ON RESTRICTED TERNARY WORDS AND INSETS MILAN JANJIC´ Department for Mathematics and Informatics, University of Banja Luka, Abstract. We investigate combinatorial properties of a kind of insets we defined in an earlier paper, interpreting them now in terms of restricted ternary words. This allows us to give new combinatorial interpretations of a number of known integer sequences, namely the coefficients of Chebyshev polynomials of both kinds, Fibonacci numbers, Delannoy numbers, asymmetric Delannoy numbers, Sulanke numbers, coordinating sequences for some cubic lattices, crystal ball sequences for some cubic lattices, and others. We also obtain several new properties of said insets. In particular, we derive three generating functions when two of three variables are constant. At the end, we state 40 combinatorial configurations counted by our words. 1. Introduction In M. Janji´cand B. Petkovi´c[2], we defined the notion of inset in the following way. Let m,k,n be nonnegative integers. Let q1, q2,...,qn be positive integers. Consider the set X consisting of blocks X1,X2,...,Xn, Y such that |Xi| = qi, (i = m,n 1, 2,...,n), |Y | = m. Then, k,Q equals the number of (n + k)-subsets of X intersecting each block Xi, (i =1 , 2,...,n ). We call such subsets the (n + k)-insets of X. m,n We investigate the properties of the function k,2 in which qi =2, (i =1, 2,...,n). m,n m,n To simplify our notation, we write k instead of k,2 . Thus, we consider the three-dimensional array m,n A = :0 ≤ m, 0 ≤ n, 0 ≤ k ≤ m + n . k, 2 m,n arXiv:1905.04465v1 [math.CO] 11 May 2019 The following two explicit formulae for k are proved in [2, Equations(8) and (11)]. n m,n n m +2n − 2i (1) = (−1)i . k i n + k Xi=0 m,n m m n (2) =2n−k 2i . k i k − i Xi=0 0,0 In particular, we have 0 = 1. Date: May 14, 2019. 1 2 MILAN JANJIC´ m,n Remark 1. We note that k are even if n > k. In the case n = k, Equation (2) becomes m,n m m n = 2i , n i i Xi=0 which is the well-known formula for the Delannoy numbers D(m,n). From [2, Eq. (11)], we obtain the following identity: m,n m,n − 1 m +1,n − 1 (3) = + , k k k Using this identity, we easily express the Fibonacci numbers in terms of insets. Corollary 1. For m ≥ 0, we have m ⌊ +1 ⌋ 2 m − i, 1 Fm = . +3 i Xi=0 Proof. From (3), we obtain m m ⌊ +1 ⌋ ⌊ +1 ⌋ 2 m − i, 1 2 m − i m − i − 1 = + , i i i Xi=0 Xi=0 and the proof follows from the well-known expression of the Fibonacci numbers in terms of the binomial coefficients. From [2, Eq. (13)], we obtain the following identity: m,n m,n − 1 m,n − 1 (4) =2 + . k k k − 1 m,n We proceed by deriving some new properties of the function k . Proposition 1. For 0 ≤ p ≤ m;0 <n, the following formula holds: p m +1,n − 1 p m − p +1,n + p − 1 − i (5) = (−1)i . k i k Xi=0 Proof. The formula is obviously true for p = 0. We next write (3) in the following form: m +1,n − 1 m,n m,n − 1 (6) = − . k k k This means that (5) holds for p = 1. Applying (6) on the right-hand side of the same formula yields m +1,n − 1 m − 1,n +1 m − 1,n m − 1,n − 1 = − 2 + . k k k k This means that (6) holds for p = 2. Repeating the same procedure, we obtain that (5) is true for each p ≤ m. Proposition 2. The following formula holds: m m +1,n n i,n (7) =2n−k−1 + . k +1 k +1 k Xi=0 ON RESTRICTED TERNARY WORDS AND INSETS 3 In particular, for n ≤ k ≤ m + n, we have m m +1,n i,n (8) = . k +1 k Xi=0 Proof. From [2, Proposition 10], it follows that m +1,n m,n m,n m,n m − 1,n m − 1,n = + = + + k +1 k k +1 k k k +1 m,n 0,n 0,n = + ··· + + . k k k +1 0,n n−k−1 n Taking m = 0 in (2), we obtain k+1 = 2 k+1 , which proves (7). The statement (8) is obvious. Remark 2. The equation (8) is an analog of the horizontal recurrence for the binomial coefficients. Proposition 3. Let p be a positive integer such that 1 ≤ p ≤ min{n, k}. Then, p m,n − p m,n m,n − i (9) − =2 · . k − p k k − i +1 Xi=1 Proof. From Equation (4), we obtain the following sequence of equalities: m,n m,n − 1 m,n − 1 − =2 , k k − 1 k m,n − 1 m,n − 2 m,n − 2 − =2 , k − 1 k − 2 k − 1 . m,n − p +1 m,n − p m,n − p − =2 . k − p +1 k − p k − p +1 The sum of the expressions on the left-hand side and the sum of the expressions on the right-hand side of the preceding equations produces Equation (9). m,n−p m,n Corollary 2. The numbers k−p and k are of the same parity. Remark 3. Note that for n = 0, the array A is the standard Pascal triangle. m,n Proposition 4. For a fixed positive n, the array { k : (m, k = 0, 1,...)} is a Pascal trapeze. The elements of the leftmost diagonal equal 2n, the rightmost n−k n diagonal consists of ones, and the first row consists of 2 k , (k =0, 1,...,m+n). m,n Proof. In the case k = 0, we have 0 , which is the number of insets of n elements. In such an inset, we must choose exactly one element from each main block. We have 2n such insets. We conclude that the leftmost diagonal of the array A is a n m,n constant sequence consisting of 2 . Furthermore, if n+k>m+2n, then k = 0, since there is no inset having more than m +2n elements. If k + n = m +2 n, that m,n is, if k = m + n, then k = 1. Hence, the rightmost diagonal consists of ones. 0,n n−k n Since the first row is obtained for m = 0, Proposition 2 yields k =2 k , (k = 4 MILAN JANJIC´ 0, 1,...,m + n). Finally, from [2, Proposition 10], the following (Pascal) recurrence holds: m,n m − 1,n m − 1,n (10) = + . k k − 1 k A combinatorial interpretation of m,n 2. k In this section, we relate the array A to restricted ternary words. We firstly 0,n 0,n consider the insets counted by k . Note that k = 0, if k>n. Also, we have 0,n n = 1. Let X be the set 0,n Proposition 5. The number k equals the number of ternary words of length n having k letters equal to 2. Proof. Assume that k ≤ n. From each (n + k)-inset of X, we form a ternary word of length n in the following way. Put 2 in the place of every main block from which both elements are in X. Then, insert 1 in the place of every main block from which the only first element is in X, and fill all other places with zeros. It is clear that this correspondence is bijective. We now consider the case m> 0. m,n Proposition 6. For k ≤ m + n, the number k is the number of ternary words of length m + n having k letters equal to 2 and none of the first m letters is equal to 0. Proof. The following formula is proved in [2, Proposition 11]: m m,n m 0,n = . k i k − i Xi=0 0,n Take i ∈{0, 1,...,m}. According to Proposition 5, k−i is the number of ternary m words of length n having k − i letters equal to 2. On the other hand, i is the m 0,n number of words of length m over {1, 2} having i letters equal to 2. Hence, i k−i is the number of ternary words of length m + n, having k letters equal to 2, and beginning with a subword of length m over {1, 2}. Summing over all i, we obtain the assertion. 1,3 Example 1. (1) For m = 1,n = 3, and k = 2, we have 2 = 18. The corresponding words counted by this number are 1022, 1122, 1202, 1212, 1220, 1221, 2200, 2211, 2210, 2201, 2020, 2121, 2021, 2120, 2002, 2112, 2012, 2102. 1,3 (2) For m =1,n = 3, and k = 3, we have 3 = 7. The corresponding words are 1222, 2122, 2022, 2212, 2202, 2221, 2220. 2,3 (3) For m = 2,n = 3, and k = 4, we have 4 = 8. The corresponding numbers are 12222, 21222, 22122, 22212, 22221, 22220, 22202, 22022. m,n We derive a property of the function k by introducing a new parameter p. ON RESTRICTED TERNARY WORDS AND INSETS 5 Identity 1. If 0 ≤ p ≤ n, then p m,n p m + i,n − p (11) = . k i k Xi=0 Proof. The result may be proved by induction on p, using Equation (3). We add a combinatorial proof.