aa r X i v : . [ m a t h . C O ] J un Some formulas for the number of gluings
A. V. Pastor ∗ O. P. Rodionova
Consider K disks D , D , . . . , D K . Let the boundary circle of D i is alsodenoted by D i . Let 2 N points be marked on the boundary circles D ,. . . , D K such that at least one point is marked on each circle. We fix on theeach circle D i a counterclockwise orientation. The marked points divide thecircle D i into several arcs (which do not contain marked points). One ofthese arcs is marked with the number i .Thus we have 2 N arcs on K circles. We split these arcs into pairs and gluetogether correspondent arcs such that two arcs in each pair are oppositelyoriented. We obtain as a result of gluing a compact orientable surface withoutboundary. This surface can be disconnected. Points marked on the circlesand arcs of these circles form a graph drawn on the obtained surface.In what follows we call disks D , D , . . . , D K by polygons . Then markedpoints are vertices and arcs of our circles by are edges of these polygons.Moreover, let a disk with exactly M marked points on its boundary circle becalled M -gon. We allow polygons with one and two vertices. For each edgewe fix the first and the last vertex (in counterclockwise orientation of theirpolygon). Definition 1.
Let a map be an ordered pair (
X, G ), where G is a graph(possibly with loops and multiple edges), embedded in a compact orientablesurface X without boundary, such that connected components of X \ G ,called faces of the map, are homeomorphic to discs.Two maps ( X, G ) and ( X ′ , G ′ ) are isomorphic , if there is an orientationpreserving homeomorphism f : X → X ′ , such that f ( G ) = G ′ . ∗ This research was supported by grant of the President of Russian FederationNSh-3229.2012.1; Russian Foundation for Basic Research, grant 11-01-00760-a and grantof Russian government (FCP). genus of the map (
X, G ) be the genus of the surface X .All necessary facts on maps (in particular, about connection betweenmaps and permutations) one can find in [3]. We recall some notations, thatare necessary for our paper. Remark 1.
Note, that several nonisomorphic maps can be correspondent toa graph G . We add to a graph G an additional construction such that themap correspondent to G would be unique up to isomorphism.Let’s assign to each edge of the graph G a pair of oppositely orientedarcs and for every vertex put in cyclic order all arcs with the beginning atthis vertex, (i.e. for every vertex let’s set a clockwise order in which outgoingarcs pass out this vertex). It is known that fixing a cyclic order of outgoingarcs for every vertex of the graph G determine up to isomorphism a mapcorrespondent to G (see, for examples, [3]).Let’s set a cyclic order of outgoing arcs for every vertex of a graph G . Weobtain as a result a permutation σ on the set A of all arcs of the graph G ( σ ( e ) is the next (in the cyclic order) arc outgoing from the beginning of anarc e ). We also define permutations ι, τ ∈ S ( A ) as follows. Let ι ( e ) be thearc opposite to e , i.e. correspondent to the same edge of G but oppositelyoriented. Let τ = σι . Since ι = 1, we also have σ = τ ι .We say that an arc e of a map ( X, G ) belongs to a face F , if e belongs tothe boundary of F and is oriented in the direction of counterclockwise walkaround F . Note, that the arc τ ( e ) also belongs to the face F and is the nextarc for e in the counterclockwise walk of the boundary of F . Hence we callthe arc τ ( e ) by the next arc for e , and the arc τ − ( e ) by the previous arcfor e . Let’s also note that cycles of the permutations σ correspond to verticesof the map ( X, G ); cycles of the permutations ι correspond to edges of themap ( X, G ); and cycles of the permutations τ correspond to faces of the map( X, G ).It is easy to see that the procedure of gluing from K polygons gives usa map with N edges and K faces. Faces of obtained map correspond topolygons we have glued, so let’s enumerate these faces with numbers from 1to K as well as correspondent polygons. Each edge of the map corresponds totwo edges of polygons. Counterclockwise orientations of edges of all polygonsgives us a bijection from the set of edges of polygons to the set of arcs ofthe map. Let’s mark all arcs of the map correspondent to marked edges ofpolygons with the same number.Thus we have obtained a map with N edges and K faces. Faces of thismap are marked with integers from 1 to K , and each face has exactly onearc marked with its number. We say that such map is marked .2e say that marked maps ( X, G ) and ( X ′ , G ′ ) are isomorphic , if there isan isomorphism of maps f : ( X, G ) → ( X ′ , G ′ ), that preserves numbers offaces and marks on arcs of maps. It is easy to see that the procedure of gluingdescribed above define a map uniquely up to isomorphism. Moreover, everymap can be glued from a collection of polygons uniquely up to isomorphism.Speaking about number of marked maps or about number of gluings of sometype we will always mean the number of maps or gluings up to isomorphism(i.e. the number of classes of isomorphic marked maps).Similarly, speaking about marked maps of some type we mean classes ofisomorphic marked maps of this type. The main object of our research is the number of gluings together K polygons,that have (together) 2 N edges, into a connected orientable surface of genus g . Remark 2.
This problem have an equivalent formulation. Let’s enumerateall arcs of the map with integers from 1 to 2 N in the following way: at firstwe enumerate arcs of the face 1 beginning at the arc marked with 1, thenwe similarly enumerate arcs of the face 2, and so on. Then the permutations ι and τ defined above act on the set { , , . . . , N } . Moreover, ι consists of N independent transpositions and τ consists of K independent cycles, suchthat elements of each cycle are consecutive integers. It is known (see [3]),that such a pair of permutations defines a map up to isomorphism. Note,that cycles of the permutation σ = τ ι correspond to vertices of the map thatwe construct. Since this map has V vertices, N edges and K faces, the genusof a surface on which our map is drawn can be calculated by Euler’s formula V − N + K = 2 − g . Thus our problem can be reduced to a problem offinding the number of pairs of permutations ( ι, τ ) of type described above,such that the group h ι, τ i generated by the permutations ι and τ is transitiveand τ ι is a composition of N − K + 2 − g independent cycles. Definition 2.
We denote by ε g ( N, K ) the number of ways to glue together K polygons, that have (together) 2 N edges, into connected orientable surfaceof genus g .For N = 0 we set ε (0 ,
1) = 1 and ε g (0 , K ) = 0 if g + K > ε g ( N, K ) can be interpreted as the number of marked mapswith N edges and K faces on a surface of genus g . In this interpretation thenumber ε (0 ,
1) corresponds to a sphere with single marked vertex.3 emark 3.
As it was mentioned above, the number of vertices of the map isequal to N − K +2 − g . Since this number must be positive, for N < K +2 g − ε g ( N, K ) = 0.Let’s consider one more object: bicolored gluings , defined as follows. Letvertices of each of polygons D , D , . . . , D K be properly colored with whiteand black colors and we can glue together only vertices of the same color.The result of such gluing is a bicolored map which vertices are properlycolored with white and black colors. The edges of polygons and correspondentarcs of the map, outgoing from white (black) vertices we call white ( black )respectively. Marked edge in every polygon must be white.Note, that bicolored polygon must have even number of vertices and weglue together edges of different colors. Definition 3.
Let marked bicolored map be a marked map, which verticesare properly colored with white and black colors, such that all marked arcsare white.A bicolored map which marked arcs are not necessarily white (i.e. can bewhite or black) is called quasimarked . Definition 4.
