COUNTING STAIRCASES IN INTEGER COMPOSITIONS

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1 COUNTING STAIRCASES IN INTEGER COMPOSITIONS AUBREY BLECHER AND TOUFIK MANSOUR Abstract The main theorem establishes the generating function F which counts the number of times the staircase m + fits inside an integer composition of n where F = k m qxm y (1 q)x (m+1 2 ) ( y ) m + xy k m = m 1 æ=0 k m 1 ( ) x mj ( j 2 ) y j ( k m qxm y k m 1 Here x and y respectively track the composition size and number of parts, whilst q tracks the number of such staircases contained ) 1 Introduction In several recent papers the notion of integer compositions of n (represented as the associated bargraph) have been used to model certain problems in physics See for example [2, 7 9] where bargraphs are a representation of a polymer at an adsorbing wall subject to several forces In a paper by a current author et al (see [1]), the x-ray process was modelled using permutation matrices as a two dimensional analogue of the object being x-rayed, where the examining rays are modelled by diagonal lines with equation x + y = n for positive integers n The current paper is based instead on integer compositions as the object analogue and where the examining rays are represented by equation x y = n for non negative integers n Since this model is essentially parameterized by the degree to which the x-rays are contained inside an arbitrary composition, it translates naturally to obtaining a generating function which tracks the number of "staircases" which are contained inside particular integer compositions of n More precisely, we will obtain a generating function which counts (with the exponent s of q as tracker) the number of times the staircase m + (m fixed) fits inside particular compositions So the term of our generating function n(a, b, s)x a y b q s indicates that there are in total n(a, b, s) compositions of a with b parts in which the staircases m + occurs exactly s times Date: October 27, Mathematics Subject Classification 05A18, 05A15, 15A06, 15A09 Key words and phrases composition, generating function 1

2 2 AUBREY BLECHER AND TOUFIK MANSOUR 11 Definitions A composition of a positive integer n is a sequence of k positive integers a 1, a 2, a k, each called a part such that n = k ß=1 a i; A staircase m + is a word with m sequential parts from left to right where for 1 i m the ith part i See for example the staircase in Figure 1 below Figure 1 The staircase Much recent work has been done on various statistics relating to compositions See, for example, [3, 5, 6] and [4] and references therein A particular composition may be represented as a bargraph (see [4] and [2]) For example the composition of 13 represented in Figure 2 as a bargraph, contains exactly one staircase, three staircases and five 1 + staircases It contains no others Figure 2 The composition containing one staircase (coloured) and three staircases In this paper, compositions (ie their associated bargraphs) are the analogue for a (2-dimensional) object to be x-rayed (as explained above) Across all possible compositions, the shapes are parameterized in a generating function by a marker variable q which tracks the number of m + staircases (again with m fixed) that fit inside a composition The generating function in question is defined as (1) F = n(a, b, s)x a y b q s, a 1;b 1;s 0 where n(a, b, s) is the number of compositions of a with b parts that contain s staircases m +

3 COUNTING STAIRCASES IN INTEGER COMPOSITIONS 3 The main theorem arrived at by the end of the paper consists in establishing a formula for the generating function F defined in equation (1) We state it here for completeness: F = k m qxm y (1 q)x (m+1 2 ) ( y ) m + xy k m 1 ( k m qxm y k m 1 ), where k m = m 1 ) j Prior to this main theorem, several lemmas present a set of recursions which are used in proving this result æ=0 xmj ( j 2 ) ( y 2 Proofs 21 Warmup: compositions containing words of the form or Consider words which are of the form ; ie, words of two parts adjacent to each other from left to right with the first being a letter > 0 and the second being a letter > 1 We let F be the generating function for all words; F a be the generating function for all words starting with the letter a and in general F a1 a 2 a n be the gf (generating function) for words starting with the letters a 1 a 2 a n So by definition (2) F = 1 + F a a 1 And we have the following recurrence: (3) F a = x a y + F a1 + F a2 + F a3 + Now F a1 = x a yf 1 and F ab = qx a yf b for b > 1 So F a = x a y(1 + F 1 + qf 2 + qf 3 + ) Thus for all a 1, we have F a = x a y(1 q)(1 + F 1 ) + qx a yf As the second part of our warmup, we now examine the pattern , ie, we focus on compositions which contain this word sequence Extracting part of the first letter, we have (4) F a = x a 1 F 1 From equation (2), (5) F = 1 + F a = F 1 a 1 Also (6) F 1 = xy + (F 11 + F 12 + F 13 + ) = xy + xy(f 1 + F 12 + xf 12 + x 2 F 12 + ) = xy + xyf F 12, Online Journal of Analytic Combinatorics, Issue 11 (2016), #7

