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Kyungpook Mathematical Journal 2016; 56(2): 409-418

Published online June 1, 2016

Copyright © Kyungpook Mathematical Journal.

Two Variants of the Reciprocal Super Catalan Matrix

E. Kiliç1, N.Ömür2, S. Koparal3, Y. Ulutas3

Department of Mathematics, TOBB University of Economics and Technology, 06560 Ankara, Turkey1
Department of Mathematics, Kocaeli University, 41380 ˙Izmit Kocaeli, Turkey2
Department of Mathematics, Kocaeli University, 41380 ˙Izmit Kocaeli, Turkey3

Received: July 27, 2015; Accepted: December 10, 2015

In this paper, we define two kinds variants of the super Catalan matrix as well as their q-analoques. We give explicit expressions for LU-decompositions of these matrices and their inverses.

Keywords: Catalan matrix, q-binomial coefficients, LU-decompositions, determinant.

For a given sequence {an}n=0, the Hankel matrix is defined by

(a0a1a2a1a2a3a2a3a4).

One can obtain a combinatorial matrix having interesting properties from a Hankel matrix. For example, the Hilbert matrix Hn = [hij ] is defined by hij=1i+j-1 (for more details, see [2, 4]) and the Filbert matrix ℱn = [fij ] is defined by fij=1Fi+j-1 (see [1, 6]). Clearly they are of the forms

Hn=(111213121314131415)         and Fn=(1F01F11F21F11F21F31F21F31F4),

where Fn is nth Fibonacci number.

Throughout this paper, we will use the q-Pochhammer symbol

(x;q)n=(1-x)(1-xq)(1-xqn-1)

and the Gaussian q-binomial coefficients

[nk]q=(q;q)n(q;q)k(q;q)n-k.

It is clearly that

limq1[nk]q=(nk),

where (nk) is the usual binomial coefficient.

The Cauchy binomial theorem is given by

k=0nq(k+12)[nk]qxk=k=1n(1+xqk),

and Rothe’s formula (see [2]) is given by

k=0n(-1)kq(k2)[nk]qxk=(x;q)n=k=0n-1(1+xqk).

Prodinger [3] consider the reciprocal super Catalan matrix M with the entries mij=i!(i+j)!j!(2i)!(2j)! and obtain explicit formulae for its LU-decomposition, the LU-decomposition of its inverse, and some related matrices. For all results, q-analogues are also presented.

We rewrite the matrix M in terms of three binomial coefficients, two of them is in the reciprocal form, as shown

mij=i!(i+j)!j!(2i)!(2j)!=i!j!(2i)!(2j)!=(2ii)-1(2jj)-1(i+ji).

By inspiring the matrix M, we consider two kinds variants of it by keeping only one binomial coefficient in reciprocal form in the first one and two binomial coefficients in reciprocal forms in the second one. Also we will add two additional parameters to each one. Now we define these matrices: The first one is the matrix A = [aij] of order n with the entries

aij=(2i+mi)(2j+tj)-1(i+ji)

and the second one is the n × n matrix B with the entries

bij=(2i+mi)-1(2j+tj)(i+ji)-1,

where m and t are nonnegative integers and all indices of these matrices start at (0, 0).

We write the matrices and ℬ which are the q-analagues of the matrices A and B, respectively. We give explicit expressions for LU-decompositions of these matrices and their inverses.

By help of a computer, LU-decompositions of these matrices were firstly obtained and then we have achieved the formulas by certain skills especially guessing skill. Using q-Zeilberger algorithm [5] and elementary matrix operations, the proofs are given as combinatorial identities. We will discuss a few of them here rather than all of them.

The matrix = [âij ] has the entries a^ij=[2i+mi]q[2j+tj]q-1[i+ji]q for 0 ≤ i, jn − 1. Now we will give expressions for LU-decompositions L1U1 = A and L2U2 = A−1, for L1-1,U1-1 and for L2-1,U2-1, by the following theorem.

Theorem 2.1

For m,t ≥ 0,

(L1)ij=[2i+mi]q[ij]q[2j+mj]q-1,(L1-1)ij=(-1)i-jq(i-j2)[2i+mi]q[ij]q[2j+mj]q-1,(U1)ij=qi2-1[2i+mi]q[ji]q[2j+tj]q-1,(Un-1)ij=(-1)i-jq(j-i2)-j2+1[2i+ti]q[ji]q[2j+mj]q-1,(L2)ij=(-1)i-jq(n-i-12)-(n-j-12)[2i+ti]q[n-j-1i-j]q[2j+1j]q×[i+j+1i]q-1[2j+tj]q-1,(L2-1)ij=q(n-i-1)(j-i)[i+jj]q[n-j-1i-j]q[2i+ti]q[2ii]q-1[2j+tj]q-1,(U2)ij=(-1)i-jq(n-j-12)-(n-i-12)-(n-i-1)(2i+1)[2i+ti]q[n+ii+j+1]q×[i+jj-i]q[i+jj]q-1[2j+mj]q-1,(U2-1)ij=q(n-j-1)(i+j+1)[2j+1j-i]q[i+ji]q[2i+mi]q[2j+tj]q-1×[n+ji+j+1]q-1,

and

det A=k=0n-1qk2[2k+mk]q[2k+tk]q-1.
Proof

To prove L1L1-1=In where In is the identity matrix of order n, consider

jki(L1)ik(L1-1)kj=(-1)j[2i+mi]q[2j+mj]q-1jki(-1)kq(k-j2)[ik]q[kj]q.

