Linear chord diagrams with long chords

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1 Liear chord diagrams with log chords Everett Sulliva Departmet of Mathematics Dartmouth College Haover New Hampshire, U.S.A. Submitted: Feb 7, 2017; Accepted: Oct 7, 2017; Published: Oct 20, 2017 Mathematics Subject Classificatios: 05C88, 05C89 Abstract A liear chord diagram of size is a partitio of the set {1, 2,..., 2} ito sets of size two, called chords. From a table showig the umber of liear chord diagrams of degree such that every chord has legth at least k, we observe that if we proceed far eough alog the diagoals, they are give by a geometric sequece. We prove that this holds for all diagoals, ad idetify whe the effect starts. 1 Itroductio A liear chord diagram is a matchig of {1, 2,..., 2}. Chord diagrams arise i may differet cotexts from the study of RNA [5] to kot theory [6]. I combiatorics chord diagrams show up i the méage problem [4], partitios [2], ad iterval orders [3]. I may of the situatios give above the objects beig paired lie o a circle ad so each pair is a chord. I this paper the focus will be o liear chord diagrams which ca be obtaied from a chord diagram by cuttig the circle at some poit. We will address diagrams where there is a specified miimum legth for each chord. From a table coutig the umber of such diagrams for, the size, ad k, the miimum legth, we observe that if we proceed far eough alog the diagoals, they are give by a geometric sequece. We prove that this holds for all diagoals, ad idetify whe the effect starts. 2 Statemet of Result A liear chord diagram of size is a partitio of the set {1, 2,..., 2} ito parts of size 2. We ca draw liear chord diagrams with arcs coectig the partitio blocks the electroic joural of combiatorics 24(4) (2017), #P4.20 1

2 Table 1: Coutig chord diagram with log chords M (1) M (2) M (3) M (4) M (5) M (6) M (7) M (8) M (9) M (10) M (11) The first four rows ca be foud i the OEIS uder the idetificatio umbers A001147, A000806, A190823, ad A190824, respectively. If c = {s c, e c } where s c < e c is a block of a liear chord diagram, we say that s c is the start poit of c ad e c is the ed poit. The legth of c is e c s c. We say that a chord c covers the iteger i if s c < i < e c. We say that a chord c covers a chord d if it covers s d ad e d. Defiitio 1 Let D deote the set of all liear chord diagrams with chords. Let M (k) deote the class of all liear chord diagrams such that every chord has legth at least k. Let M (k) deote the set of all liear chord diagrams with chords such that every chord has legth at least k. Table 1 shows the sizes of M (k) for various ad k. If k is fixed, M (k) usig o the order of 2 k 2 arithmetic operatios. For k = 1, M (1) chord diagrams, which is give by M (1) = (2)! (!2 ). ca be computed simply couts all liear For k = 2 ad k = 3, a = M (2) ad b = M (3) ca be computed usig liear recurreces: a = (2 1)a 1 + a 2 b = (2 + 2)b 1 (6 10)b 2 + (6 16)b 3 (2 8)b 4 b 5. The recurrece for M (2), ca be foud i [1]; the recurrece for M (3) is ew. Cojecturally, there is a liear recurrece for every sequece M (k) where k is fixed: We will address these matters elsewhere. the electroic joural of combiatorics 24(4) (2017), #P4.20 2

3 Here we address the diagoals of Table 1. The shaded squares highlight a patter. For each shaded square the umber i the square oe below ad oe to the right of it is exactly ( k + 1) times the umber i the curret square. This patter holds for all such squares: Theorem 2 Let ad k be positive itegers such that 3( k) ad k. The M (k+1) +1 = ( k + 1) M (k). 3 Outlie of the proof We cosider each diagoal separately. We refer to the i th diagoal as all the etries such that ( k +1) = i. For ay etry M (k) o the i th diagoal we create ( k +1) fuctios α,k,j (j {0,..., k}) which are ijective ito M (k+1) +1. We show that the images of these fuctios are disjoit ad cover M (k+1) +1. Ad so there are ( k + 1) times as may elemets i M (k+1) +1 as there are i M (k). To create the bijectio α,k,j we cosider the middle 2( k) idices. Here is a example from a elemet of M (4) 6 : Ay chords startig or edig i the middle idices are highlighted: A ew chord is iserted coverig oly the idices i the middle: the electroic joural of combiatorics 24(4) (2017), #P4.20 3

