to expand becomes tedious as thehigh number of algebra binomials multiplywere increases. confirm(multiply) this claim,the tryfollowing: to (a+b)(a+b) (a+b) = 1a + 3a b + 3ab + 1b 3 2 2 3 A helpful technique usually(alearned ``FOIL" (acronym for first, (a following b)(a b) .using this compute the the ``FOIL" method: b)(a atbthe )(a time, b)(acalled b) . Certainly and (a+b)(a+b) (a+b)(a+b)= 1a + 4a b + 6a b + 4ab + 1b 4 3 2 2 3 4 R OW S U M Here is a small experiment for the reader: choose any row, and add up all of the numbers in this row. Observe outside, madebut thisitcomputation more straightforward. This method, however, method will give ainside, correctlast), solution, is an inefficient approach. A helpful theorem, named the quickly number binomialsThe to multiply increases. To confirm thisbyclaim, binomial becomes theorem, tedious providesasa the more cannyofprocedure. following list is easily computed the try to the sum of all numbers in row three and row four: when If it is not clear why the previous expansions deserved n = 3 , the row sum is 1+3+3+1 = 8 ; when n = 4 , the row some space in this paper, observe the coefficients that sum is 1+4+6+4+1 =16 . Now notice, 8 = 23 and 16 = 24 , binomial compute theorem. the following using the ``FOIL" method: (a b)(a b)(a b)(a b) . Certainly this (a bbut ) = it1aisan 1b,inefficient approach. A helpful theorem, named the method will give a correct solution, and see the exponents here match the value of n that 2 appear in this list, Triangle. Recall from high school algebra that students were asked to expand (multiply) the following: (a b)and (a b)refer = 1a 2 back 2ab 1bto , Pascal’s binomial theorem, provides a more canny 3procedure. The2 following list is easily computed by the 2 3 ( a b ) ( a b )( a b ) = 1 a 3 a b 3 ab 1 b was chosen. In general, by selecting any single row, the The coefficients that arise in the previous Recall high from school highalgebra school that algebra students that were were to asked expand to(multiply) expand (multiply) the first, following: the following: helpful technique usually learned at thestudents time,asked called ``FOIL" (acronym for (a expansions b)(a Recall b) . A from binomial andtheorem. (a b)(a b)(a b)(a b) = 1a 4 4a 3b 6a 2 b 2 4ab 3 1b 4 . n of alltechnique numbers in row will be 2 ,forwhere are well-known as binomial coefficients, which make Aathis helpful . Atechnique helpful usually learned usuallyat learned thethis time, atmethod, the called time, ``FOIL" calledalways ``FOIL" (acronym for first, first, (a blast), )(aup (amade b) b. )( sum b)computation outside, inside, more straightforward. This however, quickly(acronym (a b) = 1some a 1bspace , If it is not clear why the previous expansions deserved in this paper, observe the n is the row index. Wow! 2 2 the numbers in Pascal’s Consider a situation outside,asinside, outside, last), inside, made thismade computation this computation more straightforward. moreTo straightforward. This This try however, method, quickly quickly (a b)(aTriangle. b) = 1a 2ab 1b , becomes tedious the number of last), binomials to multiply increases. confirm thismethod, claim, to however, coefficients that appear in this list, and refer back to Pascal's Triangle. (a b)(a b)( ) = 1a 3 3the a 2 b 3number ab 2 1b 3 in which one wanted toa bknow of ways becomes becomes tedious as tedious the number as the of number binomials of binomials to multiply to multiply increases. increases. To confirm To this confirm claim, this tryclaim, to try to compute3 the following using the ``FOIL" method: (a b)(a b)(a b)(a b) . Certainly 4 2 andin (athe b)previous (a b)(aexpansions b)(a b) are = 1awell-known 4a 3b 6as a 2 bbinomial 4ab coefficients, 1b 4 . The coefficients that arise A LT E R N AT I N G S U M three cards could be selected from a standard deck of Certainly this this compute the following the following using ``FOIL" the ``FOIL" method: method: (a b)(Aa(helpful ab)(ba)(atheorem, b)(ba)(ab)named b. )( a bthe ) . Certainly method will give