University of Twente The Netherlands

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1 Faculty of Mathematical Scieces t Uiversity of Twete The Netherlads P.O. Box AE Eschede The Netherlads Phoe: Fax: memo@math.utwete.l Memoradum No. 59 Hoede-sequeces F. Göbel November 00 ISSN

2 Hoede-sequeces F. Göbel Faculity of Mathematical Scieces Uiversity of Twete P.O. Box AE Eschede The Netherlads October 5, 00 Abstract I a attempt to prove the double-cycle-cojecture for cubic graphs, C. Hoede formulated the followig combiatorial problem. Give a partitio of {,,...,3} ito equal classes, is it possible to choose from each class a umber such that these umbers form a icreasig sequece of alteratig parity? Let a Hoede-sequece be defied as a icreasig sequece of atural umbers of alteratig parity. We determie the average umber of Hoede-sequeces w.r.t. arbitrary partitios, ad obtai bouds for the maximum ad miimum umber of Hoede-sequeces w.r.t. partitios ito equal classes. Key words: Icreasig sequeces, alteratig parity. AMS Subject Classificatios: 05A5, 05A6, 05A8. Itroductio ad summary A Hoede-sequece is a icreasig sequece of atural umbers of alteratig parity. Let P be a partitio of N,whereN = {,,...,}. A Hoede-sequece w.r.t. P is a Hoede-sequece for which a bijectio exists betwee its terms ad the parts of P such that the image of each term t cotais t.

3 Example. Let P be the followig partitio of N 8 : {, }, {3, 5, 8}, {4, 6, 7}. The,3,4 is a Hoede-sequece w.r.t. P as well as,4,5 ad three more. Cojecture. (Hoede []. Each partitio P of N 3 ito sets of cardiality 3 admits a Hoede-sequece w.r.t. P. The cojecture arose i a attempt to prove the double-cycle-cojecture for cubic graphs. To each partitio P of N, a partitio of the umber correspods, the parts of which are the cardialities of the parts P. We call this partitio of the iduced partitio of P. A partitio Q of a umber is admissible if each partitio P of N for which Q is the iduced partitio, admits a Hoede-sequece w.r.t. P ;otherwiseit is forbidde. Hoede s cojecture ca ow be reformulated as: the partitio 3 is admissible for all. Obviously, a admissible partitio cotiues to be admissible if parts are deleted. Coversely, a forbidde partitio cotiues to be forbidde if parts are supplied. So forbidde partitios are i fact forbidde subpartitios, aalogous to the cocept of forbidde subgraphs i graph theory. I Sectio ad 3 we preset some examples of forbidde ad admissible partitios, respectively. Sectio 4 is dedicated to a auxiliary result, viz. the umber of Hoede-sequeces of legth with terms i N m. This result is applied i Sectio 5 to determie the average umber of Hoede-sequeces w.r.t. arbitrary partitios, i particular partitios ito equal parts. I the fial Sectio 6 we cosider partitio ito equal parts for which the umber of Hoede-sequeces is maximum or miimum, respectively. We cojecture that the umber of Hoede-sequeces w.r.t. partitios of N 3 ito equal classes is at least the Fiboacci umber F +3. Forbidde partitios Propositio. The followig partitios are forbidde. (a k,, (k, (b k,,, (k, (c p, q,, (p, q,

4 (d p, q,r,, (p, q, r. Proof. (a Take {}, {, 3,...,k +}, {k +}. (b Take {}, {3}, {5}, {the rest}. (c Take {},P = {, 3,...,p +},Q = {p +, p +q +3},R = {p + 3,...,p +q +}. Now is forced, as well as a eve umber from P. Whether the third (odd umber is chose from Q or from R, cotiuatio is ot possible. (d Take A = {, p+}, B = {,...,p+}, C = {p+3,...,(p+q+}, D = {(p+q+3, (p+q+r+4}, E = {(p+q+4,...,(p+q+r+3}. ThefirsttwoumbersmustcomefromA B. Also,thefirstumber must be odd. The third umber must come from C ad it has to be odd. Now it is clear that a cotiuatio is ot possible. Example for case (d. Take p = q = r =. The A = {, 6}, B = {, 3, 4, 5}, C= {7, 8, 9, 0}, D= {, 6}, E= {, 3, 4, 5}. Propositio. Let k be a iteger. If P is a partitio of ito k + parts i which k parts occur with sum [ ],thep is forbidde. Proof. Let P be a partitio of N for which P of above is the iduced partitio, ad let V,...,V k be subsets i P with k V i [ ]. Now put all eve umbers i k V i. Example. Let k =, = 40. The the followig partitio of 40 is forbidde accordig to the above result:,9,6,5,5,4. 3 Admissible partitios Propositio 3. Each partitio of ito 3 parts, ot of the form k,,, is admissible. Proof. Let A, B, C be the three parts, ad let A. Letk be the largest umber for which,,...,k A, adletk + B. We distiguish two cases: k =adk. 3