We denote by B g ( N, K ) the number of ways to glue together K bicolored polygons, that have (together) 2 N edges, into connected ori-entable surface of genus g .For N = 0 we set B (0 ,
1) = 1 and B g (0 , K ) = 0 if g + K > B g ( N, K ) can be interpreted as the number of marked bi-colored maps with N edges and K faces on a surface of genus g . Remark 4.
Note, that there is a natural bijection from the set of quasi-marked bicolored maps where the arc marked with the number i is blackto the set of quasimarked bicolored maps where this arc is white: we markwith i the white arc that is the next to the marked black arc. The inversebijection is similar.Thus, the number of quasimarked maps with N edges and K faces isequal to 2 K B g ( N, K ).Gluings a surface of genus g from one polygon were for the first timeconsidered by J. Harer and D. Zagier [8]. In this paper the numbers of suchgluings (denoted by ε g ( N )) were used to calculate Euler’s characteristic ofmoduli space. The following recursion for the numbers ε g ( N ) was provedin [8]: ε g ( N ) = 2 N − N + 1 (2 ε g ( N −
1) + ( N − N − ε g − ( N − . (1)4We set ε (0) = 1.) Many proofs of formula (1) are known now. In thepaper [4] a bijective proof for this formula was given.The recursion for the numbers of bicolored gluings of a surface of genus g from one polygon (denoted by B g ( N )) is similar to (1). It was obtained inthe papers [9] and [1] independently: B g ( N ) = 1 N + 1 (2(2 N − B g ( N −
1) + ( N − N − B g − ( N − . (2)(We set B (0) = 1.) A bijective proof of the formula (2) was obtained in [10].One can find explicit formulas for the numbers ε g ( N ) and B g ( N ) forsmall g in the papers [7] and [1]. We write down two explicit formulas weneed in what follows: ε ( N ) = B ( N ) = (2 N )! N !( N + 1)! ; (3) ε ( N ) = (2 N )!12 N !( N − . (4)Gluings of a surface of genus g from two polygons were considered in thepapers [5] and [2]. In [5] the generating function for the numbers of gluingsof a surface of genus g from a p -gon and a q -gon was obtained, but it wastedious. However, no recurrence or explicit formula for ε g ( N,
2) was obtainedin this paper. A more simple formula was presented in [2]: it was provedthat for every g ≥ X N ≥ ε g ( N, z N = P [2] g ( z )(1 − z ) g +2 , where P [2] g ( z ) is a polynomial of degree at most 3 g + 1. A recursive methodfor calculating this polynomial was presented. Moreover, explicit formulasfor the numbers ε ( N, ε ( N,
2) and ε ( N,
2) were proved in [2]: ε ( N,
2) = N N − ; (5) ε ( N,
2) = 112 (13 N + 3) N ( N − N − N − ; (6) ε ( N,
2) = 1180 (445 N − N − N ( N − N − N − N − N − . (7)In this paper we give elementary proofs of formulas (5) and (6). We ob-tain explicit formulas for the numbers B ( N,
2) and ε ( N,
3) (see theorems 4and 6). 5 .3 An operation of deleting an edge
All formulas proved in our paper are proved by induction. The plan of theseproofs is as follows. We consider an arbitrary marked connected map ofgenus g with N > K faces. After deleting the edge marked with1 from this map we obtain a connected map or a pair of connected maps withless number of edges. This help us to obtain a recurrence equation for thenumbers ε g ( N, K ). After that we by induction prove the explicit formula.Consider the operation of deleting an edge in details. Let (
X, G ) bea marked map, where X is a connected surface of genus g and G is agraph with V vertices and N > D , D , . . . , D K , with m , m , . . . , m K edges respectively. Then m + m + . . . + m K = 2 N . Denote by e i the arc marked with i , by e ′ i the arc opposite to e i and by ˜ e i the edge of the graph G correspondent tothese two arcs. Let’s delete from G the edge ˜ e . For every vertex of obtainedgraph G we put in cyclic order all outgoing arcs as in the map ( X, G ) (wesimply exclude all deleted arcs from the cyclic order). As it was mentionedin remark 1, there exists a unique up to isomorphism map ( X , G ) corre-spondent to the graph G with fixed cyclic order of arcs. If the surfaces X and X are homeomorphic then we assume without loss of generality that X = X . However, these surfaces can be not homeomorphic: for example,if the graph G is disconnected, then, clearly, the surface X is disconnectedtoo.To make the map ( X , G ) marked, we must enumerate the faces andmark some arcs. Consider the following 4 cases:1. the arcs e and e ′ belong to different faces ;2. e and e ′ are successive arcs of the face number e and e ′ are unsuccessive arcs of the face number , the graph G isconnected ;4. e and e ′ are unsuccessive arcs of the face number , the graph G isdisconnected .For each of these cases we describe the obtained map or pair of maps, andcount the number of ways to obtain each map or pair of maps as a result ofapplying the operation to all maps satisfying the conditions of the case weconsider.Let’s begin the case analysis. 6 . The arcs e and e ′ belong to different faces. Let e ′ belongs to the face number j = 1. Then after deleting the edge˜ e the faces 1 and j become glued together. Hence we obtain a map ( X, G )of genus g with N − K − K − X, G ) and correct thecorrespondent marks on arcs. It remains to mark with 1 an arc of the map(
X, G ). If e j = e ′ , then we mark with 1 the arc e j . If e j = e ′ and m > e in the map ( X, G ). Otherwise e j = e ′ and m = 1, then we mark with 1 the arc next to e j in the map ( X, G ) (seefigure 1). Note, that in the last case we have m j >
1, since the map (
X, G )is connected and
N >
1. We obtain as a result a connected marked map ofgenus g with N − K − m + m j − e e ′ f = e j e e j = e ′ f e e j = e ′ f Figure 1: Operation of deleting an edge, case . In each subcase f is thearc that will be marked with 1 in the obtained map. Lemma 1.
Let’s apply the operation of deleting an edge to all marked mapssatisfying the conditions of case . Then every marked map of genus g with N − edges and K − faces, such that face number of this mapcontains M arcs, is obtained exactly ( M +1)( M +2)( K − times. Proof.
Let (
X, G ) be a marked map of genus g with N − K − F be the face number 1 and D be the M -gon, correspondent tothe face F . Denote by f the arc marked with 1 and by w the beginningof the marked edge of D . This map was obtained by deleting an edge ˜ e ofan initial map, this edge was drawn inside the face F . Hence the edge ˜ e isuniquely up to isomorphism defined by an unordered pair { a, b } of verticesof the polygon D (the vertices in this pair can coincide). The number ofsuch pairs is equal to M ( M +1)2 . 7et’s reconstruct marks on edges of the initial map. At first note, that theface F can be glued from the faces number 1 and number j of the initial map,where the number j can be chosen in K − j ,we must to point out which part of F will have number 1 and which onewill have number j . Then we must choose marked arcs in two new faces (thenumbers of all other faces and their marked arcs are reconstructed uniquely).Note, that it’s enough to choose the arc e j : then we set the enumeration ofparts uniquely and the added arc lying in the face number 1 will be markedwith 1.There are two possible cases: e j = f or e j = e ′ (see the descriptionof the operation). The first case is possible for any edge ˜ e , i.e. occurs M ( M +1)2 times. The second case is possible only if one of the vertices a and b coincides with w . (If exactly one of the vertices a, b coincides with w , thenthe direction of e j is defined uniquely. If both vertices a and b coincide with w then the direction of e j can be set in two ways.) Thus, the second caseoccurs M + 1 times. Hence, for each j we have ( M +1)( M +2)2 ways, and totally ( M +1)( M +2)( K − ways. e and e ′ are successive arcs of the face number . Since (
X, G ) is connected and
N >
1, in this case we have m > e has degree 1. Denote thisend by u . Then after deleting the edge ˜ e and the vertex u we obtain amap ( X, G ′ ) with N − K faces. We preserve the enumerationof faces and marks on all arcs with exception of the deleted arc e . Then wemark with 1 the arc of ( X, G ) which is the next to those of arcs e and e ′ ,that begins at u (see figure 2). As a result we obtain a marked connectedmap of genus g with V − K faces. u uf f e e ′ e ′ e Figure 2: The operation of deleting an edge, case . Disposition of the arcs e and e ′ . Lemma 2.