4 4 AUBREY BLECHER AND TOUFIK MANSOUR where (7) F 12 = x 3 y 2 + F F (F ) = x 3 y 2 + x 3 y 2 F 1 + x 2 yf 12 + (qx 3 yf 12 + qx 4 yf 12 + ) = x 3 y 2 + x 3 y 2 F 1 + x 2 yf 12 + qx3 y F 12 The last three equations have three unknowns F, F 1, and F 12 which we can solve for F using Cramer s rule However, instead, we try the general pattern 22 The general pattern m + As before, F a = x a 1 F 1 and (8) F = 1 + F a = F 1 a 1 Now (9) and F 1 = xy + (F 11 + F 12 + F 13 + ) = xy + xy(f 1 + F 12 + xf 12 + x 2 F 12 + ) = xy + xyf F 12 (10) Next, by a similar process (11) F 12 = x 3 y 2 + F F (F ) = x 3 y 2 + x 3 y 2 F 1 + x 2 yf 12 + (F xf x 2 F ) = x 3 y 2 + x 3 y 2 F 1 + x 2 yf F 123 F 123 = x 6 y 3 + x 6 y 3 F 1 + x 5 y 2 F 12 + x 3 yf F 1234 Proceeding in this way, we obtain in general for all j m 1 (12) with (13) F 12 j = x (j+1 2 ) y j + x (j+1 2 ) (1 2 ) y j F 1 + x (j+1 2 ) (2 2 ) y j 1 F 12 + x (j+1 2 ) (3 2 ) y j 2 F x (j+1 2 ) ( j 2 ) yf 12 j + 1 F 12 j+1 F 12 m = qx m yf 12 m 1

5 COUNTING STAIRCASES IN INTEGER COMPOSITIONS 5 To simplify the presentation we put z = 1 Now, we rewrite equations (7)-(13) in matrix form So we first define the matrix A as 1 z (2 2 ) (1 2 ) y z x (3 2 ) (1 2 ) y 2 (3 2 ) (2 2 ) y z 0 0 x (m 1 2 ) (1 2 ) y m 2 x (m 1 2 ) (2 2 ) y m 3 x (m 2 ) (3 2 ) y m 4 x (m 1 2 ) (m 2 2 ) y z 0 0 x (m 2 ) (1 2 ) y m 1 x (m 2 ) (2 2 ) y m 2 x (m 2 ) (3 2 ) y m 3 x (m 2 ) (m 2 2 ) y 2 (m 2 ) (m 1 2 ) y z qx m y 1 and C to be the vector ( x (1 2 ), x (2 2 ) y, x (3 2 ) y 2,, x (m 1 2 ) y m 2, x (m 2 ) y m 1, 0) T Then the matrix form of our equations is AX = C where it is the first entry of matrix X (the matrix of variables from equations (7)-(13)) that is our required generating function F So defining B as the matrix obtained from the above matrix A by replacing its first column with the entries from C; ie x (1 2 ) z 0 0 x (2 2 ) y (2 2 ) (1 2 ) y z 0 x (3 2 ) y 2 x (3 2 ) (1 2 ) y 2 (3 2 ) (2 2 ) y 0 x (m 1 2 ) y m 2 x (m 1 2 ) (1 2 ) y m 2 x (m 1 2 ) (2 2 ) y m 3 x (m 1 2 ) (m 2 2 ) y z 0 x (m 2 ) y m 1 x (m 2 ) (1 2 ) y m 1 x (m 2 ) (2 2 ) y m 2 x (m 2 ) (m 2 2 ) y 2 (m 2 ) (m 1 2 ) y z qx m y 1 By Cramer s rule, we obtain (14) F = det B det A 23 Equations for det A and det B in a form that can be solved recursively Define the mxm matrix N m, to be the first m rows and columns of the (m + 1)x(m + 1) matrix A, but where the first column of A has initially been replaced by the first m entries of C To simplify the notation further, we let w ij = x (i 2 ) ( j 2 ) y i j and so explicitly written out, x (1 2 ) y 0 z x (2 2 ) y 1 w 21 z 0 N m := x (3 2 ) y 2 w 31 1 w 31 z x (m 2 ) y m 1 w m1 1 w m1 By cofactor expansions (initially along the last row of B), we obtain (15) det B = det N m + zqx m y det N m 1 Online Journal of Analytic Combinatorics, Issue 11 (2016), #7