5 k =.IfCcotais a odd umber we are through. So let all odd umbers (of N beia B. SiceC is ot empty, we ca assume that the largest umber i C is q. If ay of the umbers 3, 5,...,q isia, weare through. So assume 3, 5,...,q B. If C cotais a eve umber x < q, the,x,q is admissible. If ot, the C cotais oly the umber k. Hece all eve umbers 4, 6,...,q areia B (as well as all larger eve umbers if q +. If A cotais a eve umber q, we are through. So suppose these umbers too are i B. Up to ow we kow that A,, 3, 4,...,q B, q C. If there are o more umbers, we have a forbidde partitio. I the remaiig cases, q + A or q + B, ad a Hoede-sequece exists. k. It is coveiet to use the followig otatio: k 0 idicates a atural umber of the same parity as k; k a umber of the opposite parity. If k 0 C, we are through. So assume all k 0 N are i A B. Letk be the largest elemet i C. IfA cotais a k 0 with k<k 0 <k,theweare through: take k +, k 0, k. So suppose all these k 0 are i B. Iparticular k + B. Hece or, k +,k is a Hoede-sequece. 4 The umber of Hoede-sequeces Let H m, be the umber of Hoede-sequeces of legth with terms i N m. ( m+ ( m+ Propositio 4. If m + is eve, H m, = +,ad ( m+ if m + is odd, H m, =. Proof. Let m + be odd. Cosider a Hoede-sequece of which the first term is odd. If m is odd ad eve, the the last term is eve, so all terms ca be icreased by ad we obtai a Hoede-seuqce i N m with first term eve. Coversely, i a Hoedesequece with first term eve, all terms ca be decreased by, etc. Similarly, if m is eve ad is odd, the too, the umbers of Hoede-sequeces with eve ad odd first terms are equal. So it suffices to prove (i ( the case m + m+ odd that the umber of sequeces with first term odd is. Let (i,i,...,i be a combiatio of from N m+ (with i <i <...< i. The (i, i,...,i is a Hoede-sequece i N m (i fact 4

6 i N m with odd first term. It is clear that this map is a bijectio. This proves the formula for H m, whe m + is odd. Let m + be eve ad let (i,i,...,i be a combiatio of from N m+ with i <i <... < i. The (i, i,...,i is a Hoedesequece of legth from N m with odd first term. The combiatios of from N m+ correspod to the Hoede-sequeces with eve first term: (i, i,...,i +. I Table we give H m, for m Table Propositio 5. (a H m, = H m, +H m, (m,m. (b If m + is eve, the H m, = m ( m+. m + Proof. These results follow easily from Propositio 4. Propositio 6. The umber of Hoede-sequeces with terms i N m is give by m H m, = F m+3, where F m is the m-th Fiboacci-umber, defied = by F = F =,F m = F m + F m (m 3. Proof. Defie S m =+ m H m,. = 5

7 The S =+H, =3,S =+(+=5. Itissufficiettoprove S m = S m + S m for m 3. We prove this by iductio, usig the recursio of Propositio 5. m m S m + S m =4+ H m, + H m, = m 3 = m = = H m, + H m,m + H m, H m, + H m, = = m (H m, + H m, +H m,m + H m,m + H m,, = ad this reduces to S m. Remark. Both Table ad the formula for m H m, practically force us to defie H m,0 as. This is quite uusual: the umber of empty sets is, the empty product is, to metio two examples. It seems that the parity of the first term is iherited from the ature of the empty Hoede-sequece. = 5 The average umber of Hoede-sequeces w.r.t. partitios ito equal parts Let N k be partitioed ito classes of cardiality k. The umber of such partitios is A,k = (k!!(k!. From Propositios 4 ad 5 we kow that the total umber of differet Hoedesequeces that occur i these partitios is give by ( k+ H k, = ( k k+ k + 6 if k eve, odd, otherwise.