Let’s apply the operation of deleting an edge to all marked mapssatisfying the conditions of case . Then every marked map of genus g with N − edges and K faces is obtained exactly twice. roof. Let (
X, G ′ ) be the marked map of genus g with N − K faces; F be the face number 1 in this map and v be the beginning of thearc f marked with 1 in this map. This map can be obtained as a resultof deleting an edge uv , such that the vertex u is inside the face F and thearc f was previous arc to uv before deleting this edge. The edge uv can bedrawn uniquely up to isomorphism, all numbers of faces and arcs marked withnumbers more than 1 are preserved. The only non-uniqueness is as follows:we can mark with 1 either uv or vu . Hence we have two variants. e and e ′ are unsuccessive arcs of the face number , the graph G isconnected. Consider the permutations σ , ι , τ on the set of arcs of the map ( X, G )defined above. (The permutation σ defines the cyclic order of outgoing arcsfor every vertex; ι for any arc gives the opposite arc; τ = σι for any arcgives the next one). Let’s consider the similar permutations σ , ι , τ = σ ι on the set of arcs of the map ( X , G ). As it was mentioned above, thepermutation σ is the result of excluding the arcs e and e ′ from σ and ι isthe result of excluding the cycle ( e , e ′ ) from ι . Whence it follows, that τ ( e ) = τ ( e ) , τ ( e )
6∈ { e , e ′ } ; τ ( e ′ ) , τ ( e ) = e ; τ ( e ) , τ ( e ) = e ′ . Thus in this case the face 1 of the map (
X, G ) corresponds to two faces ofthe map ( X , G ). One of these faces consists of all arcs, lying on the wayfrom e to e ′ in the face 1 of ( X, G ), and the other face consists of all arcs,lying on the way from e ′ to e of the face 1 of ( X, G ). Let’s enumerate thefirst face with 1 and the second face with K + 1 and mark in these faces thearcs, next to e and e ′ respectively (see figure 3a). We obtain a connectedmarked map with N − K + 1 faces. Since this map ( X , G )contains V vertices, by Euler’s formula it has genus g − Remark 5.
In this case we delete an edge drawn on one of handles of thesurface X . After that one connected component of the set X \ G becomeshomeomorphic to a cylinder. We cut off this cylinder and replace it by twodisks correspondent to its bases. As a result we obtain the map ( X , G ). Lemma 3.
Let’s apply the operation of deleting an edge to all marked mapssatisfying the conditions of case . Then every marked map of genus g − with N − edges and K + 1 faces is obtained exactly once. roof. Let ( X , G ) be a marked map of genus g − N − K + 1 faces. Denote by f i the arc marked with i and by v i its beginningvertex. Let’s show that initial marked map ( X, G ) can be reconstructeduniquely. Note, that the graph G is a result of adding to G the edge v v K +1 .The cyclic order of outgoing arcs for each vertex is reconstructed uniquely:for a vertex v i (where i ∈ { , K + 1 } ) the added arc is disposed just before f i ,for all other vertices the cyclic order is preserved. Thus the map ( X, G )is reconstructed uniquely. The numbers of all faces with exception of 1and K + 1 and their marked arcs are preserved too. The arc marked with 1is also reconstructed uniquely: it leads from v K +1 to v . e and e ′ are unsuccessive arcs of the face number , the graph G isdisconnected. In this case the map ( X , G ) consists of two connected components. Asin previous case, the face number 1 of the map ( X, G ) corresponds to twofaces of the map ( X , G ). One of these faces consists of all arcs, lying on theway from e to e ′ in the face 1 of ( X, G ), and the other face consists of all arcs,lying on the way from e ′ to e in the face 1 of ( X, G ). Note, that these twofaces belong to different connected components of the map ( X , G ). Denotethese components by ( X , G ) and ( X , G ) respectively. In both componentswe enumerate with 1 the face obtained from the face 1 of ( X, G ). As inprevious case, we mark in these faces the arcs, next to e and e ′ respectively(see figure 3b). All other faces of each of maps ( X , G ) and ( X , G ) weenumerate in increasing order of their numbers in the map ( X, G ) and correctthe correspondent marks on arcs. As a result we obtain an ordered pair ofmarked maps, which have together V vertices, N − K + 1 faces.Each of this two maps contains at least one edge. Let the map ( X , G ) e e ′ f f K +1 a e e ′ f f b Figure 3: The operation of deleting an edge, cases and . In case (fig. 3a) f and f K +1 are the arcs we mark with 1 and K + 1, respectively. In case (fig. 3b) f and f are the arcs we mark with 1 in the maps ( X , G ) and( X , G ), respectively. 10ave genus g , and the map ( X , G ) have genus g . By Euler’s formula wehave g + g = g . Remark 6.
As in previous case, one of the connected components of theset X \ G is homeomorphic to a cylinder. However, in the case we considerafter replacing this cylinder by two disks correspondent to its bases, thesurface becomes disconnected. Lemma 4.
Let’s apply the operation of deleting an edge to all marked mapssatisfying the conditions of case . Consider ordered pairs of marked mapswith positive number of edges in each map, such that these maps containtogether N − edges and K + 1 faces and have sum of the genera g . Thenevery such ordered pair is obtained exactly C K − K − = C K − K − times, where K is the number of faces of the first map and K is the number of faces of thesecond map. Proof.
The proof is similar to the proof of lemma 3, but in this case theadded edge connects the beginnings of arcs marked with 1 in the maps( X , G ) and ( X , G ). The difference is that the enumeration of faces canbe reconstructed in different ways. Let’s remind, that we enumerate the facesin each of the maps ( X , G ) and ( X , G ) in the increasing order of theirnumbers in the initial map. Hence the enumeration of faces of the initial mapis uniquely defined by the choice of the set of numbers that faces of ( X , G )with numbers more than 1 have in the initial map. This set can be chosenin exactly C K − K − ways. In this section with the help of techniques of the book [6] we proof someequalities for binomial coefficients and similar formulas, that we need in whatfollows.
Definition 5.
For integers i and k , such that i ≥ k > p k ( i ) asfollows: p k ( i ) = k − Y t =0 ( i + 1 − t ) . (8)We set p ( i ) = 1 for every i ≥ p k ( i ) = ( i + 2 − k ) p k − ( i ) , i ≥ , k >
0; (9) p k ( i ) = ( i + 1) p k − ( i − , i > , k >
0; (10)11 emark 7.
The numbers p k ( i ) are often called “descending degrees” of i + 1and denoted by ( i + 1) k . Definition 6.
For integers M , m , k , such that M ≥ m ≥ k ≤ m let A ( M, m, k ) = M X j = m (2 j )!4 j j !( j − k )! . (11)For m > M ≥ − A ( M, m, k ) = 0.
