Combinatorial Identities for Incomplete Tribonacci Polynomials

Transcription

1 Avalable at Appl. Appl. Math. ISSN: Vol. 10, Issue 1 (June 2015, pp Applcatons and Appled Mathematcs: An Internatonal Journal (AAM Combnatoral Identtes for Incomplete Trbonacc Polynomals Mark Shattuck Department of Mathematcs, Unversty of Tennessee, Knoxvlle, TN shattuck@math.utk.edu Receved: September 28, 2014; Accepted: February 13, 2015 Abstract The ncomplete trbonacc polynomals, denoted by T n (s (x, generalze the usual trbonacc polynomals T n (x and have been shown to satsfy several algebrac denttes. In ths paper, we provde a combnatoral nterpretaton for T n (s (x n terms of weghted lnear tlngs nvolvng three types of tles. Ths allows one not only to supply combnatoral proofs of earler denttes for T n (s (x but also to derve new ones. In the fnal secton, we provde a formula for the ordnary generatng functon of the sequence T n (s (x for a fxed s, as prevously requested. Our dervaton s combnatoral n nature and makes use of an dentty relatng T n (s (x to T n (x. Keywords: Combnatoral proof; ncomplete trbonacc polynomals; trbonacc numbers MSC 2010 No.: 05A19; 05A15 1. Introducton The trbonacc numbers t n are defned by the recurrence relaton t n = t n 1 + t n 2 + t n 3 f n 3, wth ntal values t 0 = 0 and t 1 = t 2 = 1. See sequence A n (Sloane, The trbonacc numbers are gven equvalently by the explct formula n 2 t n+1 = B(n,, n 0, (1 40

2 AAM: Intern. J., Vol. 10, Issue 1 (June ( n j where B(n, = ( j, as shown n (Barry, The number B(n, s the n-th row, -th column entry of the trbonacc trangle, see (Allad and V.E. Hoggatt, Note that B(n, = D(n,, where D(a, b denotes the Delannoy number sequence, and the reader s referred to A n (Sloane, The trbonacc polynomals T n (x were ntroduced n (V.E. Hoggatt and Bcknell, 1973 and are defned by the recurrence T n (x = x 2 T n 1 (x+xt n 2 (x+t n 3 (x f n 3, wth ntal values T 0 (x = 0, T 1 (x = 1, and T 2 (x = x 2. In analogy to (1, the trbonacc polynomals are gven by the followng explct formula (Ramírez and Srvent, 2014: T n+1 (x = n 2 ( j ( n j x 2n 3(+j. (2 The ncomplete trbonacc polynomals T n (s (x were consdered n (Ramírez and Srvent, 2014 and are defned as s ( n j T (s n+1(x = ( j x 2n 3(+j, 0 s n 2. (3 Note that the ncomplete trbonacc polynomals generalze the ordnary ones and reduce to them when s = n (s 2. The ncomplete trbonacc number, denoted by t n, s defned as the value of T n (s (x at x = 1. Incomplete Fbonacc numbers and polynomals have also been consdered and are defned n a comparable fashon; see, e.g., (Flppon, 1996 and (Ramírez, 2013b. Some combnatoral denttes for the ncomplete Fbonacc numbers were gven n (Belbachr and Belkhr, 2014 and a b-perodc generalzaton was studed n (Ramírez, 2013a. In (Ramírez and Srvent, 2014, several denttes were found for the ncomplete trbonacc numbers and polynomals usng varous algebrac methods. In ths paper, we supply combnatoral proofs of these denttes usng a weghted tlng nterpretaton of T n (s (x (descrbed n Theorem 0.1 below. In some cases, a further generalzaton of an dentty can be gven. In addton, usng our nterpretaton, one also can fnd other relatons not gven n (Ramírez and Srvent, 2014 that are satsfed by T n (s (x. In the fnal secton, we provde an explct formula for the generatng functon of T n (s (x, as requested n (Ramírez and Srvent, Our dervaton s combnatoral n nature and makes use of some denttes nvolvng T n (x. 2. Combnatoral nterpretaton for T (s n (x We wll use the followng termnology. By a square, domno, or tromno, we wll mean, respectvely, a 1 1, 2 1, or 3 1 rectangular tle. A (lnear tlng of length n s a coverng of the numbers 1, 2,..., n wrtten n a row by squares, domnos, and tromnos, where tles of the same knd are ndstngushable. Note that such tlngs may be dentfed as compostons of n wth parts of sze 1, 2, or 3; see, e.g., (Heubach and Mansour, Let T n denote the set of all tlngs of length n. It s well known that T n has cardnalty t n+1 ; see, e.g., (Benjamn and Qunn, 2003, p. 36. We wll often represent squares, domnos, and tromnos by the letters r, d,