a compute correct solution, but the it isusing an inefficient approach. which makeIf up numbers in Pascal's Triangle. Consider a situation whichinone to it isthenot clear why the previous expansions deserved someinspace thiswanted paper, observe the Let’s now consider the alternating sums, where every fifty-two. This value is 22,100. Only the binomial most theorem, patient method will method give will a correct give a solution, correct solution, but it is an but inefficient it is an inefficient approach. approach. A helpful A theorem, helpful theorem, named the named the provides a more canny procedure. The following list is easily computed by the know the coefficients number of ways three cards be selected from a standard deck of fifty-two. This that appear in thiscould list, and refer back to Pascal's Triangle. other number is subtracted from the total. For example, human being would dare to count this bybinomial hand.theorem. These binomialbinomial theorem,theorem, provides provides a more canny a more procedure. canny procedure. The following The following list is easily list computed is easily computed by the by the value is 22,100. Only the most patient beingexpansions would dareare to well-known count this by These The coefficients that arise in human the previous as hand. binomial coefficients, the alternating sum of row three is 1 – 3 + 3 – 1 = 0, and binomial coefficients enumerate combinations, binomial namely, binomial theorem.theorem. (a b) = 1a 1b, binomial which coefficients enumerate combinations, numbera situation of ways intowhich choose make up the numbers in Pascal's namely, Triangle.the Consider onekwanted to the(a alternating b)(a b) = 1a 2 sum 2ab 1of b 2 , row four is 1 – 4 + 6 – 4 +1= 0. This the number of ways to choose k elements from a set of (a b) =(1aab1)b,= 1a 1b, know the number of ways three cardscould n be selected from a standard deck of fifty-two. This (a pattern b)(a b)(a is b)indeed = 1a 3 3a 2 btrue 23ab 2for 1b 3every Every alternating (a b)(a(a b)b=)(1aa b)2=ab1a21 b 22,ab 1b 2row. , understand exactly what n (formally denoted elements from a set of n (formally denotedas as ).).ToTo understand exactly what this is counting, 4 3 2 2 k and (a b)(a b)(a b)(a b) = 1a 4a b 36a b 2 34ab 3 2 1b 4 . human (a b)(a(a b)(ba)(a b)b=)(1aa b)3= a 1ba 3ab 3a 2 b 1b33ab 2 1b 3 value is 22,100. Only the most patient being would dare to count this by hand. These this is counting, let’s dissect a smaller example. Say that row sum, within this famous triangle, is 0. 4 2 32 2 and (a band )(expansions a(a b)(ba)(a b)( ba)(a b)b=)(some 1aa4 b)4space =a 13ba 6a4this ab bpaper, 46ab a 2 b3 observe 1b44ab . 3the 1b 4 . let's dissect a smaller example. Say that we wanted to count namely, all possible 2clear from achoose If it of is of not the previous deserved in k binomial coefficients enumerate combinations, thecombinations number ways to why we wanted to count all possible combinations of 2 from a it is not If clear itinisthis not why clear the previous previous expansions expansions deserved deserved some space some in space this paper, in thisobserve paper, the observe the coefficients If that appear list, andwhy refertheback to Pascal's Triangle. 4 set of 4 objects, denoted as of label each object set as {1,2,3,4} .each Order does 2 n, and set of 4 aobjects, denoted asas inn the and labelexactly object elements from set (formally denoted ).,To understand what thisnot is counting,PA S C A L’ S R EC U R R E N C E k coefficients coefficients appear this list, in expansions and this refer list, and back refer to Pascal's back to Triangle. Pascal's Triangle. The coefficients thatthat arise in that theinappear previous are well-known as binomial coefficients, in the set as {1,2,3,4}. Order does not matter in this Pascal’s Triangle features yet another interesting situation, so we can have {(1,2), (1,3) (1,4), (2,3), (2,4), (3,4)}. property: if the entries in the previous row are given, the value appearing in Pascal’s Triangle. The patterns