Lemma 5.
Let M , m , k be integers, such that M ≥ m ≥ , k ≤ m . Then A ( M, m, k ) = 22 k + 1 (cid:18) ( M + 1 − k )(2 M + 2)!4 M +1 ( M + 1)!( M + 1 − k )! − ( m − k )(2 m )!4 m m !( m − k )! (cid:19) . (12) Proof.
Let a j ( k ) = (2 j )!4 j j !( j − k )! , where k ≤ j . It is easy to see, that a j ( k ) = 22 k + 1 (( j + 1 − k ) a j +1 ( k ) − ( j − k ) a j ( k )) . Let’s sum these equalities for j = m, . . . , M . The second term of each sum-mand cancel on the first term of the next summand. Hence we obtain theequality we want to prove. Corollary 1.
For all integers M ≥ A ( M, , −
1) = M X j =0 (2 j )!4 j j !( j + 1)! = 2 − M + 2)(2 M + 2)!4 M +1 ( M + 1)!( M + 2)! . (13) Corollary 2.
For all integers M ≥ − , k ≥ A ( M, k, k ) = M X j = k (2 j )!4 j j !( j − k )! = 22 k + 1 · p k ( M )(2 M + 2)!4 M +1 ( M + 1)!( M + 1)! . (14) Proof.
For M ≥ k this equality directly follows from the formula (12). For M < k all terms of the equality we prove are equal to zero.
Definition 7.
For integers N and k , such that N, k ≥ D k ( N ) = N X i =0 p k ( i ) C i i +1 · C N − i N − i +1 i + 1 . (15)12 emma 6. For every N ≥ the following statements hold:a) D ( N ) = C N N +2 ; (16) b) D ( N ) = 2 · N − C N +12 N +2 ; (17) c) D ( N ) = ( N + 1) (cid:18) N − C N +12 N +2 (cid:19) ; (18) d) D ( N ) = N ( N + 1) (cid:18) · N − − C N +12 N +2 (cid:19) ; (19) e) D ( N ) = ( N − N ( N + 1) (cid:18) · N − − C N +12 N +2 (cid:19) . (20) Proof. a) We get use of the following formula for convolution of a sum ofbinomial coefficients (see, for example, [6, formula (5.62)]): X k C ktk + r · C N − kt ( N − k )+ s · rtk + r = C NtN + r + s . For r = s = 1, t = 2, we obtain D ( N ) = N X k =0 C k k +1 · C N − k N − k +1 k + 1 = C N N +2 . Thus, the statement a) is proved. Before proofs of the next items let’sstate some useful formulas. Let d i ( N ) = ( i + 1) C i i +1 · C N − i N − i +1 i + 1 . (21)Clearly, for N, i > d i ( N ) = 2 · i − i · d i − ( N − . (22)It follows from the formula (9), that(2 i + 1) p k − ( i ) = 2 p k ( i ) + (2 k − p k − ( i )for i ≥ k >
0. Then taking into account the formulas (10), (21) and (22)13e obtain the following chain of equalities: D k ( N ) = N X i =0 p k ( i ) C i i +1 · C N − i N − i +1 i + 1 = p k (0) C N N +1 + N X i =1 p k ( i ) i + 1 d i ( N ) == p k (0) C N N +1 + N X i =1 p k − ( i −
1) 2(2 i − i d i − ( N −
1) == p k (0) C N N +1 + 2 N − X j =0 p k − ( j ) 2 j + 1 j + 1 d j ( N −
1) == p k (0) C N N +1 + 4 N − X j =0 p k ( j ) j + 1 d j ( N −
1) + 2(2 k − N − X j =0 p k − ( j ) j + 1 d j ( N − . Whence by formulas (15) and (21) it follows, that for all k > D k ( N ) = p k (0) C N N +1 + 4 D k ( N −
1) + 2(2 k − D k − ( N − . (23) b) By formulas (23) and (16) D ( N ) = p (0) C N N +1 + 4 D ( N − − D ( N −
1) == 4 D ( N −
1) + C N N +1 − C N − N = 4 D ( N −
1) + (2 N )! N !( N + 1)! , whence by induction we can exclude the following formula: D ( N ) = 4 N D (0) + N X i =1 N − i (2 i )! i !( i + 1)! . Note, that D (0) = 1. By formula (13) we have: D ( N ) = 4 N + 4 N N X i =1 (2 i )!4 i i !( i + 1)! = 4 N N X i =0 (2 i )!4 i i !( i + 1)! == 4 N (cid:18) − N + 2)(2 N + 2)!4 N +1 ( N + 1)!( N + 2)! (cid:19) = 2 · N − C N +12 N +2 . Thus the statement b) is proved.Note, that p k (0) = 0 and D k (0) = 0 for k > D k ( N ) for k >
1: 14 k ( N ) = 4 D k ( N −
1) + 2(2 k − D k − ( N −
1) == 4 N D k (0) + 2(2 k − N X i =1 N − i D k − ( i −
1) == 2(2 k − N N X i =1 D k − ( i − i . (24) c) By formula (24), (17) and formula (14) for k = 0, we have: D ( N ) = 2 · N N X i =1 D ( i − i = 2 · N N X i =1 · i − − C i i i == 4 N · N − N N X i =1 C i i i = ( N + 1)4 N − N N X i =0 (2 i )!4 i i ! i ! == ( N + 1)4 N − N N + 1)(2 N + 2)!4 N +1 ( N + 1)!( N + 1)! = ( N + 1) (cid:18) N − C N +12 N +2 (cid:19) . d) By formulas (24), (18) and formula (14) for k = 1 we have: D ( N ) = 2 · · N N X i =1 D ( i − i = 6 · N N X i =1 i · (cid:0) i − − C i i (cid:1) i == 6 · N − N X i =1 i − · N N X i =1 (2 i )!4 i i !( i − N ( N + 1) · · N − − · N N ( N + 1)(2 N + 2)!3 · N +1 ( N + 1)!( N + 1)! == N ( N + 1) (cid:18) · N − − C N +12 N +2 (cid:19) . e) By formulas (24), (19) and formula (14) for k = 2 we have: D ( N ) = 2 · · N N X i =1 D ( i − i = 10 · N N X i =1 ( i − i · (cid:0) · i − − C i i (cid:1) i == 30 · N − N X i =1 i − N X i =1 i ! − · N N X i =2 (2 i )!4 i i !( i − · N − ( N − N ( N + 1) − · N · N − N ( N + 1)(2 N + 2)!5 · N +1 ( N + 1)!( N + 1)! == ( N − N ( N + 1) (cid:18) · N − − C N +12 N +2 (cid:19) . Gluing together two polygons into a sphere
In this section we prove recurrence and explicit formulas for the numbers ε ( N, ε ( N,
2) can be interpreted as the number of markedmaps on a sphere, that contain N edges and 2 faces. Theorem 1.
The numbers ε ( N, for N ≥ satisfy the following recursion: ε ( N,
2) = 2 N − X i =0 ε ( i ) ε ( N − i − ,
2) + N (2 N − ε ( N − . (25) Proof.