3 42 M. Shattuck and t, respectvely. Thus, a member of T n may be regarded as a word n the alphabet {r, d, t} n whch there are n 2 3j,, and j occurrences of the letters r, d, and t, respectvely, for some and j. By a longer pece wthn a member of T n, we wll mean one that s ether a domno or a tromno. Gven 0 s n (s 2, let T n denote the subset of T n whose members contan at most s longer peces. For example, f n = 5 and s = 1, then T (1 5 = {r 5, dr 3, rdr 2, r 2 dr, r 3 d, tr 2, rtr, r 2 t}. Note that T n (s s all of T n when s = n 2. By a square-and-domno tlng, we wll mean a member of T n that contans no tromnos. Gven π T n (s, let δ(π and ν(π record the number of squares and domnos, respectvely, n π. We now provde a combnatoral nterpretaton of the polynomal T (s n+1 (x n terms of lnear tlngs. Theorem 0.1: The polynomal T (s (s n+1 (x s the dstrbuton for the statstc 2δ + µ on T n. Proof: Frst note that T n+1(x (s may be wrtten as s T n+1(x (s = B(n, (x, (4 where B(n, (x = ( ( n j j x 2n 3j. We next observe that when x = 1, the polynomal B(n, (x gves the cardnalty of the set B n, consstng of square-and-domno tlngs of length n n whch the squares come n two colors, black and whte, and contanng domnos and whte squares combned. To see ths, note that members of B n, contanng exactly j domnos are n one-to-one correspondence wth words n the alphabet {D, W, B} contanng j D s, j W s, and n j B s and thus have cardnalty ( n j (n j! = j, j, n j j!( j!(n j! = Summng over j gves B n, = ( j ( n j. ( n j (. j Gven π B n,, let δ 1 (π and δ 2 (π record the number of black and whte squares, respectvely. Thus, f π B n, has j domnos, then 2δ 1 (π + δ 2 (π = 2(n j + j = 2n 3j. Consderng all j, ths mples B(n, (x s the dstrbuton polynomal on B n, for the statstc 2δ 1 (π + δ 2 (π. Suppose now λ B n, s gven and contans j domnos for some j, where 0 s. We replace each domno of λ wth a tromno and each whte square wth a domno. The resultng tlng λ belongs to T n (s and has j tromnos, j domnos, and n 2 j squares. Thus we have 2δ(λ + ν(λ = 2δ 1 (λ + δ 2 (λ

