A study of a family of generating functions of Nelsen-Schmidt type and some identities on restricted barred preferential arrangements
aa r X i v : . [ m a t h . C O ] A p r A study of a family of generating functions ofNelsen-Schmidt type and some identities onrestricted barred preferential arrangements
S.Nkonkobe, V.Murali
Department of Mathematics (Pure & Applied)Rhodes UniversityGrahamstown 6140 South [email protected], [email protected] Abstract
A preferential arrangement of a set X n = { , , . . . , n } is an orderedpartition of the set X n induced with a linear order. Separation of blocksof a preferential arrangement with bars result in the notation of barredpreferential arrangements. Roger Nelsen and Harvey Schmidt proposedthe family of generating functions P k ( m ) = e km − e m ; which for k = 0 andfor k = 2 they have shown that the generating functions are exponentialgenerating functions for the number of preferential arrangements of aset X n and the number of chains in the power set of X n respectively. Inthis study we propose combinatorial structures whose integer sequencesare generated by members of the family for all values of k in Z + . To dothis we use a notion of restricted barred preferential arrangements. Wethen propose a more general family of generating functions P rj ( m ) = e rm (2 − e m ) j for r, j ∈ Z + . We derive some new identities on restrictedbarred preferential arrangements and give their combinatorial proofs.We also propose conjectures on number of restricted barred preferentialarrangements. Mathematics Subject Classifications:05A18,05A19,05A16, 2013
Keywords : Barred preferential arrangements, restricted barred preferential arrangements, re-stricted sections and free sections. introduction The integer sequence 1,1,3,13,75,541,... that arise for the number ofpreferential arrangements of X n can be tracked back to [1] a year 1859paper by Arthur Cayley; in connection with analytical forms calledtrees. The word preferential arrangements is due to Gross in [4]. Theinteger sequence is A000670 in Sloane [8]. Since Cayley’s paper muchwork has been done: The sequence has been given various interpreta-tions and has been connected to a number of well known combinatorialsequences (for instance in [2], [3], [5], [6]). Examples of preferentialarrangements of X = { , , } are:a) 3 12b) 2 3 1The preferential arrangement in a) constitutes of two blocks. Thefirst block of the preferential arrangement (from left to right) is theelement 3. The second block of the preferential arrangement is formedby the elements 1 and 2. The preferential arrangement in b) has threeblocks. The first block being the element 2. The second block beingthe element 3 and the third block is the element 1. The magnitude ofthe spaces between the elements tell us which elements form a singleblock.In [7] Nelsen and Schmidt have proposed the family of generatingfunctions P k ( m ) = e km − e m ; which for k = 0 and for k = 2 they haveshown that the generating functions are exponential generating func-tions for the number of preferential arrangements on X n and the num-ber of chains in the power set of X n respectively. They then asked“could there be other combinatorial structures associated with eitherthe set X n or the power set of X n whose integer sequences are gener-ated by members of the family for other values of k ?”Murali has shown that for four values of k (i.e. k = 0 , , ,
3) themembers of the family are generating functions for the number of re-stricted preferential fuzzy subsets of a set X n [3]. The main aim of this study is to propose combinatorial structures whose integer sequencesare generated by the members of the family for all values of k in Z + .From now on for a fixed k ∈ Z + we will refer to the generating function P k ( m ) = e km − e m of Nelsen and Schmidt, as the Nelsen-Schmidt generat-ing function. 3. Preliminaries
Separating blocks of a preferential arrangement with a number ofbars results in a barred preferential arrangement [6]. Examples ofbarred preferential arrangements of the set X having one bar andtwo bars are: a ) 2 | b ) 3 | | X are all forming separate blocks. Thereis a single bar separating the first and the second block. In b) the set X is preferentially arranged into two blocks. One block is the element3 and the other block is formed by the elements 1 and 2. The twobars are in-between the two blocks. With reference to bars the barredpreferential arrangement in a) has two sections one to the left of thebar (the block 2) and another section to the right of the bar (the twoblocks). We name them section zero and section one respectively. Thebarred preferential arrangement in b) has three sections. One section tothe left of the first bar. A section in-between the two bars and anothersection to the right of the second bar. We name them section zero,section one and section two respectively. In general k bars separatea preferential arrangement into k + 1 sections (possibly with emptysections) [6]. Note.