4 The ew chord the has its start poit iteratively swapped with the startig poits of the ubolded chords, startig with the oe that started last ad stoppig whe there are j uswapped ubolded chords: D α 6,4,2 (D) α 6,4,1 (D) α 6,4,0 (D) 4 Details of the proof Defiitio 3 Let C be a liear chord diagram. We defie L,k = {1, 2,..., k}, M,k = {k + 1, k + 3,..., 2 k}, ad R,k = {2 k + 1, 2 k + 1,..., 2}. Let C,k deote the set of all chords c C such that s c M,k or e c M,k, ad S C deote the set of all chords c C such that c / C,k. Lemma 4 Give ay liear chord diagram i M (k) such that 3( k) ad k, there is o chord c such that s c, e c M,k. Proof. If a chord has both its start poit ad ed poit iside M,k, the the largest legth it could have is whe it starts at k + 1 ad eds at 2 k. So the maximum legth ay such chord could have is 2 2k 1. But 3( k) which is equivalet to 3k 2. Thus the maximum legth ay such chord could have is 2 2k 1 3k 2k 1 = k 1. But every chord must have legth at least k. Thus there is o chord such that its idices of the start poit ad ed poit lie iside M,k Lemma 5 Give ay liear chord diagram i M (k) such that 3( k) ad k, C,k cotais exactly k chords that start i M,k ad k chords that ed i M,k. Proof. We first observe that o chord has its ed idex i L,k, sice it if did, its maximum legth would be k 1. Similarly, o chord has its start idex i R,k sice it if did, its the electroic joural of combiatorics 24(4) (2017), #P4.20 4

5 maximum legth would be 2 (2 k + 1) = k 1. Thus every idex i L,k is a start idex, ad every idex i R,k is a ed idex. We also observe that L,k = R,k. Cosider all chords i S C. Sice they either start or ed i M,k, they must start i L,k ad ed i R,k. Thus L,k S C chords start i L,k ad ed i M,k, ad R,k S C chords ed i R,k ad start i M,k. By Lemma 4, every chord i M either starts i L,k or eds i R,k. Thus M,k has the same umber of start idices as ed idices, ad that umber is k. Lemma 6 Give ay liear chord diagram C M (k+1) +1 such that 3( k) ad k, let a be the chord whose ed idex is 2 k + 2 (i.e. the smallest elemet i R +1,k+1 ). Let m be the umber of chords b S C such that s b < s a. The m < k + 1. Proof. Let M be the ordered set of all chords c C +1,k+1 such that e c M +1,k+1. We say k < c for k, c M if e k < e c. Observe that M is completely ordered. By Lemma 5, we have M = k. We may relabel the chords i M to be {c 1, c 2,..., c k }. Observe that by Lemma 5, e ci (k + 1) + ( k) + i = + i + 1. Sice l ci = e ci s ci k + 1 we have s ci + i + 1 (k + 1) = k + i. Let m i be the umber of chords a S C such that s a < s ci. The m 1 < k + 1. The largest umber of start idices to the left of s c2 is k + 1, but if it were that large, oe of them must be the start of c 1. Thus m 2 < k + 1. By iductio we have m i < k + 1 for all i. Now suppose m k + 1. The s ci < s a for all i sice otherwise there would exists a i such that m i k + 1. Thus s a ( k + 1) + ( k) + 1 = 2 2k + 2. Thus l a is bouded above by 2 k + 2 (2 2k + 2) = k < k + 1 which cotradicts that fact that every chord has legth at least k + 1. Thus m < k + 1. Defiitio 7 We defie α,k,i for i {0,..., k}, k, ad 3( k) to be a map from M (k) to D as follows. Give a diagram C, we isert a ew chord c with start poit right before M,k ad ed poit right after M,k to get diagram C. We the swap the start idex of the ew chord with the closest start idex of a chord i S C to its left. We cotiue to swap util there are i start idices of chords i S C to its left. Observe that sice 3( k), the umber of chords i S is at least (2 2k) 3( k) 2( k) = k. Thus every α exists ad is well defied. Example 1 Obtaiig C from C is show below: C C the electroic joural of combiatorics 24(4) (2017), #P4.20 5

6 Here is α 3,2,0 applied to a elemet of M (2) 3 : Here is α 4,3,1 applied to a elemet of M (3) 4 : I these diagrams, the thick lies are chords i C,k, the thi chords are i S C ad the greyed dashed chord is the ew iserted oe. Defiitio 8 We defie β,k for k, ad > 3( k) to be a map from M (k) to D 1 as follows. Give a diagram C, we deote c to be the chord with ed poit right after M,k. We the swap the start idex of the ew chord with the closest start idex of a chord i S C to its right. We cotiue to swap util there are o more start idices of chords i S C to its right. We the remove chord c. Lemma 9 α,k,i ( M (k) ) M (k+1) +1. Proof. We see that the result will have + 1 chords, so it suffices to show that every chord has legth at least k + 1. Cosider a chord c i C,k. If it has s c M,k ad e c R,k, its legth is icreased by 1, sice we iserted a idex betwee M,k ad R,k. Otherwise e c M,k ad s c L,k, i which case its legth is icreased by 1, sice we iserted a idex betwee M,k ad L,k. Sice the legth of such a chord had to be at least k to begi with, it must have at least legth k + 1 after applyig α,k,i. Cosider the chord we just iserted. It will cover all the idices i M,k, ad every time we swap, aother idex will be covered. Sice there are a total of chords before isertig, of which M,k cotais 2 2k of them, ad it swaps util there are i chords to its left i S, it swapped with at least (2 2k) i. Recall that the legth of the chord will be the umber of idices it covers plus 1. Thus its legth is at least 1 + (2 2k) + ( (2 2k) i) = 1 + i 1 + ( k) = k + 1, as desired. Now cosider a chord i S C. There are two cases, either it had its start idex swapped at some poit or it did t. If it did t, the it covers the ew chord c, ad has legth greater the c s legth. Thus the chord has legth at least k + 1 as desired. If it did swap, the either its startig idex icreased by 1 or more. the electroic joural of combiatorics 24(4) (2017), #P4.20 6