recurrence relation). For insight, let’s understand exactly found in Pascal’s Triangle may come as a surprise when what this means. When n = 3 and k = 2, the binomial The coefficients arisethat in the arise previous in the previous expansions are well-known well-known as binomial as binomial coefficients, coefficients, let's dissect a smaller example. Say that we wanted to count all possible combinations 2 from athat which makeThe up coefficients theofnumbers in Pascal's Triangle. Consider a expansions situation in are which one wanted to which upmake the numbers up the could numbers in Pascal's in Triangle. Pascal's Triangle. Consider Consider a situation a situation in whichinone which wanted one to wanted to know the which numbermake of ways three cards be selected from a standard deck of fifty-two. This 4 set of 4 objects, as , and combinations, label each object in thewhich set as {1,2,3,4} . Order does notit is possible to obtain any number in this triangle (a There are denoted six possible is exactly 2 know the know number themost of number ways of three ways cards three could cards be could selected be selected from a standard from a standard deck fifty-two. deck of fifty-two. This This value is 22,100. Only the patient human being would dare to count this by hand.ofThese is value 22,100. is Only 22,100. thecombinations, Only mostthe patient mostnamely, human patientbeing human would beingdare would to count dare this count by hand. this These hand. These k by binomial value coefficients enumerate the number of ways to to choose binomialbinomial coefficients coefficients enumerate enumerate combinations, namely, namely, the number the number of ways ofto ways choose to kchoose k n combinations, denoted as ).=To exactlyin what thisprevious is counting, row when n elements from a set of n (formally 3.understand Now look the coefficient introduced, but indeed a valid proof is well-established k n n ).+ ).=To2 +1= na set (formally (formally denoted as Tounderstand understand exactly exactly thiswhat is counting, this is counting, elements from a setfrom of of n see 3.what Coincidence? = 2, and thatasdenoted for each property that will be mentioned. let's dissectelements k all possible k combinations of 2 from a a smaller example. Say that we wanted to count let's dissect let'sa dissect smaller Say that we Saywanted that wetowanted count all to count possible all possible combinations combinations of 2 fromofa 2 from a a4example. smaller example. set of 4 objects, denoted as , and label each object in the set as {1,2,3,4}. Order does not 2 4 4 , and set of 4 objects, set of 4 denoted objects, denoted as , as and label each label object eachinobject the setinasthe{1,2,3,4} set as {1,2,3,4} . Order does . Order not does not 2 2 — 169 — http://via.library.depaul.edu/depaul-disc/vol1/iss1/13 2 The coefficients The coefficients that arise in that thearise previous in theexpansions previous expansions are well-known are well-known as binomialascoefficients, binomial coefficients, The coefficients that arise in the previous expansions are well-known as binomial coefficients, which make which up the make numbers up theinnumbers Pascal's inTriangle. Pascal's Consider Triangle. situationainsituation which one in which wantedone to wanted to which make up the numbers in Pascal's Triangle. Consider a situation in which one wanted to aConsider Gwetta et al.: A Generalization of Pascal’s Triangle Ecould PA UThis Lfrom D I SaCstandard O Vfrom E R IaEdeck S ( 2ofOfifty-two. 1 2deck ) of This know the offrom number waysa three of ways cards three could beDselected be selected standard fifty-two. This know the number of ways three cardsknow couldthebenumber selected standard deck ofcards fifty-two. value is 22,100. value 22,100. the most Only patient the most human patient being human would being darewould to count darethis to count by hand. this These by hand. These value is 22,100. Only the most patient human beingisOnly would dare to count this by hand. These binomial