4 AAM: Intern. J., Vol. 10, Issue 1 (June for all λ B n,. By (4, t follows that T n+1(x (s s the dstrbuton on s B n, for 2δ 1 + δ 2, equvalently, for the dstrbuton of 2δ + ν on T (s Remark: Takng x = 1 n the pror theorem shows that the cardnalty of T (s n s = n 2 shows that Tn+1 (x s the dstrbuton polynomal for 2δ + µ on all of T n. n. s t (s n+1. Takng Usng our nterpretaton for T (s n (x, one obtans the followng recurrence formula from (Ramírez and Srvent, 2014 as a corollary. Corollary 0.2: If n 2s + 1, then T (s n+3(x = x 2 T (s n+2(x + xt (s n+1(x + T (s n (x (xb(n s, s(x + B(n 1 s, s(x. (5 Proof: We wll show that the rght-hand sde of (5 gves the weghted sum of all the members of T (s n+2 wth respect to the statstc 2δ +ν by consderng the fnal pece. The frst term clearly accounts for all tlngs endng n a square. On the other hand, f a member of T (s n+2 ends n a longer pece, then there can be at most s 1 addtonal longer peces. From the proof of Theorem 0.1 above, we have for each m that B(m s, s(x gves the weght of all members of T m (s contanng exactly s longer peces. Note that addton of a longer pece to the end of a tlng already contanng s longer peces s not allowed. Thus, by subtracton, the total weght of all members of T (s n+2 endng n a domno s gven by x(t n+1(x (s B(n s, s(x and the weght of those endng n a tromno by T n (s (x B(n 1 s, s(x, whch completes the proof. 3. Some denttes of T (s n (x In ths secton, we frst generalze some prevous denttes for T n (s, whch were shown by algebrac methods, usng our combnatoral nterpretaton for T n (s (x. We also consder some further denttes for T n (s (x that can be obtaned usng Theorem 0.1. In ths secton and the next, by the weght of a subset S of T n or T n (s, we wll mean the sum λ S x2δ(λ+ν(λ. The x = 1 case of the followng dentty was shown n (Ramírez and Srvent, 2014 algebracally usng nducton. Identty 0.3: If h 1 and n 2s + 2, then h 1 x 2(h 1 T (s n+ (x = x 3 ( T (s+1 n+h+2 (x x2h T (s+1 Proof: We show, equvalently, T (s+1 n+2 (x + x 2h+1 T n (s (s (x xt n+h (x. (6 n+h+2 (x = (1 + h 1 x3 x 2(h 1 T (s n+ (x + x2h T (s+1 n+2 (x + xt (s n+h (x x2h+1 T n (s (x. For ths, we ll argue that the rght-hand sde gves the total weght of all the members of T (s+1 n+h+1. Frst note that x 2h T (s+1 n+2 (x gves the weght of the members of T (s+1 n+h+1 n whch postons n+2 through n+h+1 are covered by squares (.e., the rght-most longer pece ends at poston n+1 or before. On the other hand, the weght of all members of T (s+1 n+h+1 whose rght-most longer pece

5 44 M. Shattuck starts at poston n + 1 for some 0 h 1 s gven by ( x 2(h +1 + x 2(h 1 T (s n+ (x snce such tlngs λ are of the form λ = λ dr h or λ = λ tr h 1 for some tlng λ of length n + 1, where r m denotes a sequence of m squares. Note that λ T (s n+ 1 snce the number of longer peces n λ s lmted to s. Summng over 0 h 1 gves the ndexed sum on the rght-hand sde. Next, the term xt (s (s+1 n+h (x accounts for all members of T n+h+1 whose fnal pece s a domno whch were mssed n the sum. Fnally, members of T (s+1 n+h+1 of the form λ dr h, where λ has length n 1, were accounted for by both the x 2h T (s+1 n+2 (x term and by the = 0 term of the ndexed sum; hence, we must subtract ther weght, x 2h+1 T n (s (x, to correct for ths double count. Combnng all of the cases above completes the proof. The followng dentty from (Ramírez and Srvent, 2014 gves a formula for the sum of all the ncomplete trbonacc polynomals of a fxed order. Identty 0.4: If n 1, then l s=0 where l = n 2. T (s n+1(x = (l + 1T n+1 (x l ( ( n j x 2n 3(+j, (7 j Proof: Let λ T n and suppose that t contans exactly k longer peces, where 0 k l. Then the weght of λ s counted by each summand of l s=0 T n+1(x (s such that s k. That s, the tlng λ s counted l + 1 k tmes by ths sum. The proof of Theorem 0.1 above shows that the total weght of all members of T n contanng exactly k longer peces s gven by k ( ( k n k j x 2n 3(k+j, j k upon consderng the number j of domnos. Thus, the only nner sum n the double sum on the rght-hand sde of (7 n whch λ s counted occurs when = k and here t s counted k tmes (due to the extra factor of = k. Snce λ s clearly counted l + 1 tmes by the term (l + 1T n+1 (x, we have by subtracton that λ s counted l + 1 k tmes by the rght-hand sde of (7 as well. Snce λ was arbtrary, the dentty follows. The x = 1 case of the next dentty was conjectured n (Ramírez and Srvent, 2014 and follows from the generatng functon proof gven n (Klç and Prodnger, Identty 0.5: If n 1, then where l = n 2. l s=0 T (s n+1 (x = (l + 1T n 1 n+1(x (xt j (x + T j 1 (xt n j (x, (8 Proof: Suppose λ T n has exactly k longer peces. By the proof of the precedng dentty, we need only show that the weght of λ s counted k tmes by the sum on the rght-hand sde of (8. Note that xt j (xt n j (x gves the weght of all members of T n n whch a domno covers postons j and j +1, whle T j 1 (xt n j (x gves the weght of those n whch a tromno covers j=1