7 Suppose that its startig idex icreased by 1. The the umber of idices that lie i betwee its edpoits has icreased by 1. Whe we iserted c, it was icreased by 2, but the we moved the startig idex forward by 1, causig it to lose 1. Thus its legth icreased by exactly 1. Sice it must have had legth k to begi with it ow has legth at least k + 1. Suppose that its startig idex icreased by more the 1. Let a be its origial startig idex after isertig c ad b be its startig idex after isertig ad swappig c. The the idex b 1 is the startig idex of some poit i M +1,k+1, sice b a > 1 ad otherwise b would have occurred sooer. Thus the chord with startig idex b 1 has legth at least k + 1. Sice the edig idex of our chord lies i R which is at least 1 more the the edig idex of the chord at b 1, the legth of our chord after swappig is at least k + 1. Thus α,k,i ( M (k) Lemma 10 β,k ( ) M (k) M (k+1) +1 as desired. ) M (k 1) 1. Proof. We see that the result will have 1 chords, so it suffices to show that every chord has legth at least k 1. Cosider a chord r i C,k. If it has s r M,k ad e r R,k, its legth is decreased by 1, sice we removed the first idex i R,k. Otherwise e r M ad s r L, i which case its legth is decreased by 1, sice we removed the last idex from L,k. Sice the legth of such a chord had to be at least k to begi with, it must have at least legth k 1 after applyig β,k. Now cosider a chord r i S C. We break ito two cases: Case 1: s r was swapped with s c at some poit. The s r has decreased by at least 1, which meas that l r icreased by at least 1. But whe we remove s c at the ed, l r is deceased by 2. Thus l r ever deceases by more the 1. Sice l r k, the legth of r must be at least legth k 1 after applyig β,k. Case 2: s r did ot swap with s c at some poit. The s r < s c, which meas that l r is at least 2 + l c = k + 2 sice l c has legth at least k. Whe we remove s c a the ed, l r is deceased by 2. Thus l r ever deceases by more tha 2. Sice l r k + 2, the legth of r must be at least ( legth ) k after applyig β,k. Thus β,k M (k) M (k 1) 1 as desired. Proof (of theorem 2). We shall proceed by costructig ( k + 1) ijective fuctio from M (k) to M (k+1) +1 such that their images partitio M (k+1) +1. Let C M (k). Let E,k,i be the set of all liear chord diagrams i M (k) such that the chord c with e s = 2 k + 1 (i.e. the first idex after M,k ) has i start poits of chords i S C to its left. The by lemma 6 the collectio {E +1,k+1,0,..., E +1,k+1, k 1 } partitios M (k+1) +1. By costructio we see that Im(α,k,i ) E +1,k+1,i. We also see that both β +1,k+1 E+1,k+1,i α,k,i ad the electroic joural of combiatorics 24(4) (2017), #P4.20 7

8 α,k,i β +1,k+1 E+1,k+1,i are the idetity map. Thus there is a bijectio betwee M (k) ad E,k,i for every i. Thus as desired. k M (k+1) ( ) +1 = α,k,i M (k) = ( k + 1) M (k), i=0 Refereces [1] M. Hazewikel ad V. Kalashikov. Coutig iterlacig pairs o the circle. Departmet of Aalysis, Algebra ad Geometry: Report AM. Stichtig Mathematisch Cetrum, [2] W. N. Hsieh. Proportios of irreducible diagrams. Studies i Applied Mathematics, pages , [3] E. S. J. Justicz ad P. Wikler. Radom itervals. America Mathematical Mothly, 97(10): , [4] E. Lucas. Théorie des ombres. Gauthier-Villars, Paris, [5] C. M. Reidys. Combiatorial Computatioal Biology of RNA. Spriger-Verlag, New York, [6] S. D. S. Chmutov ad J. Mostovoy. Itroductio to Vassiliev Kot Ivariats. Uiversity Press, Cambridge, the electroic joural of combiatorics 24(4) (2017), #P4.20 8