coefficients binomial enumerate combinations, namely,k the namely, number theofnumber ways to of choose ways tok choose k binomial coefficients enumerate combinations, namely,coefficients the numberenumerate of ways combinations, to choose Absolutely not. Take another value, say n6 and SY M M E T RY n n n elements is counting, ). n denoted (formally (formally as denoted ). To as understand exactly what exactly this isproperty what counting, this is counting, elements a To set understand of from set5 ofexactly A similar that is n (formally denoted what this elements from a set ofand 15n a 10 . To understand k2, notice thatas from k ). k k immediately identifiable is the symmetry in every row. The reason behind this follows a In order to obtain a particular entry, add the following let'swedissect let's a smaller aallsmaller Say example. that weSay wanted thatofwe to count wanted count allcombinations possible combinations of 2 from aof 2 from a let's dissect a smaller example. Say that wanted todissect countexample. possible combinations 2 from a alltopossible similar concept used to explain the symmetry of Pascal’s two consecutive numbers that are in the previous row: 4 4 4 of 4each objects, set of 4denoted objects,asset denoted as label and each and label object in the object set as in the set as. {1,2,3,4} Order does . Order not does not {1,2,3,4} set of 4 objects, denoted as , andsetlabel doeseach not as ,{1,2,3,4} 2 . , Order Triangle . We will illustrate the entry directly above,object andin the the entry immediately to 2 2 another reason when we the left of this value. The following table is modified discuss the recurrence relation. Other similarities are not to illustrate the recurrence relation of the binomial as direct, however, the reader should take a moment to coefficients. discover patterns on their own. Hopefully this triangle is becoming more and more beautiful to the reader! The following section introduces a new type of triangle not so well-known in mathematics. n 0 1 2 3 4 5 6 .. . k 1 1 1 1 1 1 1 .. . = 0, . . . , n 1 2 1 3 3 1 4 5 6 6 4 1 10 10 5 1 15 20 15 6 1 Recall from TA B L E Table 2 : G EO N O2. MIA L C O E F F I C I Ecoefficients N TS W H E N r when 2 Geonomial n 0 1 2 3 4 5 6 .. . k 1 1 1 1 1 1 1 .. . r=2 = 0, . . . , 2n 1 2 3 4 5 6n 1 3 2 1 6 7 6 3 1 10 16 19 16 10 4 1 Table Geonomial coefficients 15 30 2.45 51 45 30 when 15 r 5= 2 1 21 50 90 126 141 126 90 50 21 6 1 k = 0, . . . , 2n 0 1 1 1 1 1 2 1 2 3 2 1 3 1 3 6 7 6 3 1 4 1 4the 10 previous 16 19 16 section 10 4 that 1 Recall from the coefficients 5 1 5 15 30 45 51 45 30 15 5 1 6 1 section 6 21 50 90 126from 141 126 90 in 50Pascal’s 21the 6 product 1 Triangle theinprevious that the coefficients Pascal’s Triangle arise expanding .. .. . . arise from ex of binomials. thegeonomial geonomial panding the product of binomials.Interestingly Interestingly enough, enough, the coefficients becom coefficients become explicit through a similar process explicit through a similar process of expansion. The following list will illustrate the coeffi pefully this triangle is becoming more and more beautiful to the reader! The following The following list will illustrate the T R I A N G L E O F G EO N O M I A L C O E F F I C I E N TS Recallof fromexpansion. the previous section that the coefficients in Pascal’s Triangle arise from excients connected to the values in Table 2. connected to the values Table 2. coefficients become n introduces a In new type of triangle so well-known in mathematics. this section, we willnot refer to the numbers in the new panding coefficients the product of binomials. Interestingly enough,inthe geonomial explicit through a similar process of expansion. The following list will illustrate the coeffitriangles as geonomial coefficients. These numbers (1 + a + a2 )1 = 1a2 + 1a + 1 2 1 3. Triangle of Geonomial Coefficients (1aa ) in Table 1a21a1 cients connected to the values 2. satisfy a definition stated by Brenti and Welker and 4 4 3 3 2 2a + 1 + 2a2a + 3a 3a2 + (1 (1aa + a + a2 )22)2= 1a 1a 2a1 appear in an identity central to the paper [2]. The (1 + a + a2 )1 = 1a2 + 1a + 1 his section, we will refer to the numbers in the new triangles as geonomial coefficients. 