Let’s consider all marked maps on a sphere that contain
N > g = 0, case from the description of this operation is impossible. Hencewe have as a result of operation one of these cases: • a marked map on a sphere with N − • a marked map on a sphere with N − • an ordered pair of two marked maps on spheres, that contain together N − ε ( N −
1) marked maps on a sphere with N − N (2 N −
1) times, and each of ε ( N − , N − N − i edges. Clearly, the number of such pairs is equal to2 ε ( i ) ε ( N − i − , ε ( N,
2) = N (2 N − ε ( N −
1) + 2 ε ( N − ,
2) + 2 N − X i =1 ε ( i ) ε ( N − i − , . To complete the proof of lemma it remains to note that ε (0) = 1. Theorem 2.
For N ≥ ε ( N,
2) = N · N − . (26)16 roof. We proof this statement by induction. The base for N = 1 is obvious.Let’s prove the induction step. Let ε ( k,
2) = k · k − for all k < N . We willshow that ε ( N,
2) = N · N − .By formulas (25), (3), (13) and by formula (14) for k = 0 we have thefollowing: ε ( N,
2) = 2 N − X i =0 ε ( i ) ε ( N − i − ,
2) + N (2 N − ε ( N −
1) == 2 N − X i =0 (2 i )! i !( i + 1)! ( N − i − N − i − + N (2 N − ε ( N −
1) == 2 · N − N N − X i =0 (2 i )!4 i i !( i + 1)! − N − X i =0 (2 i )!4 i i ! i ! ! + N (2 N − ε ( N −
1) == 2 N − (cid:18) N − N (2 N − N − ( N − N ! − N − N − N − ( N − N − (cid:19) ++ N (2 N −
1) (2 N − N − N ! = N · N − . In this section we prove recurrence and explicit formulas for the numbers B ( N, B ( N,
2) can be interpreted as the number of markedbicolored maps on a sphere, that contain N edges and 2 faces. Theorem 3.
The numbers B ( N, for N ≥ satisfy the following recur-sion: B ( N,
2) = 2 N − X i =0 B ( i ) B ( N − i − ,
2) + N ( N − B ( N − . (27) Proof.
The result of applying an operation of deleting an edge to a bicol-ored marked map is either a bicolored quasimarked map or a ordered pairof bicolored quasimarked maps. Let as apply this operation to all markedbicolored maps on a sphere with
N > Two arcs of the deleted edge ˜ e belong to different faces. e is white and the opposite arc e ′ is black. Hence e ′ couldn’t be marked with the number 2. Consequently, the arc marked with 1in the obtained map was marked with 2 in the initial map, and this arc iswhite. Thus in this case we obtain a marked bicolored map on a spherewith N − X, G ) be a bicolored marked map on a sphere with N − F be its only face; D be the (2 N − F and f be the arc marked with 1. This map can be obtainedby deleting from the initial map an edge ˜ e drawn inside the face F . Thisedge is defined uniquely up to isomorphism by its ends — a pair of vertices ofdifferent colors of the polygon D . The marks are also reconstructed uniquely:the arc f was marked with the number 2, and the white arc correspondentto the edge ˜ e was marked with 1.It remains to count the number of ways to draw the edge ˜ e , such that itswhite arc e and the arc f lie in different parts of the face F . Let’s enumeratethe vertices of D with integers from 1 to 2 N − D . Then white verticeshave odd numbers and black vertices have even numbers. Let a and b bethe beginning and the end of the arc e . Then a is odd and b is even, a, b ∈ { , . . . , N − } . The arcs e and f lie in different parts of the face F if and only if the arcs e ′ and f lie in the same part of F . The lattermeans that b > a (see figure 4). Now it is easy to see that the number ofsuch ordered pairs a, b is N ( N − .Hence each of B ( N −
1) marked bicolored maps on a sphere with N − N ( N − times, i.e. the case occurs for N ( N − B ( N −
1) initial maps. e ′ e f ab Figure 4: the operation of deleting an edge for a bicolored map, case . Two arcs of the deleted edge are successive arcs of the face number .
18n this case we obtain a quasimarked bicolored map with N − B ( N − , B ( N − ,
2) quasimarked maps, satisfying the con-ditions of case , occurs exactly once. Indeed, by the same arguments as inthe proof of lemma 2 we have that the deleted edge uv can be reconstructeduniquely up to isomorphism. Exactly one of two arcs uv and vu is white,this arc is to be marked with 1. Thus, case occurs for N ( N − B ( N − Two arcs of the deleted edge are unsuccessive arcs of the face number . Since g = 0, the graph becomes disconnected, i.e. the case we consider corre-sponds to the case of the description of the operation of deleting an edge.In this case we obtain an ordered pair of two quasimarked bicolored mapson a sphere, that contain together N − e (which is black) and in the second map it is the arc next to e ′ (whichis white). By remark 4 we have that the number of quasimarked maps weconsider is equal to the number of ordered pairs of marked bicolored mapson a sphere, that contain together N − occurs exactly once. Hence, case occurs for the number of initialmaps equal to the number of ordered pairs of marked bicolored maps on asphere, that contain together N − i edges is equal to 2 B ( i ) B ( N − i − , B (1 ,
2) = 0 case occurs for2 P N − i =1 B ( i ) B ( N − − i,
2) initial maps.Since each of B ( N,
2) initial maps satisfies the condition of one of thecases , and , we obtain the following formula: B ( N,
2) = N ( N − B ( N − B ( N − , N − X i =1 B ( i ) B ( N − i − , . Since B (0) = 1 we have B ( N,
2) = 2 N − X i =0 B ( i ) B ( N − i − ,
2) + N ( N − B ( N − . heorem 4. The numbers B ( N, for N ≥ satisfy the following formula: B ( N,
2) = ε ( N − ,
2) = ( N − · N − . (28) Proof.
We prove by induction, that B ( N,
2) = ε ( N − , N = 1 is obvious. For N = 2 it follows from formulas (27), (3) and (26) that B (2 ,
2) = ε (1 ,
2) = 1.Let’s prove the induction step. Let the equality B ( i,
2) = ε ( i − , i ≤ N . We will prove that this equality holds for N + 1. Note,that by formula (3) for all i ≥ ε ( i ) = B ( i ). Moreover, it followsfrom formula (3), that( N + 1) N B ( N ) = ( N + 1) N ε ( N ) = N (2 N − ε ( N − . Then by formulas (27), (25) and by induction assumption we have, that B ( N + 1 ,
2) = 2 N − X i =0 B ( i ) B ( N − i,
2) + ( N + 1) N B ( N ) == 2 N − X i =0 ε ( i ) ε ( N − i − ,
2) + N (2 N − ε ( N −
1) == ε ( N, . Remark 8.
It would be interesting to find a bijective proof of the identity B ( N,
2) = ε ( N − , In this section we prove recurrence and explicit formulas for the num-bers ε ( N, ε ( N,
3) is the number of marked maps on asphere, that contain N edges and three faces. In this section we need an auxiliary notion.
Definition 8.
Denote by ˜ ε ( N, M, K ) the number of marked maps on asphere with N arcs (i. e. N/ K faces, such that the face number 1contains exactly M edges. 20learly, ˜ ε ( N, M, K ) is also the number of ways to glue a sphere from K polygons, that contain together N edges where the first polygon contains M edges.Note, that ˜ ε ( N, M, K ) > N, M, K ≥ N is even;(iii) N ≥ M + K − M arcs and each of other faces must have at least one arc. Lemma 7.
Let numbers
N > , K = 2 and M satisfy the conditions (i)-(iii) .Then the following recursion holds: ˜ ε ( N, M,
2) = 2 ⌊ M − ⌋ X i =0 ε ( i )˜ ε ( N − i − , M − i − , N − M ) ε (cid:18) N − (cid:19) . (29) Proof.
