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Catalan number is defined by $C_{n}=\frac{1}{n+1}\binom{2n}{n}.$ Two natural $q$-analogs of Catalan numbers are (see Carlitz and Scoville, A note on weighted sequences, Fibonacci Quarterly, 13 (1975), 303-306)

  1. $C_n(q)=\frac{1}{[n+1]_q}{2n\brack n}_q.$

  2. $C_{n+1}(q)=\sum\limits_{k=0}^{n}q^kC_k(q)C_{n-k}(q).$

However definitions (1) and (2) are not equivalent. Is there any known recurrence relation like (2) in the literature that (1) satisfies?

138 Aspen
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J. Yomcha
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1 Answers1

1

[toc]

Good question, they are not compatible.

case1----[MacMahon] Mirror symmetrical with self

QInteger[n_, q_] := Sum[q^i, {i, 0, n - 1}];
QCatalan[n_, q_] := QBinomial[2 n, n, q]/QInteger[n + 1, q];
QCatalan[#, q] &   /@ Range[0, 5] // FunctionExpand // 
  FullSimplify // Expand

$ C_n(q)=\frac{1}{[n+1]_q}\left[\begin{array}{c} 2 n \\ n \end{array}\right]_q $ produces the follows.

$$ \left\{1,1,q^2+1,q^6+q^4+q^3+q^2+1,q^{12}+q^{10}+q^9+2 q^8+q^7+2 q^6+q^5+2 q^4+q^3+q^2+1,q^{20}+q^{18}+q^{17}+2 q^{16}+2 q^{15}+3 q^{14}+2 q^{13}+4 q^{12}+3 q^{11}+4 q^{10}+3 q^9+4 q^8+2 q^7+3 q^6+2 q^5+2 q^4+q^3+q^2+1\right\} $$

case2----[Carlitz and Riordan], Mirror symmetrical with case3

qc[0] := 1;
qc[1] := 1;
qc[n_] := Sum[qc[k]*qc[n - 1 - k]*q^(k), {k, 0, n - 1}];
qc /@ Range[0,5] // Expand

However, $C_{n+1}(q)=\sum\limits_{k=0}^n q^k C_k(q) C_{n-k}(q)$ produces

$$ \left\{1,1,q+1,q^3+q^2+2 q+1,q^6+q^5+2 q^4+3 q^3+3 q^2+3 q+1,q^{10}+q^9+2 q^8+3 q^7+5 q^6+5 q^5+7 q^4+7 q^3+6 q^2+4 q+1\right\} $$

case3----[Carlitz] Mirror symmetrical with case2

qc[0] := 1;
qc[1] := 1;
qc[n_] := 
  Sum[qc[k]*qc[n - 1 - k]*q^((k + 1) (n - 1 - k)), {k, 0, n - 1}];
qc /@ Range[0,5] // Expand

$C_{n+1}=\sum\limits_{k=0}^n C_k C_{n-k} q^{(k+1)(n-k)}$ produces $$ \left\{1,1,q+1,q^3+2 q^2+q+1,q^6+3 q^5+3 q^4+3 q^3+2 q^2+q+1,q^{10}+4 q^9+6 q^8+7 q^7+7 q^6+5 q^5+5 q^4+3 q^3+2 q^2+q+1\right\} $$

I dont't konw such recurrence (1.2) that, \eqref{1.2} is compatible with \eqref{1.1}

$$ C_n(q)=\frac{1}{[n+1]_q}\left[\begin{array}{c} 2 n \\ n \end{array}\right]_q \label{1.1} \tag{1.1} $$

$$ ?\tag{1.2} \label{1.2} $$

ref

https://homepage.univie.ac.at/josef.hofbauer/85jct_catalan.pdf https://mathoverflow.net/questions/263164/generating-q-catalan-numbers?rq=1 https://www2.math.upenn.edu/~jhaglund/books/qtcat.pdf https://www.symmetricfunctions.com/q-analogues.htm#prelimQanalogsCatalan

138 Aspen
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