Since

[ik]q[kj]q=[ij]q[i-jk-j]q,

we have

jki(L1)ik(L1-1)kj=(-1)j[2i+mi]q[2j+mj]q-1[ij]qjki(-1)kq(k-j2)[i-jk-j]q=[2i+mi]q[2j+mj]q-1[ij]q0ki-j(-1)kq(k2)[i-jk]q.

Using Rothe’s formula, we see that (1; q)ij is equal to 1 if i = j and 0 otherwise. Then we get

jki(L1)ik(L1-1)kj=δi,j,

as claimed, where δi,j is Kronecker delta. For U1 and U1-1, we write

ikj(U1)ik(U1-1)kj=qi2[2i+mi]q[2j+mj]q-1ikj(-1)k-jq(j-k2)-j2[ki]q[jk]q=(-1)jqi2-j2[2i+mi]q[2j+mj]q-1[ji]qikj(-1)kq(j-k2)[j-ik-i]q=(-1)i+jqi2-j2+(j2)+(i+12)-ij[2i+mi]q[2j+mj]q-1[ji]q×0kj-iq(k+12)[j-ik]q(-qi-j)k.

By the Cauchy binomial theorem, for ij, we get

0kj-iq(k+12)[j-ik]q(-qi-j)k=k=1j-i(1-qi-j+k)=0.

Then

ikj(U1)ik(U1-1)kj=δi,j.

For LU-decomposition, we will show

0kmin{i,j}(L1)ik(U1)kj=a^ij,

where A = [âij ]. Then

0kmin{i,j}(L1)ik(U1)kj=[2i+mi]q[2j+tj]q-10kmin{i,j}qk2-1[ik]q[jk]q=q-1[2i+mi]q[2j+tj]q-1(q;q)i(q;q)j×0kmin{i,j}qk21(q;q)i-k(q;q)j-k(q;q)k(q;q)k.

Denote the last sum in the above equation by SUMk. The Mathematica version of the q-Zeilberger algorithm [5] produces the recursion

SUMi=(1-qi+j)(1-qi)2SUMi-1.

Since SUM0=1(q;q)i(q;q)j, we obtain

SUMi=[i+ji]q1(q;q)i(q;q)j.

Then we write

0kmin{i,j}(L1)ik(U1)kj=a^ij.

For L2 and L2-1, we have

jki(L2)ik(L2-1)kj=(-1)iq(n-i-12)[2i+ti]q[2j+tj]q-1×jki(-1)kq-(n-k-12)+(n-k-1)(j-k)[n-k-i-k]q[2k+1k]q×[i+k+1i]q-1[k+jj]q[n-j-1k-j]q[2kk]q-1=(-1)iq(n-1)(j+1)-(n2)[2i+ti]q[2j+tj]q-1(q;q)i(q;q)n-j-1(q;q)j(q;q)n-i-1+jki(-1)kq(k2)-kj(q;q)2k+1(q;q)k+j(q;q)i-k(q;q)i+k+1(q;q)k-j(q;q)2k.

By the q-Zeilberger algorithm for the second sum in the last equation, we obtain that it is equal to 0 provided that ij. If i = j, it is obvious that (L2)ik(L2-1)kj=1. Thus

jki(L2)ik(L2-1)kj=δi,j,

as claimed. Similarly we have

ikj(U2)ik(U2-1)kj=δi,j.

For the LU –decomposition of , we should that = L2U2 which is same as A=U2-1L2-1. So it is sufficient to show that

max{i,j}kn-1(U2-1)ik(L2-1)kj=a^ij.

Hence

max{i,j}kn-1(U2-1)ik(L2-1)kj=[2i+mi]q[2j+tj]q-1×max{i,j}kn-1q(n-k-1)(i+k+1)(n-k-1)(j-k)[2k+1k-i]q[i+ki]q×[n+ki+k+1]q-1[k+jj]q[n-j-1k-j]q[2kk]q-1=qn(i+j-1)[2i+mi]q[2j+tj]q-1×max{i,j}kn-1q(n-k-1)(i+j+1)[2k+1k-i]q[i+ki]q×[k+jj]q[n-j-1k-j]q[2kk]q-1[n+ki+k+1]q-1.

If we take (n + 1) instead of n, we write

jknq(n-k)(i+j+1)[2k+1k-i]q[i+ki]q[k+jj]q[n-jk-j]q[2kk]q-1[n+k+1i+k+1]q-1.