Consider all marked maps on a sphere with
N > N/ M arcs. Let’s apply toeach of this maps the operation of deleting an edge. Consider three followingcases. e ′ e e = f ab a e = e ′ e f a = 1 2 b b e = e ′ e f b = a = 1 2 c Figure 5: The operation of deleting an edge. Fixed number of arcs of theface 1 of the obtained map. Case . Two arcs of the deleted edge ˜ e belong to different faces.
21n this case we obtain a marked map on a sphere with N/ − X, G ) be a marked map with N/ − F be itsonly face; D be a ( N − F and f be thearc, marked with 1. Similarly to the proof of lemma 1 let’s note, that theinitial map (from which the map ( X, G ) was obtained) is uniquely definedby the deleted edge ˜ e (i.e. by a pair of vertices of the polygon D ) and bythe arc e , marked with 2 in the initial map.Let’s enumerate vertices of the polygon D with numbers from 1 to N − e isdefined uniquely up to isomorphism by a pair of numbers a, b ∈ { , . . . , N − } , where b ≥ a . We consider two subcases. e = f . In this case any pair a, b such that b − a = M − e (see figure 5a). There are N − M − e is an arc of the edge ˜ e . In this case a = 1. Let M >
1. Then b = N − M and the arc marked with 1 in the initial map must precede thearc f (see figure 5b). These conditions define the arc e uniquely. If M = 1,then b = a = 1 and in the initial map the arc e precedes the arc f , i.e. alsois uniquely defined (see figure 5c).Thus each of ε (cid:0) N − (cid:1) marked maps on a sphere with N/ − N − M times. Hence, the case occurs for ( N − M ) ε (cid:0) N − (cid:1) initial maps. Two arcs of the deleted edge are successive arcs of the face number . In this case we obtain a marked map on a sphere with N/ − M − occurs for 2˜ ε ( N − , M − , Two arcs of the deleted edge are unsuccessive arcs of the face number 1.
Since g = 0, the graph becomes disconnected, i.e. the case we consider corre-sponds to the case of the description of the operation of deleting an edge.In this case we obtain an ordered pair of two marked maps on a sphere, thatcontain together N/ − M − i edges is equal to 2 ε ( i )˜ ε ( N − i − , M − i − , occurs for 2 P ⌊ M − ⌋ i =1 ε ( i )˜ ε ( N − i − , M − i − ,
2) initial maps.Summing these results and taking into account that ε (0) = 1, we obtainthe desired formula. 22 heorem 5. For N ≥ the numbers ε ( N, satisfy the following recursion: ε ( N,
3) = 2 N − X i =0 (cid:16) ε ( i ) ε ( N − i − ,
3) + ε ( i, ε ( N − i − , (cid:17) ++ N − X i =1 ( i + 1)( i + 2)˜ ε (2 N − , i, . (30) Proof.
Consider all marked maps on a sphere, that contain
N > Two arcs of the deleted edge ˜ e belong to different faces. In this case we obtain a marked map on a sphere with N − i arcs (Clearly, 1 ≤ i ≤ N − ε (2 N − , i,
2) such maps is obtained exactly ( i + 1)( i + 2) times.Thus, case occurs for P N − i =1 ( i + 1)( i + 2)˜ ε (2 N − , i,
2) initial maps. Two arcs of the deleted edge are successive arcs of the face number . In this case we obtain a marked map on a sphere with N − ε ( N − ,
3) such maps is obtained exactly twice.Thus, case occurs for 2 ε ( N − ,
3) initial maps. Two arcs of the deleted edge are unsuccessive arcs of the face number . Since g = 0, the graph becomes disconnected, i.e. the case we consider corre-sponds to the case of the description of the operation of deleting an edge.In this case we obtain an ordered pair of marked maps on a sphere, thatcontain together N − One map of the pair contains one face, the other map contains threefaces.
Let the map with one face contains i edges. By remark 3 the map with threefaces contains at least 2 edges, hence, 1 ≤ i ≤ N −
3. By lemma 4 each of2 ε ( i ) ε ( N − i − ,
3) such pairs of maps occurs once. Hence, case 3.1 occursfor 2 P N − i =1 ε ( i ) ε ( N − i − ,
3) initial maps.
Both maps of the pair contain two faces.
Let the first map of this pair contains i edges. Clearly 1 ≤ i ≤ N −
2. Bylemma 4 each of ε ( i, ε ( N − i − ,
2) such pairs of maps occurs twice.Hence, case 3.2 occurs for 2 P N − i =1 ε ( i, ε ( N − i − ,
2) initial maps.Summing these results and taking into account that ε (0) = 1 and ε (0 ,
2) = ε (1 ,
3) = 0, we obtain the desired formula.23 .2 Explicit formula
Before deducing an explicit formula for the numbers of gluing together threepolygons into a sphere we prove some auxiliary lemmas.
Lemma 8.
For any N ≥ the following equality holds: N − X i =1 ˜ ε (2 N, i,
2) = ε ( N,
2) = N N − . (31) Proof.
This statement follows immediately from the definitions of˜ ε (2 N, i,
2) and ε ( N,
2) and from formula (26).
Lemma 9.
For any N ≥ the following equality holds: N − X i =1 i ˜ ε (2 N, i,
2) = N N − . (32) Proof.
Let T ( N ) = P N − i =1 i ˜ ε (2 N, i, T ( N ) = N N − .The base for N = 1 is obvious. Let’s prove the induction step. Let T ( k ) = k k − for each k < N . We will show, that then T ( N ) = N N − .By formula (29) we have: T ( N ) = N − X i =1 i ˜ ε (2 N, i,
2) =2 N − X i =1 i ⌊ i − ⌋ X j =0 ε ( j )˜ ε (2 N − j − , i − j − , N − X i =1 i (2 N − i ) ε ( N −
1) = T ( N ) + T ( N ) , where T ( N ) and T ( N ) are the first and the second summands of the ob-tained expression, respectively.At first we calculate T ( N ). Note, that T ( N ) = 2 N − X j =0 ε ( j ) N − X i =2 j +3 i ˜ ε (2 N − j − , i − j − , . Let k = i − j − N j = N − j −
1. By the induction assumption and24ormula (31) we have: T ( N ) = 2 N − X j =0 ε ( j ) N j − X k =1 ( k + 2 j + 2)˜ ε (2 N j , k,
2) == 2 N − X j =0 ε ( j ) (cid:16) N j − X k =1 k ˜ ε (2 N j , k,
2) + (2 j + 2) N j − X k =1 ˜ ε (2 N j , k, (cid:17) == 2 N − X j =0 ε ( j ) (cid:0) N j N j − + (2 j + 2) N j N j − (cid:1) == 2 N − N − X j =0 ε ( j )4 j ( N − ( j + 1) ) . Applying formulas (3), (13) and formula (14) for k = 0 ,
1, we have: T ( N ) = 2 N − N − X j =0 (2 j )!4 j j !( j + 1)! ( N − ( j + 1) ) == 2 N − (cid:18) N N − X j =0 (2 j )!4 j j !( j + 1)! − N − X j =0 (2 j )!4 j j ! j ! − N − X j =1 (2 j )!4 j j !( j − (cid:19) == 2 N − (cid:18) N (cid:18) − N (2 N − N − ( N − N ! (cid:19) − N − N − N − ( N − N − −− · ( N − N − N − N − ( N − N − (cid:19) == N N − − ε ( N − N (4 N − . Now let’s calculate T ( N ): T ( N ) = N − X i =1 i (2 N − i ) ε ( N −
1) = ε ( N − N (4 N − . Summing the expressions for T ( N ) and T ( N ) we obtain the desiredformula. Lemma 10.