2 3 6 5 4 ) 1a 3a 6a 7a336a22 3a1 (1aa geonomial coefficients count the number of ways we can 6 5 4 (1 + a (1 ++aa2 )+3 a=2 )21a 4 3a 3+ 6a2 + 7a + 6a + 3a + 1 = 1a+ + 2a + 3a + 2a + 1 numbers satisfy a definition stated by Brenti and Welker and appear in an identity write an integer k as a sum of n digits whose values are = 1a + 3a + 6a 7a +left-hand 6a + 3a + side 1 (1 + a the + a )polynomials Notice that on+the are zerocoefficients and r . Thecount geonomial coefficients l to the paper integers [2]. Thebetween geonomial the number waysthe we can write Noticeofthat polynomials on the left-hand side are raised to some exponent, whic raised to some exponent which happens to be their row different families, on a parameter Notice polynomials the left-hand sidetable are raised to asome exponent, eger k as a sumfall of ninto digits whose values aredepending integers between zero and r. Thethat geonomial happens to be their rowtheindex in our on table. The next shows different casewhich of geonomi index in our table. The next table shows a different case r 1. If r 1, we recover the entries in Pascal’s Triangle. happens to be their row index in our table. The next table shows a different case of geonomial ients fall into different families, depending on a parameter r ≥ 1. If .r = 1, we recover of .geonomial coefficients. The tables for r 2 and r 3 are illustratedcoefficients below. The coefficients ntries in Pascal’s Triangle. = 2 and rcoefficients = 3 are illustrated below. patterns that The occurtables with for ther geonomial 2 3 6 5 4 3 2 TA B L E 3 : G EO N O M I A L C O E F F I C I E N TS W H E N r 3 Table 3. Geonomials coefficients coefficients when r = 3r = 3 are true values of coefficients r , and the following cases Table 3. Geonomials when atterns that occur withfor theallgeonomial are true for all are values of r, and the k = 0, . . . , 3n n selected for visual reference. For now, let us focus on the = 0, . . . , 3n ing cases are selected for visual reference. For now, let us focus nonkthe 0mysterious 1 0 1 mysterious similarities between Pascal’s Triangle and 1 1 1 1 1 1 1 1 2 1 1 21 3 4 3 2 1 rities between Pascal’s Triangle and the following triangles. the following triangles. 2 1 2 3 3 1 34 6 310 4 1 4 10 20 1 3symmetry 35 5 6 1 51015 12 ymmetry. A similar property that is immediately identifiable is3 the in 4 1 4 ... 10... 20 31 5 the 1 5symmetry 15 35 65 row. The reason behind this follows a similar concept used to explain of .. .. . . 212 12 1 10 6 3 1 31 40 44 40 31 20 10 4 1 12 3 135 1 101 65 35 15 5 1 65 10110135 6155 155 40 44 40 31 20 10 4 1 101 135 155 155 135 101 65 35 15 5 1 ’s Triangle . We will illustrate another reason when we discuss the recurrence relation. er similarities are not as direct, however, the reader should take a moment to discover ns on their own. — 170 — Published by Via Sapientiae, 2012 3 DePaul Discoveries, Vol. 1 [2012], Iss. 1, Art. 13 A G E N E R A L I Z AT I O N O F PA S C A L ' S T R I A N G L E To be consistent with the format used above, here A LT E R N AT I N G S U M are some products of polynomials whose coefficients The previous section proved that the alternating sum of represent values from Table 3. the binomial coefficients will always be 0. For geonomial coefficients, there are two possible outcomes. When (1aa2a3)1 1a31a2 1a1 r is an even number, the alternating sum in each row (1aa2a3)2 1a6 2a53a4 4a33a22a1 (1aa2a3)3 1a9 3a86a7 10a6 12a5 12a410a36a23a1 In general, we were able to prove that the geonomial coefficients appear in the following expansions: (1aa2…ar)n , which is a repeated product of a finite geometric series 1aa2…ar (hence our choice for the name geonomial). The proof, however, is too technical to include in this paper, and we invite the reader to email us for specific details if