6 6 AUBREY BLECHER AND TOUFIK MANSOUR And let C m 1 be the (m 1)x(m 1) matrix obtained by deleting the first row and column of N m So, for example, 1 w 21 z 0 0 C 4 = w 31 1 w 32 z 0 w 41 w 42 1 w 43 z w 51 w 52 w 53 1 w 54 By employing cofactor expansions (also, initially along the last row of A), we see that (16) det A = det C m 1 + zqx m y det C m 2 Again, by employing co-factor expansions along the last row of C 4, we see that det C 4 = (1 w 54 ) det C 3 + zw 53 det C 2 w 52 z 2 det C 1 + w 51 z 3 det C 0, where det C 0 := 1 In general, a cofactor expansion along the last row of C m yields for m 1 det C m = (1 w m+1m ) det C m 1 + m 1 æ=1 ( 1) m 1 j w m+1j z m j det C j 1 Once again making the replacement w ij = x (i 2 ) ( j 2 ) y i j, we have for m 1 (17) det C m = ( m y) det C m 1 + m 1 æ=1 ( 1) m 1 j x (m+1 2 ) ( j 2 ) y m+1 j z m j det C j 1 Dropping m by 1 and multiplying this equation by x m yz, we obtain (18) x m yz det C m 1 = x m yz( m 1 y) det C m 2 + By subtracting (18) from (17), we obtain det C m + x m yz det C m 1 Simplifying, m 2 æ=1 ( 1) m 1 j x (m+1 2 ) ( j 2 ) y m+1 j z m j det C j 1 = ( m y) det C m 1 + x m yz( m 1 y) det C m 2 + x 2m 1 y 2 z det C m 2 (19) det C m = ( m y(1 + z)) det C m 1 + x m yz det C m 2, where det C 1 := 1; det C 0 = 1; det C 1 = y = 1 w 21 For ease of notation in the remainder of the paper, we abbreviate det C m as C m, and define the generating function C(t) = m 0 C m t m By multiplying equation (19) by t m and then summing from 1 to infinity, we obtain C(t) 1 = tc(t) (1 + z)xytc(xt) + x 2 yt 2 zc(xt) + xyzt

7 COUNTING STAIRCASES IN INTEGER COMPOSITIONS 7 Therefore (20) C(t) = 1 + xyzt 1 t xytc(xt) 1 + z(t) 1 t Again to simplify the notation, substitute f (t) := 1+xyzt 1 t and iterate the previous equation to obtain: (21) and ϕ(t) := xyt 1+z(t) 1 t, C(t) = f (t) + ϕ(t)c(xt) = f (t) + ϕ(t) f (xt) + ϕ(t)ϕ(xt)c(x 2 t) Repeatedly iterating (assuming x < 1), we obtain Recall that z = 1 C(t) = = f (x j t) j 1 ß=0 ϕ(x i t) ( 1) j 1 + xj+1 yzt j t x (j+1 2 ) y j t j j 1 ß=0 which implies 1 + z = x Therefore, 1 + z( i+1 t) i t 1+z ) C(t) = ( 1) j (1 + x j+1 yzt)x (j+1 2 ) y j t j j ß=1 (1 zxi t j ß=0 (i t) (1 + z)j = ( 1) j (1 + x j+1 yzt)x (j+1 2 ) y j t j ( x j 1 )j ß=0 (i t) j ß=0 (i t) (1 + x = j+1 yzt)x j(j+3) 2 y j t j () j ( j t) For further notational simplification, we let f j = (1 + xj+1 yzt)x j(j+3) 2 y j t j () j ( j t) Finally, substituting for the remaining z as above and using partial fractions f j = = j(j+3) x1+ 2 y j+1 t j () j+1 + x j(j+3) 2 y j ( xy)t j () j+1 ( j t) Hence the mth coefficient of C(t) is given by So, we obtain the following lemma j(j+3) x1+ 2 y j+1 t j () j+1 + x j(j+3) 2 y j ( xy)t j () j+1 x jk t k k 0 x(m+2 2 C m = )ym+1 m () m+1 + x j2 +3j 2 j 2 +jm y j ( xy) j=0 () j+1 Online Journal of Analytic Combinatorics, Issue 11 (2016), #7