For any N ≥ the following equality holds: N − X i =1 ( i + 1)( i + 2)˜ ε (2 N, i,
2) = 4 N − N ( N + 1)(5 N + 7) . (33)25 roof. Denote the left part of the formula (33) by P ( N ) and the right partby ˜ P ( N ). We will prove by induction that P ( N ) = ˜ P ( N ).The base for N = 1 can be immediately verified. Let’s prove the inductionstep. The induction assumption is that P ( k ) = ˜ P ( k ) for all k < N . Let’sshow that then P ( N ) = ˜ P ( N ). By formula (29) we have: P ( N ) = N − X i =1 ( i + 1)( i + 2)˜ ε (2 N, i,
2) == N − X i =1 ( i + 1)( i + 2) · ⌊ i − ⌋ X j =0 ε ( j )˜ ε (2 N − j − , i − j − , N − X i =1 ( i + 1)( i + 2)(2 N − i ) ε ( N −
1) = P ( N ) + P ( N ) , where P ( N ) and P ( N ) are the first and the second summands of the ob-tained expression, respectively.At first we calculate P ( N ). Note, that P ( N ) = 2 N − X j =0 ε ( j ) (cid:18) N − X i =2 j +3 ( i + 1)( i + 2)˜ ε (2 N − j − , i − j − , (cid:19) . As in lemma 9 we set k = i − j − N j = N − j −
1. Then P ( N ) = 2 N − X j =0 ε ( j ) (cid:18) N j − X k =1 ( k + 1 + 2 j + 2)( k + 2 + 2 j + 2)˜ ε (2 N j , k, (cid:19) == 2 N − X j =0 ε ( j ) N j − X k =1 ( k + 1)( k + 2)˜ ε (2 N j , k, · N − X j =0 ε ( j )( j + 1) N j − X k =1 k ˜ ε (2 N j , k, · N − X j =0 ε ( j )( j + 1)(2 j + 5) N j − X k =1 ˜ ε (2 N j , k, . Applying induction assumption and formulas (32) and (31) we have: P ( N ) = 2 N − X j =0 ε ( j )4 N j − R j ( N ) , R j ( N ) = N j ( N j + 1)(5 N j + 7) + 16( j + 1) N j + 8( j + 1)(2 j + 5) N j == N ( N + 1)(5 N + 7) + ( N − N − j + 1) −− ( N + 27)( j + 1) j − j + 1) j ( j − . Applying formulas (3), (13) and formula (14) for k = 0 , , P ( N ) = 2 N − X j =0 ε ( j )4 N − j − R j ( N ) = 2 N − N − X j =0 (2 j )!4 j j !( j + 1)! R j ( N ) == 2 N − N ( N + 1)(5 N + 7) N − X j =0 (2 j )!4 j j !( j + 1)! ++ 2 N − ( N − N − N − X j =0 (2 j )!4 j j ! j ! −− N − ( N + 27) N − X j =1 (2 j )!4 j j !( j − − N − · N − X j =2 (2 j )!4 j j !( j − N − N ( N + 1)(5 N + 7) (cid:18) − N (2 N − N − ( N − N ! (cid:19) ++ 2 N − ( N − N −
24) 2( N − N − N − ( N − N − −− N − ( N + 27) 2( N − N − N − · N − ( N − N − −− N − · N − N − N − N − · N − ( N − N − N − N ( N + 1)(5 N + 7) − (2 N − N − N ! R ( N ) , where R ( N ) = N ( N + 1)(5 N + 7) − ( N − N − N − N ++ 13 ( N + 27)( N − N − N + ( N − N − N − N == 4 N (cid:18) N + 4 N + 113 N − (cid:19) . By formula (3) we finally obtain: P ( N ) = 4 N − N ( N + 1)(5 N + 7) − ε ( N − N (cid:18) N + 4 N + 113 N − (cid:19) . P ( N ). P ( N ) = N − X i =1 ( i + 1)( i + 2)(2 N − i ) ε ( N −
1) == ε ( N − N − X i =1 (cid:0) N + (6 N − i + (2 N − i − i (cid:1) == ε ( N − N (cid:18) N + 4 N + 113 N − (cid:19) . Summing the expressions for P ( N ) and P ( N ) we obtain the desiredformula. Theorem 6.
For N ≥ the numbers ε ( N, satisfy the following equality: ε ( N,
3) = (8 N + 5)( N − N ( N + 1)210 C N N +1 . (34) Proof.
Let ˜ C ( N ) = (8 N +5)( N − N ( N +1)210 C N N +1 . We will prove that ε ( N,
3) =˜ C ( N ) by induction.The base for N = 1 is clear. For N = 2 it follows from the formulas (30)and (29) that ε (2 ,
3) = 6 = ˜ C (2).Let’s prove the induction step. Let ε ( k,
3) = ˜ C ( k ) for k ≤ N . We willshow, that then ε ( N + 1 ,
3) = ˜ C ( N + 1). By formula (30) we have: ε ( N + 1 ,
3) = 2 N − X i =0 ε ( i ) ε ( N − i,
3) + 2 N − X i =0 ε ( i, ε ( N − i, N − X i =1 ( i + 1)( i + 2)˜ ε (2 N, i,
2) = S ( N ) + S ( N ) + S ( N ) , where S ( N ), S ( N ) and S ( N ) are the first, the second and the third sum-mands of the obtained expression, respectively.By formula (33) we conclude: S ( N ) = N − X i =1 ( i + 1)( i + 2)˜ ε (2 N, i,
2) = 4 N − N ( N + 1)(5 N + 7) . (35)Applying formula (26) we obtain: S ( N ) = 2 N − X i =0 ε ( i, ε ( N − i,
2) = 2 N − X i =1 i i − · ( N − i )4 N − i − == 2 · N − (cid:18) N ( N − − ( N − N (2 N − (cid:19) = 4 N − N − N ( N + 1) . (36)28ow consider S ( N ). Since ε ( N − i,
3) = 0 for i = N , we have: S ( N ) = 2 N − X i =0 ε ( i ) ε ( N − i,
3) = 2 N X i =0 ε ( i ) ε ( N − i, . Applying the induction assumption for ε ( N − i,
3) and formula (3) weobtain: S ( N ) = N X i =0 i )!(8( N − i ) + 5)( N − i − N − i )( N − i + 1) i !( i + 1)! · C N − i N − i )+1 == 1105 N X i =0 R i ( N ) C i i +1 · C N − i N − i +1 i + 1 , where R i ( N ) = (8( N − i ) + 5)( N − i − N − i )( N − i + 1) == N ( N + 1)( N + 2)(8 N + 13) − ( i + 1) N ( N + 1)(32 N + 31)++ i ( i + 1) N (48 N + 15) − ( i − i ( i + 1)(32 N − i − i − i ( i + 1) . Applying formula (15), and, after that, formulas (16-20), we can writethe following:105 S ( N ) = N ( N + 1)( N + 2)(8 N + 13) D ( N ) −− N ( N + 1)(32 N + 31) D ( N ) + N (48 N + 15) D ( N ) −− (32 N − D ( N ) + 8 D ( N ) == N ( N + 1)( N + 2)(8 N + 13) C N N +2 −− N ( N + 1)(32 N + 31) (cid:18) · N − C N +12 N +2 (cid:19) ++ N (48 N + 15)( N + 1) (cid:18) N − C N +12 N +2 (cid:19) −− (32 N − N ( N + 1) (cid:18) · N − − C N +12 N +2 (cid:19) ++ 8( N − N ( N + 1) (cid:18) · N − − C N +12 N +2 (cid:19) . Let’s substitute in the first summand C N N +2 by the following expression: C N N +2 = 12 C N +12 N +3 − · C N +12 N +2 N + 2 . C N +12 N +2 cancel on.Transforming other summands we get the following formula: S ( N ) = 1210 N ( N + 1)( N + 2)(8 N + 13) C N +12 N +3 − N − N ( N + 1)(4 N + 5) . (37)Summing the expressions for S ( N ), S ( N ) and S ( N ) (formulas (37),(36) and (35)) we obtain the following: ε ( N + 1 ,
3) = S ( N ) + S ( N ) + S ( N ) == 1210 N ( N + 1)( N + 2)(8 N + 13) C N +12 N +3 = ˜ C ( N + 1) . In this section we prove recurrence and explicit formulas for numbers ε ( N, ε ( N,
2) is the number of marked maps on a torus (surface ofgenus 1), that contain N edges and 2 faces. Theorem 7.