Denote sum in (2.1) by SUMn. For in and jn, the q-Zeilberger algorithm gives the following recursion

SUMn=SUMn-1.

Thus, SUMn=SUMj=[i+ji]q which completes the proof except the case (i, j) = (n − 1, n − 1), which could be easily checked. The proof is obtained.

In this section, the matrix ℬ = [ij] is defined with entries

b^ij=[2i+mi]q-1[2j+tj]q[i+ji]q-1

for 0 ≤ i, jn−1. Now we will give expressions for LU-decompositions L3U3 = ℬ and L4U4 = ℬ−1, for L3-1,U3-1 and for L4-1,U4-1 without proof by the following theorem

Theorem 3.1

For m, t ≥ 0 and 0 ≤ i, jn − 1,

(L3)ij=[2jj]q[2j+mj]q[ij]q[i+jj]q-1[2i+mi]q-1,(L3-1)ij=(-1)i-jq(i-j2)[i+j-1j]q[2j+mj]q[ij]q[2i-1i-1]q-1[2i+mi]q-1,(U3)ij=(-1)iqi2+(i2)[2j+tj]q[i+j-1j-i]q[i+ji]q-1[i+j-1j]q-1[2i+mi]q-1,(U3-1)ij=(-1)iq(j-i2)-(j2)-j2[ji]q[2j+mj]q[2jj]q[i+j-1j]q[2i+ti]q-1,(L4)ij=(-1)i-jq-(n-j-12)+(n-i-12)[n+i-1i]q[2j+tj]q[n-j+1i-j]q×[2i+ti]q-1[n+j-1j]q-1,(L4-1)ij=q(n-i-1)(j-i)[2j+tj]q[n-j-1i-j]q[n+i-1i-j]q[2i+ti]q-1[ij]q-1,(U4)ij=(-1)n-j-1q(n-j-12)+(i+n-1)(i-n+1)[n+j-1i]q[n-i+j-1j]q×[n-i-1j-i]q-1[2j+mj]q[2i+ti]q-1.(U4)ij-1=(-1)jq(n2)-(j+12)+i(n-j-1)[2j+tj]q[n-i-1j-i]q[n+i-j-1i]q-1×[n+i-1j]q-1[2i+mi]q-1

and

det k=0n-1(-1)kqk(3k-1)/2[2k+tt]q[2k+mk]q-1[k+tk]q-1[2k-1k]q-1.

In this section, we have the following results without proof by using the results of Theorem 2.1 with the fact given in (1.1). For 0 ≤ i, j < n − 1,

(L1)ij=(2i+mi)(ij)(2j+mj)-1,(L1-1)ij=(-1)i-j(2i+mi)(ij)(2j+mj)-1,(U1)ij=(2i+mi)(ji)(2j+tj)-1,(U1-1)ij=(-1)i-j(2i+ti)(ji)(2j+mj)-1,(L2)ij=(-1)i-j(2i+ti)(n-j-1i-j)(2j+1j)(i+j+1i)-1(2j+tj)-1,(L2-1)ij=(i+jj)(n-j-1i-j)(2i+ti)(2ii)-1(2j+tj)-1,(U2)ij=(-1)i-j(2i+ti)(n+ii+j+1)(i+jj-i)(i+jj)-1(2j+mj)-1,(U2-1)ij=(2j+1j-i)(i+ji)(2i+mi)(2j+tj)-1(n+ji+j+1)-1,det A=k=0n-1(2k+mk)(2k+tk)-1.

In this section, we have the following results without proof by using the results of Theorem 2.2 with the fact given in (1.1). For 0 ≤ i, j < n − 1,

(L3)ij=(2jj)(2j+mj)(ij)(i+jj)-1(2i+mi)-1,(L3-1)ij=(-1)i-j(i+j-1j)(2j+mj)(ij)(2i-1i-1)-1(2i+mi)-1,(U3)ij=(-1)i(2j+tj)(i+j-1j-i)(i+ji)-1(i+j-1j)-1(2i+mi)-1,(U3-1)ij=(-1)i-1(ji)(2j+mj)(2jj)(i+j-1i)(2i+ti)-1,(L4)ij=(-1)i-j(n+i-1i)(2j+tj)(n-j+1i-j)(2i+ti)-1(n+j-1j)-1,(L4-1)ij=(2j+tj)(n-j-1i-j)(n+i-1i-j)(2i+ti)-1(ij)-1,(U4)ij=(-1)n-j(n+j-1i)(n-i+j-1j)(n-i-1j-i)(2j+mj)×(2i+ti)-1,(U4-1)ij=(-1)j(2j+tj)(n-i-1j-i)(n+i-j-1i)-1(n+i-1j)-1×(2i+mi)-1,det B=k=0n-1(-1)k(2k+tk)(2k+mk)-1(k+tk)-1(2k-1k)-1.
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