The elements within each section of a preferential arrangementare preferentially arranged among themselves.
We denote the set of all barred preferential arrangements of the set X n having k bars by Q kn and the number of these barred preferential arrangements by J kn i.e | Q kn | = J kn . A closed form, a recurrence relationand generating function for the numbers J kn are given respectivel in thefollowing theorems; Theorem 1. [6] for all n, k ≥ we have; J kn = n P s =0 (cid:8) ns (cid:9) s ! (cid:0) k + ss (cid:1) Where (cid:8) ns (cid:9) are Stirling numbers of the second kind. Theorem 2. [6] for n, k ≥ we have; J kn = n P s =0 (cid:0) ns (cid:1) J s J k − n − s Theorem 3. [6] for k ≥ we have; q k ( m ) = ∞ P n =0 J kn m n n ! = − e m ) k +1 Restricted barred preferential arrangements
Whether we view barred preferential arrangements as a result offirst placing bars and then distributing elements on each section orfirst preferentially arranging elements into blocks and then introducebars in-between the blocks of the preferential arrangements; the result-ing number of barred preferential arrangements is the same. Here wechoose the former.In finding the total number of barred preferential arrangements of X n having k bars we generalise equation 24 of [5] from one bar to k bars using the same kind of argument as in [5]; we argue as follows: Wefirst place the k bars. There are k + 1 sections before, in-between andafter the k bars. In a distribution of the n elements of X n , say thereare w i elements on the i th section. There are n ! ways of permuting the n elements among the k + 1 sections. There are w i ! ways of permutingelements of section i among themselves. There are J w i ways of prefer-entially arranging elements of section i among themselves. Hence thetotal number J kn of barred preferential arrangements of X n having k bars is;(1) J kn = X w + ··· + w k +1 = n n ! w ! × w ! · · · w k +1 ! J w × J w × · · · × J w k +1 Where the summation is taken on all solutions of the equation w + · · · + w k +1 = n in non-negative integers.Equation 1 is a dual of the closed form in theorem 1 in section 3above. In (1) the number of barred preferential arrangements is ob-tained using an argument where bars are first placed and then elementsbeing distributed on the sections. Whereas in theorem 1 the number ofbarred preferential arrangements is obtained using an argument; wherethe elements of X n are first preferentially arranged into blocks, thenbars are introduced in-between the blocks. ◦ Definition 1.
A free section of a barred preferential arrangement is asection whose elements can be preferentially arranged among themselvesin any possible way.
What definition 1 means is that; when elements of a free sectionare preferentially arranged, the preferential arrangement can have anypossible number of blocks. So for w i elements in a free section thereare J w i possible ways of preferentially arranging the elements amongthemselves. Definition 2.
A restricted section of a barred preferential arrangementis a section that can have a maximum of one block.