For N ≥ the numbers ε ( N, satisfy the following recursion: ε ( N,
2) = 2 N − X i =0 (cid:16) ε ( i ) ε ( N − i − ,
2) + ε ( i, ε ( N − i − (cid:17) ++ N (2 N − ε ( N −
1) + ε ( N − , . (38) Proof.
Consider all marked maps on a torus, that contain
N > Two arcs of the deleted edge ˜ e belong to different faces. In this case we obtain a map on a torus with N − ε ( N −
1) such maps is obtained N (2 N −
1) times. Thus,case occurs for N (2 N − ε ( N −
1) initial maps. Two arcs of the deleted edge are successive arcs of the face number . In this case we obtain a marked map on a torus with N − ε ( N − ,
2) such maps occurs twice. Thus, case occurs for 2 ε ( N − ,
2) initial maps. Two arcs of the deleted edge are unsuccessive arcs of the face number .The obtained graph is connected.
30n this case we obtain a marked map on a sphere with N − ε ( N − ,
3) such maps occurs once. Thus, case occurs for ε ( N − ,
3) initial maps. Two arcs of the deleted edge are unsuccessive arcs of the face number .The obtained graph is disconnected. In this case we obtain an ordered pair of maps, that contain together N − The map of genus 0 has one face.
Let this map has i edges. Byremark 3 the other map contains at least 3 edges, consequently, 1 ≤ i ≤ N − ε ( i ) ε ( N − i − , occurs for 2 P N − i =1 ε ( i ) ε ( N − i − ,
2) initial maps.
The map of genus 0 has two faces.
Let this map has i edges. Byremark 3 the other map contains at least 2 edges, consequently, 1 ≤ i ≤ N − ε ( i, ε ( N − i − occurs for 2 P N − i =1 ε ( i, ε ( N − i −
1) initial maps.Summing the formulas for all these cases and taking into account that ε (0) = 1 and ε (0 ,
2) = ε (2 ,
2) = 0, we obtain the desired formula.
Theorem 8.
For N ≥ the numbers ε ( N, satisfy the following formula: ε ( N,
2) = 4 N − N ( N − N − N + 3)3 . (39) Proof.
Let F ( N ) = 4 N − N ( N − N − N +3)3 . We will prove the statement ε ( N,
2) = F ( N ) by induction.The base for N = 1 , ε (1 ,
2) = F (1) = ε (2 ,
2) = F (2) = 0. Let’s prove the induction step. Let for all k < N we have ε ( k,
2) = F ( k ). Then we will show that ε ( N,
2) = F ( N ).By formula (38) we have: ε ( N,
2) = 2 N − X i =0 ε ( i ) ε ( N − i − ,
2) + 2 N − X i =0 ε ( i, ε ( N − i − (cid:16) N (2 N − ε ( N −
1) + ε ( N − , (cid:17) = S ( N ) + S ( N ) + S ( N ) , where S ( N ), S ( N ), S ( N ) — are the first, the second and the third sum-mands of the obtained expression respectively.31t first let’s calculate S ( N ). By formulas (4) and (34) we have: S ( N ) = N (2 N − ε ( N −
1) + ε ( N − ,
3) == N (2 N − N − N − N − N −
1) + 5)( N − N − N C N − N − == (51 N − N − N − N − . (40)Now, taking into account ε (0 ,
2) = 0, let’s modify the sum S ( N ).Let j = N − − i . By formulas (26), (4) and formula (14) for k = 2 , S ( N ) = 2 N − X i =1 ε ( i, ε ( N − i −
1) = 2 N − X j =2 ε ( N − − j, ε ( j ) == 2 N − X j =2 ( N − − j )4 N − − j − (2 j )!12 j !( j − N − ( N − N − X j =2 (2 j )!4 j j !( j − − N − X j =3 (2 j )!4 j j !( j − ! == 2 N − (cid:18) ( N −
3) 2( N − N − N − N − · N − ( N − N − −− N − N − N − N − N − · N − ( N − N − (cid:19) == ( N − N − N − N − . (41)Now let’s modify S ( N ). By the induction assumption: ε ( N − − i,
2) == ( N − i − N − i − N − i − N − i −
10) 4 N − i − s i ( N ) 2 N − · i , where s i ( N ) = ( N − i − N − i − N − i − N − i −
10) == N ( N − N − N + 3) − ( N − N − N − i + 1)++ 3( N − N − i ( i + 1) − (52 N − i − i ( i + 1)++ 13( i − i − i ( i + 1) . ε ( N − − i,
2) = 0 for i = N −
3, we obtain: S ( N ) = 2 N − X i =0 ε ( i ) ε ( N − i − ,
2) = 2 N − N − X i =0 (2 i )!4 i i !( i + 1)! s i ( N ) . By the formula for s i ( N ), formula (13) and formula (14) for k = 0 , . . . , S ( N )2 N − = N ( N − N − N + 3) N − X i =0 (2 i )!4 i i !( i + 1)! −− ( N − N − N − N − X i =0 (2 i )!4 i i ! i ! ++ 3( N − N − N − X i =1 (2 i )!4 i i !( i − −− (52 N − N − X i =2 (2 i )!4 i i !( i − N − X i =3 (2 i )!4 i i !( i − N ( N − N − N + 3) (cid:18) − N − N − N − ( N − N − (cid:19) −− ( N − N − N −
30) 2( N − N − N − ( N − N − N − N −
36) 2( N − N − N − · N − ( N − N − −− (52 N − N − N − N − N − · N − ( N − N − N − N − N − N − N − · N − ( N − N − N ( N − N − N + 3) − N − N − N − · N − ( N − N − R ( N ) , where R ( N ) = 35 N ( N − N − N + 3)++ 35( N − N − N − N − −− N − N − N − N − N − N − N − N − −− · N − N − N − N −
6) == 4(52 N − N − N − N − . S ( N ) = 4 N − N ( N − N − N + 3) − (52 N − N − N − N − . (42)Summing the expressions for S ( N ), S ( N ) and S ( N ) (formulas (42),(41) and (40)), we obtain: ε ( N,
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