6 AAM: Intern. J., Vol. 10, Issue 1 (June postons j 1, j, and j + 1. Thus, for each longer pece of λ, there s a term n the sum that counts the weght of λ, whch mples that λ s counted k tmes by the sum, as desred. Remark: Comparng the x = 1 cases of the precedng two denttes, t follows that a n z n = n 1 z 2 + z 3 (1 z z 2 z 3 2, where a n = n 2 ( ( n j, j whch can also be shown drectly usng the methods of (Wlf, 2005, Secton 4.3, as was done n (Klç and Prodnger, The followng three denttes do not occur n (Ramírez and Srvent, 2014 but are a consequence of the combnatoral nterpretaton of T (s n (x gven n Theorem 0.1. Identty 0.6: If n 2s + 1, then T (s s n+1 (x = (x +2 T (s n 2 (x + x T (s 1 n 2 2 (x. (9 Proof: Suppose a member of T n (s ends n exactly domnos, where 0 s. If the rght-most pece that s not a domno s a square, then the tles comng to the left of ths square consttute a member of T (s (s n 2 1 and thus the weght of the correspondng subset of T n s x +2 T (s n 2 (x. On the other hand, f the rght-most non-domno pece s a tromno, then the tles to the left of ths tromno form a member of T (s 1 n 2 3 and thus the weght of the correspondng subset s x T (s 1 n 2 2 (x. Consderng all possble gves (9. Our next formula relates the ncomplete trbonacc polynomals to the trnomal coeffcents. Identty 0.7: If n 3s + 1, then T (s n (x = s s ( s, j, s j x 2s 2j T (s j n s 2j(x. (10 Proof: Suppose that there are domnos and j tromnos among the fnal s tles wthn a member of T (s ( s n 1, where n 3s + 1. Then there are,j,s j ways to arrange these tles, whch contrbute x 2(s j+ towards the weght, wth the remanng tles formng a member of T n s 2j 1. (s j Consderng all possble and j gves (10. The ncomplete Fbonacc polynomals ntroduced n (Ramírez, 2013a are gven as s ( n r 1 n 1 F n (s (x = x n 2r 1, 0 s. r 2 r=0 Our next dentty relates the ncomplete Fbonacc and trbonacc polynomals.

7 46 M. Shattuck Identty 0.8: If n 2s, then by n 2 T n+1(x (s = x n/2 F n+1(x (s 3/2 + =1 x ( 1/2 s 1 (T (j n 1 Proof: Frst note that the weght of all members of T (s n s ( n r x 2n 3r = x n/2 F (s n+1 r (x3/2. r=0 (j 1 (s j 1 (x T n 1 (xf (x 3/2. (11 that contan no tromnos s gven So assume a member of T n (s contans at least one tromno and that the left-most tromno covers postons through +2. Suppose further that there are exactly r domnos to the left of the leftmost tromno. Then the weght of all such members of T n (s s gven by ( r 1 r x 2 3r 2 T (s r 1 n 1 (x. Summng over the possble and r mples that the total weght of all the members of T n (s contanng at least one tromno s n 2 s 1 ( r 1 x 2 3r 2 T (s r 1 n 1 (x. r =1 r=0 To obtan the expresson n (11, we wrte T (s r 1 n 1 0. We then obtan a total weght formula of n 2 s 1 ( r 1 whch gves (11. =1 r=0 =1 r x 2 3r 2 s r 1 as s r 1 (T (j (j 1 ( 1 n 1 T n 1, where T n 1 = n 2 s 1 s j 1 = (T (j n 1 T (j 1 n 1 n 2 s 1 = (T (j n 1 T (j 1 =1 r=0 (T (j n 1 T (j 1 n 1 ( r 1 n 1x ( 1/2 F (s j 1 r x 2 3r 2 (x 3/2, 4. Generatng functon formula for T (s n (x The generatng functon formula for the ncomplete trbonacc numbers was found n (Ramírez and Srvent, 2014 and a formula was requested for the correspondng polynomals. The next result provdes such a formula. We remark that our method s more combnatoral than that used n (Ramírez and Srvent, 2014 n the case x = 1 and thus supples an alternate proof n that case. Theorem 0.9: Let Q s (z be the generatng functon for the ncomplete trbonacc polynomals T n (s (x, where n 2s + 1. Then Q s (z z = T ( 2s+1(x + (T 2s 1(x + xt 2s(xz + T 2s(xz 2 z 2 x+z s+1 1 x 2 z. (12 2s+1 1 x 2 z xz 2 z 3

