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14 votes
1 answer
2k views

When does a Catalan number equal a Fibonacci number?

The $n=3$'rd Catalan number (A000108) is $1,1,2,5$ : $\frac{\binom{2n}{n}}{n+1}=\frac{\binom{6}{3}}{4}=\frac{20}{4}=$ 5. The $n=4$'th Fibonacci number (A000045) is $1,1,2,3,5,...$ : 5. Q. Which ...
Joseph O'Rourke's user avatar
9 votes
2 answers
709 views

Oddity of generalized Catalan numbers: Part I

The famous (classical) Catalan numbers $C_{1,n}=\frac1{n+1}\binom{2n}n$ satisfy the following well-known arithmetic property: $$\text{$C_{1,n}$ is odd iff $n=2^j-1$ for some $j$}.\tag1$$ Consider the "...
T. Amdeberhan's user avatar
9 votes
3 answers
1k views

Intuition Behind a Decimal Representation with Catalan Numbers

From $0 = 0.5 - 0.5 = 0.5 - \sqrt{0.25}$, we can adjust the subtrahend slightly to obtain $$0.5 - \sqrt{0.249} = 0.001\ 001\ 002\ 005\ 014\ 042\ldots$$ where the decimal representation contains the ...
Benjamin Dickman's user avatar
9 votes
1 answer
470 views

Products of Catalan numbers

Let $c(n)=\frac{1}{n+1}\binom{2n}{n}$ be the Catalan number. It seems that a product $\prod_{n\in I} c(n)$, where $I\subset\mathbb N_{>1}$, is never a Catalan number. Is this a (known) fact?
Martin Rubey's user avatar
  • 5,792
8 votes
2 answers
810 views

Modular congruences related to sums of Catalan numbers

I am curious if somebody can be helpful concerning the following experimental observation: There exist two rational sequences $\alpha_0,\alpha_1,\dots$ and $\beta_0,\beta_1,\dots$, both with values ...
Roland Bacher's user avatar
7 votes
4 answers
3k views

A generalization of Catalan numbers

It is well-known that the $n$th Catalan number is equal to $(n+1)^{-1}\binom{2n}{n}$. A long time ago I had wondered what happens if you look at the sequence generated by $(n+k)^{-1}\binom{pn}{n}$ - ...
Felix Goldberg's user avatar
6 votes
1 answer
281 views

Another generalization of parity of Catalan numbers

Recently, a question by T. Amdeberhan gathered up many enjoyable proofs that a Catalan number $C_n$ is odd if and only if $n=2^r-1$. Noam D. Elkies' answer considered $F=\sum_{n=0}^\infty C_n x^{n+1}$....
user196574's user avatar
5 votes
1 answer
613 views

generating $q$-Catalan numbers

An $n$-Dyck path (or a Catalan path) is a lattice path $P$, unit East and North steps, in an $n\times n$ square grid which stays (weakly) above the main diagonal. Let $\square_n$ denote all such paths....
T. Amdeberhan's user avatar
3 votes
1 answer
193 views

Sequences that sums up to second differences of Bell and Catalan numbers

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Let $g(n)$ be A025480, $g(2n) = n$...
Notamathematician's user avatar
3 votes
1 answer
224 views

"Oddity" of $q$-Catalan polynomials: Part II

This question extends my earlier MO post for which I'm grateful for answers and useful comments. The Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ satisfy: $\text{$C_{1,n}$ is odd iff $n=2^j-1$ for ...
T. Amdeberhan's user avatar
2 votes
1 answer
147 views

$R$-recursion for Fibonacci numbers using signed Catalan numbers

Let $F_n$ be A000045 (i.e., Fibonacci numbers). Here $$ F_n = F_{n-1} + F_{n-2}, \\ F_0 = 0, F_1 = 1. $$ Let $C_n$ be A000108 (i.e., Catalan numbers). Here $$ C_n = \frac{1}{n+1}\binom{2n}{n}. $$ Let $...
Notamathematician's user avatar
1 vote
1 answer
257 views

A divisibility problem involving Catalan numbers

The Catalan numbers in combinatorics are given by $$C_n=\frac1{n+1}\binom{2n}n=\binom{2n}n-\binom{2n}{n+1}\ \ (n=0,1,2,\ldots).$$ In 2014 I formulated the following conjecture. Conjecture. For each $...
Zhi-Wei Sun's user avatar
  • 15.6k