ENUMERATING (2+2)-FREE POSETS BY THE NUMBER OF MINIMAL ELEMENTS AND OTHER STATISTICS

Transcription

1 ENUMERATING 2+2)-FREE POSETS BY THE NUMBER OF MINIMAL ELEMENTS AND OTHER STATISTICS SERGEY KITAEV AND JEFFREY REMMEL Abstract. A poset s sad to be 2 + 2)-free f t does not contan an nduced subposet that s somorphc to 2 + 2, the unon of two dsjont 2-element chans. In a recent paper, Bousquet-Mélou et al. found the generatng functon for the number of 2 + 2)-free posets: P t) = 1 1 t) ). n 0 We extend ths result by fndng the generatng functon for 2 + 2)-free posets when four statstcs are taken nto account, one of whch s the number of mnmal elements n a poset. We also show that n a specal case when only mnmal elements are of nterest, our rather nvolved generatng functon can be rewrtten n the form P t, z) = n,k 0 p n,k t n z k = 1 + n 0 zt 1 zt) n t) ) where p n,k equals the number of 2 + 2)-free posets of sze n wth k mnmal elements. An alternatve way to wrte the last generatng functon s P t, z) = 1 1 t) 1 1 zt)) n 0 whch was conjectured by us and proved recently by several authors.

2 2 SERGEY KITAEV AND JEFFREY REMMEL 1. Introducton Defnton. A poset s 2 + 2)-free f t does not contan an nduced subposet that s somorphc to 2 + 2, the unon of two dsjont 2-element chans. P s the set of all 2 + 2)-free posets. Fshburn [7] showed that a poset s 2 + 2)-free precsely when t s somorphc to an nterval order. Bousquet-Mélou et al. [1] showed that the generatng functon for the number p n of 2 + 2)-free posets on n elements s P t) = n 0 p n t n = n 0 ) 1 1 t). Defnton. The number of ascents of an nteger sequence x 1,..., x ) s ascx 1,..., x ) = { 1 j < : x j < x j+1 }. Defnton. A sequence x 1,..., x n ) N n s an ascent sequence of length n f x 1 = 0 and x [0, 1+ascx 1,..., x 1 )] for all 2 n. A s the set of all ascent sequences. Example. 0, 1, 0, 2, 3, 1, 0, 0) s an ascent sequence, whereas 0, 0, 2, 1, 3, 6) s not. Bousquet-Mélou et al. [1] gave a bjecton between 2 + 2)-free posets and ascent sequences, whch play the key role n the recent study of 2 + 2)-free posets. Our man result s an explct form of the generatng functon Gt, u, v, z, x) = p P t szep) u levelsp) v mnmaxp) z mnp) x ldsp) = t lengthw) u ascw) v lastw) z zerosw) x runw) w A where sze = number of elements, levels = number of levels, mnmax = level of mnmum maxmal element, mn = number of mnmal elements, and lds = sze of non-trval last down-set, length = the number of elements n the sequence, last = the rghtmost element of the sequence, zeros = the number of 0 s n the sequence, run = the number of elements n the leftmost run of 0 s = the number of 0 s to the left of the leftmost non-zero element.

3 ENUMERATING 2+2)-FREE POSETS BY THE NUMBER OF MINIMAL ELEMENTS 3 2. Man result For r 1, let G r := G r t, u, v, z) denote the coeffcent of x r n Gt, u, v, z, x). Thus G r t, u, v, z) s the g.f. of those ascent sequences that begn wth r 0 s followed by 1. Clearly, snce the sequence has no ascents and no ntal run of 0 s by defnton), we have that the generatng functon for such sequences s 1 + tz + tz) 2 + = 1 1 tz where 1 corresponds to the empty word. Thus, G = 1 1 tz + G r x r. r 1 For k 1, we let δ k = u 1 t) k u 1) and we set δ 0 = γ 0 = δ 0 = γ 0 = 1. γ k = u 1 zt)1 t) k 1 u 1) δ k = δ k u=uv = uv 1 t) k uv 1) γ k = γ k u=uv = uv 1 zt)1 t) k 1 uv 1) Theorem 1. For all r 1, G r t, u, v, z) = tr+1 z r u vv 1) + t1 u)zv 1) v) vδ 1 1 s 0 +uv 3 t1 uv) ) uv) s 1 t) s δ s+1 s 0 s δs+1 γ. u s 1 t) s δ s δ s+1 s+1 γ Our man result s the followng theorem. Theorem 2. Gt, u, v, z, x) = + t1 u)zv 1) v) s tz) + t 2 zxu vv 1) 1 tzx)vδ 1 1) u s 1 t) s δ s δ s+1 s+1 γ + uv 3 t1 uv) s 0 ) uv) s 1 t) s δ s+1 s δs+1 γ.