8 AAM: Intern. J., Vol. 10, Issue 1 (June Proof: Let r n = r n (x be gven by s ( ( s n + s j 2 r n = x 2n+s 3j 3 + j s wth r 0 = r 1 = 0 and r 2 = x s+1. s ( s j ( n + s j 3 s x 2n+s 3j 6, n 3, We clam that r (x gves the total weght wth respect to the statstc 2δ +ν of all the members of T +2s contanng exactly s+1 longer peces and endng n a longer pece, the subset of whch we wll denote by A. To show ths, frst note that r (x evaluated at x = 1 s seen to gve the number of square-and-domno tlngs of length + s 2 or + s 3 n whch squares are black or whte and havng exactly s whte squares and domnos combned. We then ncrease the length of each whte square and each domno by one and add a domno to the end f the orgnal tlng had length + s 2 and add a tromno to the end f t had length + s 3. Ths yelds all members of A n a bjectve manner and thus mples r (x at x = 1 gves the cardnalty of A. Note that members of A endng n a domno contan j 2 squares, s j + 1 domnos, and j tromnos for some 0 j s, whle members of A endng s a tromno contan j 3 squares, s j domnos, and j + 1 tromnos for some j. Summng over j then mples that r (x s the dstrbuton for the statstc 2δ + ν on A, as clamed. By the nterpretaton for r (x just descrbed, the product r (xt n 2s (x gves the total weght of all members of T n 1 contanng at least s+1 longer peces n whch the (s+1-st longer pece ends at poston +2s snce the fnal n 2s 1 postons of such a member of T n 1 may be covered by any tlng. Summng over all possble then gves the total weght of all members of T n 1 contanng strctly more than s longer peces. Subtractng from T n (x thus gves the weght of all members of T n 1 contanng at most s longer peces and mples the followng dentty: n 2s 1 T n (s (x = T n (x r (xt n 2s (x, n 2s + 1. (13 In order to fnd a closed form expresson for Q s (z usng (13, we express T n = T n (x as follows: T n = T n 2s T 2s+1 + T n 2s 1 (T 2s 1 + xt 2s + T n 2s 2 T 2s, n 2s + 1. (14 We provde a combnatoral proof of (14 as follows. Note that (14 s clearly true f s = 0 or f n = 2s + 1 snce T 0 = T 1 = 0, so we may assume s 1 and n 2s + 2. Observe frst that the T n 2s T 2s+1 term gves the weght of all members of T n 1 n whch there s no pece coverng the boundary between postons 2s and 2s + 1. On the other hand, the total weght of the members of T n 1 n whch a domno covers ths boundary s gven by xt n 2s 1 T 2s. Fnally, f a tromno covers the boundary between postons 2s and 2s + 1, then that tromno covers ether postons 2s 1, 2s, and 2s + 1 or postons 2s, 2s + 1, and 2s + 2. In the former case, the weght of the correspondng members of T n 1 s T n 2s 1 T 2s 1, whle n the latter t would be T n 2s 2 T 2s. Combnng all of the cases above gves (14. Multplyng both sdes of the equaton T (s n = T n 2s T 2s+1 + T n 2s 1 (T 2s 1 + xt 2s + T n 2s 2 T 2s n 2s 1 r T n 2s