What definition 2 means is that; when elements of a restricted sectionare preferentially arranged; they can have a maximum of one block. Sofor w j elements on a restricted section; there is 1 way of preferentiallyarranging the elements among themselves i.e. into one block.In constructing a barred preferential arrangement of X n having k barsby first placing bars and then distributing elements on sections; we require that one fixed section to be a free section and all the other k sections to be restricted sections. The chosen fixed section can beany of the sections but must be fixed. In getting the number of barredpreferential arrangements with the restriction we argue as follows: Letssay there are w i elements on section i . We assume the free section isthe j th section (from left to right). There are n ! ways of permutingthe n elements of X n among the k + 1 sections. There are w i ! ways ofpermuting elements of section i among themselves. There are J w j waysof preferentially arranging elements of the chosen free section amongthemselves. There is one way of preferentially arranging elements ofeach of the k restricted sections. Hence the total number of these re-stricted barred preferential arrangements (denoted by p kn ) is given by; (2) p kn = X w + ··· + w k +1 = n n ! w ! × w ! × · · · × w j ! × · · · × w k +1 ! h (1) × (1) × · · · × J w j × · · · × (1) i Where the summation is taken on all solutions of the equation w + · · · + w k +1 = n in non-negative integers. In the product of the k + 1 terms in the bracket in (2); all the entries are ones except the j th entry which is J w j . This is due to the fact that the j th section isthe chosen free section. The other k sections are all restricted sections.We denote the set of these restricted barred preferential arrangementsby G kn so | G kn | = p kn . The set G kn ⊆ Q kn where Q kn is the set of all barredpreferential arrangements of X n having k bars without any restrictionsconsidered in section 3 above.We now seek the exponential generating function for the numbers p kn in (2). In finding the generating function of the numbers we definea convolution of exponential generating functions: Given generatingfunctions X = ∞ P n =0 x n m n n ! ,. . . , X s = ∞ P n s =0 x ns m ns n s ! their convolution is; (3) X × X × · · · × X s = ∞ X n =0 (cid:18) P n + ··· + n s = n n ! n ! × n ! ··· n s ! x n × x n · · · x sn s (cid:19) m n n ! Where the summation is taken on all solutions of the equation n + · · · + n s = n in non-negative integers. By (3) the generatingfunction of the sequence p kn in (2) is; (4) Pk ( m ) = ∞ P n mn n ! × ∞ P n mn n ! × · · · × ∞ P nj =0 J wj mnjnj ! × · · · × ∞ P nk +1=0 mnk +1 nk +1! ! = ∞ X n =0 pknmnn ! We observe that each term in the product in (4) is a generating func-tion. Where k of them are e m and one of them is − e m . Hence thegenerating function in (4) is; (5) P k ( m ) = ∞ X n =0 p kn m n n ! = e m × e m × · · · × − e m × · · · × e m = e km − e m The term − e m in the generating function in (5) is due to the freesection (see theorem 3 of section 2 above: this is the case k = 0). Theterms e m are due to the restricted sections.We recognise the generating function in (5) as the Nelsen-Schmidtgenerating function. So the Nelsen-Schmidt generating function fora fixed k in Z + ; is the generating function for the number of barredpreferential arrangements of X n having k bars. In which one fixedsection is a free section and the other k sections are restricted sections.The idea of restricted barred preferential arrangements with one bar,where one section is a free section and the other is a restricted sectionappear in [11] in a different context. The idea in the paper appears ina campus security problem, where the author seeks the total possiblenumber A n of possible combination locks having n button. In the paperthey derived the exponential generating function of A n as e x − e x . Henceour work above can be viewed as a generalisation of the idea to multiplebars. Theorem 4. for n ≥ , k ≥ we have; p kn = n P s =0 (cid:0) ns (cid:1) p s k n − s We recall p kn is the number of barred preferential arrangements of X n having k bars, in which one fixed section is a free section; andthe other k sections are restricted sections. The set of these restricted barred preferential arrangements being G kn . In proving the theoremwe view the number of elements in G kn being obtained