desired. is 1. On the other hand, when r is an odd number, the alternating sum is 0. To test this, consider row three from Table 2: 13676–311. Now use row three of Table 3 for confirmation. The alternating sum is 13610121210631 0 . The reader can prove this property by setting a –1 in the polynomials above. R EC U R R E N C E R E L AT I O N When Whenrrisisan aneven even number, number, the alternating alternating sum suminineach each row row isis1.1. On Onthe the other otherthe hand, hand,when when Just as thethe binomial coefficients are obtainable given rrisisan anodd oddnumber, number, the thealternating alternating sum sumisis0.0.Triangle, To Totest testthis, this, consider considerrow row three threefrom fromTable Table2:2: previous row of Pascal’s geonomial coefficients R OW S U M 11−−33++66−−77++66−− 33++11= 1.Now Nowuse use row rowthree threeofof Table Table33for for confirmation. confirmation. The Thealternating alternating have a=1. similar recurrence relation. Recall from Pascal’s Recall from Pascal’s Triangle that the sum of the numbers every can bereader obtained by adding sum sumisis11−−33++66−Triangle −10 10++12 12−−that 12 12++10 10−−66+number +33−−11==0. 0.The The reader can canprove prove this thisproperty propertyby by in any single row was 2n (where n represented the row consecutive numbers in the previous row. For these setting settingaa==−1 −1inintwo the thepolynomials polynomialsabove. above. index). Let’s attempt the same experiment! By arbitrarily new triangles, the number of consecutive values required selecting row three, when n 3 from Table 2,3.4. the row 3.4. Recurrence Recurrence Relation. Relation. Just Justasasthe thebinomial binomialcoefficients coefficientsare areobtainable obtainablegiven giventhe thepreviprevito add is r 1. When r 2 , the number of consecutive sum is 1367631 27. Remember that Table ous ousrow rowofofPascal’s Pascal’sTriangle, Triangle,geonomial geonomialcoefficients coefficientshave haveaasimilar similarrecurrence recurrencerelation. relation.Recall Recall values to add in order to obtain a geonomial coefficient, 5 (21) . Triangle 2 is the case when r = 2 , and notice that 27 33 from from Pascal’s Pascal’s Trianglethat thatevery everynumber numbercan canbe beobtained obtainedby byadding addingtwo twoconsecutive consecutivenumnumis three. Comparably, when r 3, by adding four From the same table, let n 5 so we obtainbers the sum bersininthe theprevious previousrow. row.For Forthese thesenew newtriangles, triangles,the thenumber numberofofconsecutive consecutivevalues valuesrequired required consecutive values in the previous row of the selected 5 15 1530455145301551 (21) totoadd addisisr.r++1.1. When Whenrr==2,2,the thenumber numberofofconsecutive consecutivevalues valuestotoadd addininorder ordertotoobtain obtainaa row, a new geonomial coefficient is obtained. Therefore, Let’s now switch to Table 3 to see if the same pattern coefficient,isisthree. geonomial geonomialcoefficient, three. Comparably, Comparably,when whenrr==3,3,by byadding addingfour fourconsecutive consecutivevalues since the entries are initially symmetric, the entries in values occurs. Add up all values in row two, when n = 2,inso we previous get inthe theprevious row row of of the the selected selected row, row, a a new new geonomial geonomial coefficient coefficient is is obtained. obtained. Therefore, Therefore, the following rows will abide to the same property. The the sum 1+2+3+4+3+2+1 = 16 = 42 = (3+1)2. Also, when since sincethe theentries entriesare areinitially initiallytables symmetric, symmetric, the theentries entriesto ininview the thefollowing following rows rowswill willabide abidetotothe the following are modified this property. n = 5, we have 1+5+15+35+65+101+135+155+155+135 same sameproperty. property.The Thefollowing followingtables tablesare aremodified modifiedtotoview viewthis thisproperty. property. +101+65+35+15+5+1=1024=45 =(3+1)5. More generally, TA B L E 4 : r 4 Table Table4.4. rr==22 another property of the geonomial coefficients is that the sum in each row can be obtained by (r1)n, where r represents a particular case and n is the row index. We can verify this by revisiting our polynomial and setting a 1, where we get (1112 )n 3n (21)n, and (111213 )n 4n (31) n, and, more generally, . nn 00 11 22 33 44 55 66 .. .. .. 