8 8 AUBREY BLECHER AND TOUFIK MANSOUR Lemma 21 The determinants C m of the matrices obtained from N m+1 (see equation (23)) by deleting its first row and column are given by ( ) (22) C m = x (m+2 2 ) y m+1 + xy m j=0 ( ) x (m+1)j ( j 2 ) y j For initial cases, we have det N 1 = 1 and det N 2 = y zxy By a cofactor expansion along the last row, we obtain for m 2 (23) det N m = ( m 1 y) det N m 1 + m 2 æ=1 ( 1) m j x (m 2 ) ( j 2 ) y m j z m 1 j det N j + ( 1) m 1 x (m 2 ) y m 1 z m 1 Dropping m by 1 and multiplying this equation by x m 1 yz (a similar process to that used in a previous section), we obtain for m 3 x m 1 yz det N m 1 = x m 1 yz( m 2 y) det N m 2 (24) + m 3 æ=1 Subtracting (24) from (23), we obtain ( 1) m j x (m 2 ) ( j 2 ) y m j z m 1 j det N j + ( 1) m 1 x (m 2 ) y m 1 z m 1 det N m + x m 1 yz det N m 1 = ( m 1 y) det N m 1 + x m 1 yz( m 2 y) det N m 2 + x 2m 3 y 2 z det N m 2 = ( m 1 y) det N m 1 + x m 1 yz det N m 2 Hence for m 2, (25) det N m = ( m 1 y(1 + z)) det N m 1 + x m 1 yz det N m 2 with det N 0 = 0 and det N 1 = 1 For the rest of the paper we simplify matters by abbreviating N m := det N m and now define the generating function N(t) = m 0 N m t m By multiplying equation (25) by t m, summing from 1 to infinity, we obtain with N 1 := 0 Hence N(t) t = tn(t) y(1 + z)tn(xt) + xyzt 2 N(xt) (26) N(t) = t 1 t + xyzt2 y(1 + z)t N(xt) 1 t

9 COUNTING STAIRCASES IN INTEGER COMPOSITIONS 9 Repeatedly iterating (26) on t (while recalling that z = 1, and assuming x < 1), we obtain N(t) = = Thus, we have our final lemma x j t j t x j t j t j 1 yx i t( xi+1 t ß=0 j 1 ß=0 yx i t x j2 +3j 2 y = j t j+1 ( j t)() j Lemma 22 With N m := det N m (see (23)) (27) N m = [t m ]N(t) = m 1 j=0 + x ) i t ( ) x mj ( j 2 ) y j 24 The generating function F Finally, apply (15) and (16) to (14) Then, use lemma 21 and lemma 22, to obtain: Theorem 23 The generating function F = a 1;b 1;s 0 n(a, b, s)x a y b q s for the number of staircases m + (tracked by the exponent of variable q) contained in particular compositions (of a with b parts) is given by (28) F = N m qxm y (1 q)x (m+1 2 ) ( y ) m + xy N m 1 ( N m qxm y N m 1 ) For example, Theorem 23 with q = 1 yields F q=1 = y, which is the generating function for the number of compositions of n with exactly m parts (see [4]) By differentiating the generating function F with respect to q and then substituting q = 1, we obtain df dq q=1 = = x (m+1 2 ) ( y ( xy) 2 () 2 ( m 1 j=0 xmj ( j 2 ) ( y x (m+1 2 ) y m ( xy) 2 () m 2 = x(m+1 2 ) () m (j + 1) xj y m+j () j ) m ) j m 1 j=1 xmj ( j 2 ) ( y ) j ) Online Journal of Analytic Combinatorics, Issue 11 (2016), #7

10 10 AUBREY BLECHER AND TOUFIK MANSOUR Next, we extract coefficients; firstly of [y l ] to obtain (l m + 1) xl+(m 2 ) () l = (l m + 1) and then of [x n ] which leads to the following result ( l + j 1 j ) x l+j+(m 2 ), Corollary 24 The total number of staircases m + in all compositions of n with exactly l parts is given by ( n 1 ( m (l m + 1) 2 ) ) l 1 References [1] C Bebeacua, T Mansour, A Postnikov, and S Severini, On the X-rays of permutations, Elect Notes Discr Math 20 (2005) [2] A Blecher, C A Brennan and A Knopfmacher, Levels in Bargraphs, Ars Mathematica Contemporaneae 9 (2015) [3] C Brennan and AKnopfmacher, The first descent of size d or more in compositions, DMTCS proc AG (2006) [4] S Heubach, T Mansour, Combinatorics of Compositions and Words, CRC Press, Boca Raton, 2009 [5] A Knopfmacher, Some properties of dihedral compositions, Utilitas Mathematica 01 (2013) 92 [6] A Knopfmacher and A Munagi, Smallest parts in compositions, DOI 01 (2013) 3 11 [7] J Osborn and T Prellberg, Forcing adsorption of a tethered polymer by pulling, J Stat Mech-Theory E (2010) P09018 [8] A Owczarek, Exact solution for semi-flexible partially directed walks at an adsorbing wall, J Stat Mech: Theor and Exp, (2009) P11002 [9] A Owczarek, Effect of stiffness on the pulling of an adsorbing polymer from a wall: an exact solution of a partially directed walk model, J Phys A: Math Theor 43 (2010) Except where otherwise noted, content in this article is licensed under a Creative Commons Attribution 40 International license School of Mathematics, University of Witwatersrand, Johannesburg, South Africa address: AubreyBlecher@witsacza Department of Mathematics, University of Haifa, Haifa, Israel address: tmansour@univhaifaacil