8 The umber of times a fixed Hoede-sequece occurs i the above partitios is B,k = (k!!((k!! So the avarage umber of Hoede-sequeces per partitio is H k,b,k A,k which reduces to H k,k. Substitutig the formulas for H ( k,k, we obtai the followig result. Propositio 7. If N k is partitioed ito classes of cardiality k, the the average umber Q,k of Hoede-sequeces w.r.t. a partitio is give by Q,k = ( k+ ( k ( k+ k k + (k+( k if k eve, odd, otherwise. Example Take =3,k =. The umber of partitios of N 6 ito equal classes of cardiality is A 3, = 5. The umber of Hoede-sequeces is H 6,3 = 8. See Table below. Obvioulsy, for a give Hoede-sequece there are 3! partitios i which it occurs. So i this simple case B 3, =6. The total umbers of ticks i Table is 8 6, so the average per partitio is 48/5 = 3.. 7

9 /34/56 /35/46 /36/45 3/4/56 3/5/46 3/6/45 4/3/56 4/5/36 4/6/35 5/3/46 5/4/36 5/6/34 6/3/45 6/4/35 6/5/34 Table Remark. The average ca be calculated also for arbitrary partitios of N m, followig the above method. Propositio 8. Q, Q,3 ( Proof. Apply Stirlig s formula. ( ( =s +, ( ( =s 4 (. 6 Extreme values Let M,k be the maximum umber of Hoede-sequeces over all partitios of N k ito classes of cardiality k. Propositio 9. If is eve, the ( k + M,k + ( k. 8

10 Proof. First we partitio N k ito classes of k successive umbers each: A = {,...,k}, A = {k +,...,4k}, etcetera. The take A i = B i C i where B i ad C i cosist of the odd ad the eve umbers, respectively, from A i. It is easily see that the umber of Hoede-sequeces is ( k+ or ( k, depedig o whether the first term of the sequece is odd or eve. Cojecture. If is eve, the M,k = ( k+ + ( k. We do ot have a upper boud for M,k which is substatially better tha the trivial k. Let m,k be the miimum umber of Hoede-sequeces over all partitios of N k ito classes of cardiality k. From Propositio, case d, we kow that m m, =0for 5. It is a easy puzzle to show that m, =for 4. This deals with the case k =. Hoede s cojecture is equivalet with m,3 > 0. Propositio 0. m,3 F +3. Proof. Trivially m,3 =3,m,3 = 5. For the geeral case, cosider the series of partitios suggested by the followig list. =: 3, =: 6/3 45, =3: 9/3 78/4 56, =4: /3 0/4 59/6 78, =5: 5/3 34/4 5/6 0/7 89. Fix. Cosider the Hoede-sequece t i (i =,..., w.r.t. to a partitio as above with t =ort =. From the partitio delete the set {,, 3} ad replace the other umbers t i N 3 by 3 t. This gives the partitio for ad we have a bijectio to Hoede-sequeces w.r.t. our partitio for. Now cosider a Hoede-sequece t i (i =,...,witht 3. Note that oly odd umbers for t are possible ad that t =3. From the partitio of N 3 delete the sets {,, 3} ad {3, 3, 3 }, ad replace the other umbers t i N 3 by t 3. This gives the partitio for. If the reduced 9

11 Hoede-sequece starts with 0, the delete 0 ad the fial term 3 3. If it starts with a umber greater tha 0, the delete the two fial terms (which must be 3 4ad3 3. Agai we have a bijectio ad the well-kow recursio for the Fiboacci-sequece holds. Cojecture 3. m,3 = F +3. Remark. Note that Cojuecture 3 is substatially stroger tha Hoede s Cojecture. The explicite form of our cojectured lower boud may provide a idicatio for a method of proof. Refereces [] C. Hoede, Private commuicatio. 0

Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is

Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of