9 48 M. Shattuck by z n and summng over n 2s + 1 yelds ( Q s (z = T z 2s 2s+1 + (T 2s 1 + xt 2s z + T 2s z 2 r z T n z n. 0 The proof s completed by notng and 0 T n z n = n 1 ( s+1 x + z r z = z 2 1 x 2 z z 1 x 2 z xz 2 z 3, the former beng computed by the methods gven n (Wlf, 2005, Secton 4.3. Takng x = 1 n the pror theorem yelds the followng result. Corollary 0.10: Let q s (z be the generatng functon for the ncomplete trbonacc numbers t (s n. Then n 1 ( 1+z s+1 1 z q s (z z = t 2s+1 + (t 2s 1 + t 2sz + t 2sz 2 z 2. (15 2s+1 1 z z 2 z 3 Remark: Corollary 0.10 appears as (Ramírez and Srvent, 2014, Theorem 8. We note however that there was a slght msstatement of ths theorem; n partcular, the t 2s 2 factor multplyng z 2 n the numerator on the rght-hand sde of ther formula should just be t 2s. 5. Concluson In ths paper, we have provded a combnatoral nterpretaton for the ncomplete trbonacc polynomals T n (s (x n terms of weghted lnear tlngs nvolvng three types of tles. Ths not only allows one to supply combnatoral proofs of pror denttes shown by algebrac methods but also to establsh new ones. Furthermore, combnatoral reasonng provdes a way of dscoverng new denttes, some of whch may be harder to fnd by purely algebrac means. We have also made use of our combnatoral nterpretaton for T n (s (x n determnng an explct formula for ts ordnary generatng functon, as prevously requested. The generalzed k-fbonacc numbers f (k n f (k n = consdered n (Knuth, 1973 satsfy the recurrence k =1 f (k n, n k, wth ntal values f (k = 0 for 0 k 2 and f (k k 1 = 1. Note that the Fbonacc numbers correspond to the case k = 2 of f n (k, and the trbonacc numbers to the case k = 3. Perhaps the results of ths paper (and pror ones can be extended once an approprate analogue of the ncomplete trbonacc number (or polynomal n the generalzed k-fbonacc settng has been dentfed.

10 AAM: Intern. J., Vol. 10, Issue 1 (June Acknowledgement I wsh to thank Nel Sloane for pontng out to me the relaton between the B(n, coeffcents and the Delannoy numbers. REFERENCES Allad, K. and V.E. Hoggatt, J. (1977. On trbonacc numbers and related functons. Fbonacc Quarterly, 15: Barry, P. (2006. On nteger-sequence-based constructons of generalzed pascal trangles. Journal of Integer Sequences, 9(Artcle Belbachr, H. and Belkhr, A. (2014. Combnatoral expressons nvolvng fbonacc, hyperfbonacc, and ncomplete fbonacc numbers. Journal of Integer Sequences, 17(Artcle Benjamn, A. and Qunn, J. (2003. Proofs that Really Count: The Art of Combnatoral Proof. Mathematcal Assocaton of Amerca, Washngton DC. Flppon, P. (1996. Incomplete fbonacc and lucas numbers. Rendcont del Crcolo Matematco d Palermo, 45: Heubach, S. and Mansour, T. (2009. Combnatorcs of Compostons and Words. Chapman & Hall/CRC, Boca Raton, FL. Klç, E. and Prodnger, H. (2014. A note on the conjecture of ramírez and srvent. Journal of Integer Sequences, 17(Artcle Knuth, D. (1973. The Art of Computer Programmng: Sortng and Searchng, Volume 3. Addson- Wesley, Readng, MA. Ramírez, J. (2013a. B-perodc ncomplete fbonacc sequences. Annales Mathematcae et Informatcae, 42: Ramírez, J. (2013b. Incomplete generalzed fbonacc and lucas polynomals. pre-prnt. Ramírez, J. and Srvent, V. (2014. Incomplete trbonacc numbers and polynomals. Journal of Integer Sequences, 17(Artcle Sloane, N. (2010. On-lne encyclopeda of nteger sequences. V.E. Hoggatt, J. and Bcknell, M. (1973. Generalzed fbonacc polynomals. Fbonacc Quarterly, 11: Wlf, H. (2005. generatngfunctonology, thrd edton. CRC Press, Boca Raton, FL.