in the followingway: On each B ∈ G kn there are (cid:0) ns (cid:1) ways of selecting elements to gointo the free section. There are p s ways of preferentially arranging the s elements among themselves. Since the other k sections other than thechosen free section can have a maximum of one block; then the numberof possible ways of distributing the remaining n − s elements amongthe k restricted sections is k n − s . Taking the product and summing over s we obtain the result of the theorem. Theorem 5. for n, k ≥ we have; p k +1 n = n P s =0 (cid:0) ns (cid:1) p ks We recall p k +1 n is the number of barred preferential arrangements an n -element set having k + 1 bars in-which k + 1 sections are restrictedsections and one section is a free section. The set of these barredpreferential arrangements being denoted by G k +1 n . The prove of thetheorem is similar to that of theorem 4. In proving the theorem webase our argument on a fixed restricted section, say its the m th sectionon each element of G k +1 n . Lets say there are s elements which are notto go to the m th section on elements of G k +1 n . The s elements can bechosen in (cid:0) ns (cid:1) number of ways. The s elements can be preferentiallyarranged on the k + 1 other sections in p kn ways. The remaining n − s elements can be preferentially arranged on the m th section in one way.Taking the product and summing over s we obtain the result of thetheorem. Lemma 1. for k, n ≥ we have p kn = k n + n − P s =0 (cid:0) ns (cid:1) k s p on − s We recall p kn is the number of restricted barred preferential arrange-ments of an n element set having k bars in-which k fixed sections arerestricted sections and one fixed section is a free section. Where the setof these barred preferential arrangements is denoted by G kn . In proving the lemma we base our argument on a fixed free section on all barredpreferential arrangements in G kn . For argument sake lets say the chosenfree section is the first section on each element of G kn . On all barredpreferential arrangements from G Kn either the first section is empty ornon-empty. When the first section is empty that means the n elementsare preferentially arranged among the other k sections of which k allof them are restricted sections so the number of elements of G kn in thiscase is k n .The other case is when the first section has at least one element. Ingetting the number of elements of G kn in this case we argue as follows;There can be a maximum of n − s elements not in the first section in this case. There are (cid:0) ns (cid:1) of selectingelements which are not in the first section. There are k s ways of pref-erentially arranging the s elements among the other k sections. Theremaining n − s elements can be preferentially arranged in the firstsection in p n − s ways. Taking the product and summing over s we havethe number of elements of G kn in this case as n − P s =0 (cid:0) ns (cid:1) k s p n − s .Combining the two cases we obtain the result of the lemma. Lemma 2. for n, k ≥ we have p kn = p k − n + n − P s =0 (cid:0) ns (cid:1) p k − s Lemma 2 can be proved in a similar way to lemma 1. We will gen-eralise lemma 2 in ◦ as theorem 7 give a proof of the theorem. Conjecture. for k, s, n ≥ we have p kn = ∞ P s =0 ( k + s ) n s For the case k = 0 the identity in the conjecture is equation 21 of [5]. ◦ We ask:
What sort of a generating function would restricted barredpreferential arrangements having more than one free section have? Asopposed to the restricted barred preferential arrangements considered in ◦ above. Also how would the generating function for a fixed numberof restricted sections; and fixed number of free sections compare to theNelsen-Schmidt generating function we considered above? In constructing a barred preferential arrangement of X n having k barsby first placing bars and then distributing elements on sections; fromthe k + 1 sections we require that j fixed sections to be free sectionsand remaining r = k + 1 − j sections to be restricted sections. Wedenote the number of these restricted barred preferential arrangementsby p rj ( n ) and the set of these restricted barred preferential arrange-ments by G rj ( n ); so | G rj ( n ) | = p rj ( n ). In a similar way to the way weobtained the number p kn in (2); the number p rj ( n ) of these restrictedbarred preferential arrangements is; (6) p rj ( n ) = X w ··· + wk +1= n n ! w ! × w ! · · · w m ! · · · w k +1 ! (1) × (1) ×· · ·× (1) × J wm × J wm +1 ×· · ·× J wk +1 Where on the product of k + 1 terms in (6); r terms are (1)’s for the r restricted sections and j = k + 1 − r terms are the J w i ′ s for the freesections.In a similar way to the way the generating function for the numbers p kn is obtained in (5); the generating function (denoted by P rj ( m )) forthe numbers p rj ( n ) is;(7) P rj ( m ) = ∞ X n =0 p rj ( n ) m n n ! = e rm (2 − e m ) j We observe that Nelsen-Schmidt generating function is a special caseof the generating function P rj ( m ) when j = 1.We generalise lemma 1 of ◦ above to the following theorem. Theorem 6. for r, j, n ≥ we have p rj ( n ) = p rj − ( n ) + n − P s =0 (cid:0) ns (cid:1) p rj − ( s ) p ( n − s )We recall p rj ( n ) is the number of restricted barred preferential ar-rangements of an n element set having r + j − r fixedsections are restricted sections and j fixed sections are free sections.Where the set of these barred preferential arrangements is denoted by G rj ( n ). In proving the theorem we base our argument on a fixed freesection on all barred preferential arrangements in G rj ( n ). For argumentsake lets say the chosen free section is the first section of each elementof G rj ( n ) (the chosen fixed free section could have been any of the j freesections as long would be fixed). On all barred preferential arrange-ments from G rj ( n ) either the first section is empty or non-empty. Sowe have two cases.When the first section is empty that means the n elements are pref-erentially arranged among the other r + j − r ofthem are restricted and j − G rj ( n ) in this case is p rj − ( n ) (by definition of p rj − ( n )).The other case is when the first section of each element of G rj ( n ) has atleast one element. In getting the number of elements of G rj in this casewe argue as follows; There can be a maximum of n − s elements not in the first section in this case. Thereare (cid:0) ns (cid:1) of selecting elements which are not in the first section. Thereare p rj − ( n ) ways of preferentially arranging the s elements among theother r = j − n − s elements can be pref-erentially arranged in the first section in p ( n − s ) ways. Taking theproduct and summing over s we have the number of elements of G rj ( n )in this case as n − P s =0 (cid:0) ns (cid:1) p rj − ( n ) p ( n − s ).Combining the two cases we obtain the result of the lemma. We generalise lemma 2 of ◦ to the following theorem. Theorem 7. for n, r, j ≥ we have p rj ( n ) = p r − j ( n ) + n − P s =0 (cid:0) ns (cid:1) p r − j ( s )The number p rj ( n ) is the number of barred preferential arrangementsof an n -element set having r + j − r sections are re-stricted sections and j sections are free sections. Where the set of thesebarred preferential arrangements is denoted by G rj ( n ). The proof of thistheorem is similar to that of theorem 6 above. We base our argumenton a fixed restricted section. We say the chosen restricted section isthe first section of each element of G rj ( n ) (the chosen restricted sectioncould have been any of the r restricted sections as long it would befixed). All barred preferential arrangements of the set G rj ( n ) have theproperty that the chosen restricted section is either empty or has a cer-tain number of elements. We partition the set G rj ( n ) into two disjointsubsets w and w . Where the sets contain those elements of G rj ( n )in-which the chosen restricted section is empty and those in-which therestricted section has a number of elements respectively. In obtainingthe cardinality of the set w we argue as follows: On those elements of G rj ( n ) in which the chosen restricted section is empty; that means the n elements are distributed among the other r + j − r − j of the sections arefree sections. Hence the cardinality of w is p r − j ( n ) (by definition of p r − j ( n )).In finding the number of elements of the set w we argue as follows;The maximum number of elements which are not in the chosen sectionin this case is n − s elements which are not in the restricted section. There are (cid:0) ns (cid:1) ways of selecting the elements. There are p r − j ( s ) ways of preferentiallyarranging the s elements among the r + j − p ( n − s ) ways of preferentially arranging the remaining n − s elements on the restricted section (note p ( n − s ) = 1 ). Taking the product andsumming over s we obtain n − P s =0 (cid:0) ns (cid:1) p r − j ( s ) = w .Combining w and w we obtain the result of the theorem. Theorem 8. for n, r, ≥ and j ≥ we have; p r +1 j ( n ) = 2 p rj ( n ) − p rj − ( n )In proving the theorem we consider the set G rj ( n ); which is the setof all barred preferential arrangements of a set X n having k bars inwhich r fixed sections are restricted sections and j fixed sections arefree sections. By choice we choose the j free sections (see definition 1above) of each element B ∈ G rj ( n ) to be the last (from left to right) j sections of B . We want to construct the set G r +1 j ( n ) using elements ofthe set G rj ( n ).We argue as follows: We add an extra bar ∗ | to the far right of eachelement B ∈ G rj ( n ) to form the set D rj ( n ). We observe that addingthe extra bar to each element of G rj ( n ) does not affect counting; hence | D rj ( n ) | = | G rj ( n ) | = p rj ( n ). We note that on each element C ∈ D rj ( n )to the left of the bar ∗ | is a free section and the section to the right of ∗ | is empty. We define the set R rj ( n ) = { , } × D rj ( n ) as containing thesame elements as the set D rj ( n ) but with an indexing on the bar ∗ | whichis 0 and separately 1. Hence R rj ( n ) has twice the number of elements as D rj ( n ) (of which half of them have the index 0 and the other half havethe index 1). Also each element of R rj ( n ) has k + 2 sections due to theintroduction of the bar ∗ | . We now use elements of R rj ( n ) to constructthe set G r +1 j ( n ). We construct as follows:I. If the index on the bar ∗ | on an element O ∈ R rj ( n ) is 0 then such anelement will be interpreted in G r +1 j ( n ) as an element of G r +1 j ( n ) whose( k + 2) th section is empty.We collect all such elements to form the set W . The set W has p rj ( n )elements (since half of the elements in R rj ( n ) have the index 0)II. If the index on the bar ∗ | on O ∈ R rj ( n ) is 1; then we shift the last block of the ( k + 1) th section of O to be the only block to the right of ∗ | (i.e the block closest to the bar ∗ | ). We collect all such elements to formthe set K . There are p rj ( n ) elements having index 1 in R rj ( n ) (half ofthe elements in R rj ( n ) have the indexing 1).When discarding the index on the bar ∗ | in the construction of theset K ; some elements are common elements between the set K and theset W . The common elements occur when the free section closest tothe bar ∗ | is empty. In that case there is no block to put to the right ofthe bar ∗ | when constructing the set K . The common elements between K and W occur when the n elements of X n are distributed on the first k sections of each element of G rj ( n ). The k other sections are composedof r restricted sections and j − X n are distributed on these sections is p rj − ( n );so | K ∩ W | = p rj − ( n ). Hence | K ∪ W | = | K | + | W | − | K ∩ W |⇒| K ∪ W | = p rj ( n ) + p rj ( n ) − p rj − ( n ). The elements of both sets K and W have k + 2 sections; Of which for a fixed X ∈ K ∪ W , r + 1 sectionsare restricted sections and j sections are free sections. By definition K ∪ W = G r +1 j ( n ). This completes the proof.The statement of theorem 8 for the case j = 1 is p r +11 ( n ) = 2 p r − r n .Which is equation 14 of [7], proposed by Nelsen and Schmidt. Conjecture. for n, r ≥ we have; p r ( n ) = n P s =1 (cid:0) ns (cid:1) p r ( n − s ) + r n The identity in the conjecture when r = 0 appears in [4] as equa-tion 9. For r = 2 the identity in the conjecture occurs in [9] as anidentity for the number of preferential fuzzy subsets of a set X n ; itoccurs as equation 3.4 in [9]. Future work
MacMahon in [10] has shown a way on how the generating function P k ( m ) = − e m can be interpreted in a graph theoretic context usingthe idea of yokes and chains. What could be done is to take up theargument to the general generating function P k ( m ) = e km − e m of Nelsenand Schmidt. We suspect when you glue k of these yoke-chain graphsthe generating function for the number of yoke-chains is related to theNelsen-Schmidt generating function P k ( m ) = e km − e m .6. Acknowledgements
Both authors acknowledge the support from Rhodes University. Thefirst author would like also to acknowledge financial support from theDAAD-NRF scholarship of South Africa, the Levenstein Bursary ofRhodes University and the NRF-Innovation doctoral scholarship ofSouth Africa.
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