2n kk==0,0,. . . ., ,2n 11 11 11 11 11 22 33 22 11 33 66 77 10 16 16 11 44 10 15 30 30 11 55 15 21 50 50 11 66 21 .. .. .. 11 66 19 19 45 45 90 90 33 11 16 16 10 10 44 11 51 45 45 30 30 1515 55 11 51 126 141 141 126 126 90 90 50 50 21 21 66 11 126 We Wehave haveobtained obtainedaageneral generalproof prooffor forthese theserecurrence recurrencerelations relationsasaswell, well, and andwould wouldbe be thrilled thrilledfor forthe thereader readertotocontact contactus usfor formore moredetails. details. 4.4. Final FinalDiscussion Discussion The Theidentity identitypresented presentedby byBrenti Brentiand andWelker Welkerisiscombinatorial combinatorialininnature, nature,however howeverasasthey they themselves themselvesstate, state,there thereisisno noknown knowncombinatorial combinatorialexplanation explanationfor forit. it.The TheEulerian Euleriannumbers numbers — 171 — http://via.library.depaul.edu/depaul-disc/vol1/iss1/13 4 Gwetta et al.: A Generalization of Pascal’s Triangle D E PA U L D I S C O V E R I E S ( 2 O 1 2 ) TA B L E 5 : G EO N O M I A L C O E F F I C I E N TS W H E N r 3 Table 5. Geonomials coefficients when r = 3 n 0 1 2 3 4 5 .. . k 1 1 1 1 1 1 .. . = 0, . . . , 3n 1 2 3 4 5 1 1 3 4 3 2 1 6 10 12 12 10 6 3 1 10 20 31 40 44 40 31 20 10 4 1 15 35 65 101 135 155 155 135 101 65 35 15 5 1 which are contained in this identity have many well-known properties, but the properties We have obtained a general proof for these recurrence of the geonomial coefficients are less well-known. We hope that our research will provide relations as well, and would be thrilled for the reader to insight for developing a combinatorial explanation of this curious identity. contact us for more details. References [1] D. E. Knuth, The art of computerFprogramming. Volume INAL DIS C U S3.SSorting I O N and searching. Addison-Wesley Series The identity presented by Brenti and Welker is in Computer Science and Information Processing. Addison-Wesley Publishing Co., Reading, Mass.London-Don Mills, Ont., (1973). combinatorial in nature, however as they themselves [2] F. Brenti and V. Welker, The Veronese construction for formal power series and graded algebras, Ad- there is no known combinatorial vancesstate, in Applied Mathematics, 42 (2009), 545–556. explanation for it. The Eulerian numbers which are contained in this identity have many well-known properties, but the properties of the geonomial coefficients are less wellknown. We hope that our research will provide insight for developing a combinatorial explanation of this curious identity. ACKNOWLEDGEMENTS I would like to thank my research advisors, Dr. Kyle Petersen and Dr. Bridget Tenner, for their continuous support and dedication to enhancing the quality of our research. I would also like to thank DePaul University for funding this research project and allowing for us to have such a unique experience as undergraduate researchers. — 172 — Published by Via Sapientiae, 2012 5 DePaul Discoveries, Vol. 1 [2012], Iss. 1, Art. 13 A G E N E R A L I Z AT I O N O F PA S C A L ' S T R I A N G L E REFERENCES [1] D. E. Knuth, The art of computer programming. Volume 3. Sorting [2] F. Brenti and V. Welker, The Veronese construction for formal power and searching. Addison-Wesley Series in Computer Science and series and graded algebras, Advances in Applied Mathematics, 42 Information Processing. Addison-Wesley Publishing Co., Reading, (2009), 545—556. Mass.-London-Don Mills, Ont., (1973). Sanjay Cherala, Stephanie Lyngaas Matt Hartzell and Lihua Jin, chair of the Chemistry Department, enjoy the 9th Annual Natural Sciences, Mathematics, and Technology Showcase. — 173 — http://via.library.depaul.edu/depaul-disc/vol1/iss1/13 6