4 4 SERGEY KITAEV AND JEFFREY REMMEL One can use, e.g., Mathematca to compute Gt, u, v, z, x) = 1 + zt + uvxz + z 2) t 2 + uvxz + u 2 v 2 xz + uxz 2 + uvx 2 z 2 + z 3) t 3 + uvxz + u 2 vxz + 2u 2 v 2 xz + u 3 v 3 xz + uxz 2 + u 2 xz 2 + u 2 vxz 2 + +u 2 v 2 xz 2 + uvx 2 z 2 + u 2 v 2 x 2 z 2 + uxz 3 + ux 2 z 3 + uvx 3 z 3 + z 4) t 4 uvxz + 3u 2 vxz + u 3 vxz + 3u 2 v 2 xz + 2u 3 v 2 xz + 4u 3 v 3 xz + u 4 v 4 xz +uxz 2 + 3u 2 xz 2 + u 3 xz 2 + 3u 2 vxz 2 + u 3 vxz 2 + 2u 2 v 2 xz 2 + 2u 3 v 2 xz 2 + 3u 3 v 3 xz 2 +uvx 2 z 2 + u 2 vx 2 z 2 + 2u 2 v 2 x 2 z 2 + u 3 v 3 x 2 z 2 + uxz 3 + 3u 2 xz 3 + u 2 vxz 3 + u 2 v 2 xz 3 +ux 2 z 3 + u 2 x 2 z 3 + u 2 vx 2 z 3 + u 2 v 2 x 2 z 3 + uvx 3 z 3 + u 2 v 2 x 3 z 3 + uxz 4 +ux 2 z 4 + ux 3 z 4 + uvx 4 z 4 + z 5) t For nstance, the 3 sequences correspondng to the term 3u 2 v 2 xzt 5 are 01112, and Countng 2 + 2)-free posets by sze and number of mnmal elements For n 1, let H a,b,l,n denote the number of ascent sequences of length n wth a ascents and b zeros whch have last letter l. Then we frst wsh to compute Hu, z, v, t) = H a,b,l,n u a z b v l t n. Theorem 3. Hu, 1, z, t) = n 1,a,b,l 0 zt1 u)u s 1 t) s δ s+1 s γ. s=0 We would lke to set u = 1 n the power seres above, but the factor 1 u) n the seres does not allow us to do that n ths form. Thus our next step s to rewrte the seres n a form where t s obvous that we can set u = 1 n the seres: Hu, 1, z, t) = zt1 u) γ 1 + n 0 n ) n 1) n m 1 u 1) n m zt m 1 zt) m+1 m=0 u 1)m+1 1 zt) m+1 γ 1 m m u 1) j 1 zt) j u m j 1 1 t) ). j=0 =j+1

5 ENUMERATING 2+2)-FREE POSETS BY THE NUMBER OF MINIMAL ELEMENTS 5 There s no problem n settng u = 1 n ths expresson to obtan that H1, 1, z, t) = n 0 zt 1 zt) n t) ). Our defntons ensure that 1 + H1, 1, z, t) = P t, z) and we have the followng: Theorem 4. P t, z) = n,k 0 p n,k t n z k = 1 + n 0 We have used Mathematca to compute zt 1 zt) n t) ). P t, z) = 1 + zt + z + z 2) t 2 + 2z + 2z 2 + z 3) t 3 + 5z + 6z 2 + 3z 3 + z 4) t z + 21z z 3 + 4z 4 + z 5) t z + 84z z z 4 + 5z 5 + z 6) t Dervaton of the orgnal enumeratve result by Bousquet-Mélou et al. [1]. Concludng remark. P t) = n 0 = n 0 p n t n = 1 t P t, z) z z=0 1 1 t) ). Results n [1, 2, 3] show that 2 + 2)-free posets of sze n wth k mnmal elements are n bjecton wth the followng objects whch thus are also enumerated by Theorem 4): ascent sequences of length n wth k zeros; permutatons of length n avodng certan pattern whose leftmost-decreasng run s of sze k; regular lnearzed chord dagrams on 2n ponts wth ntal run of openers of sze k; upper trangular matrces whose non-negatve nteger entres sum up to n, each row and column contans a non-zero element, and the sum of entres n the frst row s k.

6 6 SERGEY KITAEV AND JEFFREY REMMEL References [1] M. Bousquet-Mélou, A. Claesson, M. Dukes, S. Ktaev: Unlabeled 2+2)-free posets, ascent sequences and pattern avodng permutatons. J. Combn. Theory Ser. A, to appear. [2] A. Claesson, M. Dukes, and S. Ktaev, A drect encodng of Stomenow s matchngs as ascent sequences, preprnt. [3] M. Dukes and R. Parvanen, Ascent sequences and upper trangular matrces contanng nonnegatve ntegers, Elect. J. Combn. 171) 2010), #R53 16pp). [4] M. H. El-Zahar, Enumeraton of ordered sets, n: I. Rval Ed.), Algorthms and Order, Kluwer Academc Publshers, Dordrecht, 1989, [5] P. C. Fshburn, Interval Graphs and Interval Orders, Wley, New York, [6] P. C. Fshburn, Intranstve ndfference n preference theory: a survey, Oper. Res ) [7] P. C. Fshburn, Intranstve ndfference wth unequal ndfference ntervals, J. Math. Psych ) [8] P. E. Haxell, J. J. McDonald, and S. K. Thomasson, Countng nterval orders, Order ) [9] S. M. Khams, Heght countng of unlabeled nterval and N-free posets, Dscrete Math ) [10] A. Stomenow, Enumeraton of chord dagrams and an upper bound for Vasslev nvarants, J. Knot Theory Ramfcatons 7 no ) [11] J. Wmp and D. Zelberger, Resurrectng the asymptotcs of lnear recurrences, J. Math. Anal. Appl., 111 no ) [12] D. Zager, Vasslev nvarants and a strange dentty related to the Dedekng eta-functon, Topology, )