Combinatorial sequences whose ratios $a_{n+1}/a_{n}$ are integers I have a proof technique in search of examples. I'm looking for combinatorially meaningful sequences $\{a_n\}$ so that $a_{n+1}/a_n$ is known or conjectured to be an integer, such that there is a relation between the $n$th case and $n+1$st, but not an obvious $a_{n+1}/a_n\to 1$ map. This means $a_n$ is the $n$th partial product of an infinite sequence of integers, but there isn't an obvious product structure.


*

*The prototype was an enumeration of
domino tilings of an Aztec
diamond of order $n$, $a_n =
   2^{n(n+1)/2}$, so $a_{n+1}/a_n =
   2^{n+1}$. (There is a nice $2^{n+1}$
to 1 map unrelated to my technique,
but it isn't obvious.)  

*Another application was a proof that
$\det \{B_{i+j}\}_{i,j=0}^n =
   \prod_{i=1}^n i! $  where $B_n$ is
the $n$th Bell number, equation
25 in the linked page.

*The counts of alternating sign matrices 1, 2, 7, 42, ... are not an example, since 
$ASM(n+1)/ASM(n) = \frac{ (3n+1)!n!}{2n! (2n+1)!}$ which is not always an integer, e.g,     7/2 is not.

What are some other interesting combinatorial families whose ratios $a_{n+1}/a_n$ are known or (preferably) conjectured to be integers?

Thanks.
 A: There are 345 sequences in the OEIS qith the word “quotient” in their names, does that help?
A: Let an be the largest power of 2 that divides Rn, the number of reduced Latin squares of order n.  We know the value of an for n≤11 (see this for example).  The sequence begins (1,1,1,22,23,26,210,217,221,228,235,...) for n≥1.
I wouldn't conjecture that an+1/an is always an integer (although, it seems plausible).  However, we do know that an+1/an is an integer for 1≤n≤10.
A: The number of $n$-block domino towers is $3^{n-1}$. The simplest bijective proof uses the fact that for the Motzkin generating function $M=M(z)$ we have $$\frac{M}{(1-zM)^2}=\frac{1}{1-3z}.$$ See pages 19-21 of this paper (and the references there): 
http://math.sfsu.edu/federico/Clase/AC/algmethods.pdf
(This is Chapter 1 in the Handbook of Enumerative Combinatorics.)
Likewise, the number of pairs of grand Dyck paths of combined semilength $n$ is $4^n$.
A: The number of pairs $(P,Q)$ of standard Young tableaux of the same shape and with $n$ squares is $n!$.
The number of oscillating tableaux of length $2n$ and empty shape is $1\cdot 3\cdot 5\cdots (2n-1)$.
The number of leaf-labeled complete (unordered) binary trees with $n$ leaves is $1\cdot 3\cdot 5\cdots (2n-3)$ (Schröder's third problem).
The number of compact-rooted directed animals of size $n$ is $3^n$. See MathSciNet MR0956559 (90c:05009).
Let $f(n)$ be the number of $n\times n$ matrices $M=(m_{ij})$ of nonnegative integers with row and column sum vector $(1,3,6,\dots,{n+1\choose 2})$ such that $m_{ij}=0$ if $j>i+1$. Then $f(n)=C_1C_2\cdots C_n$, where $C_i$ is a Catalan number. No combinatorial proof of this result is known. See Exercise 6.C12 on page 38 (solution on page 84) of http://math.mit.edu/~rstan/ec/catadd.pdf
A: The number of acyclic orientations on the complete graph on $n$ vertices is $n!$, and the number of acyclic orientation on a unit-interval graph on $n$ vertices is given by a product with $n$ factors, so you can easily construct many sequences from this.
A: If you take a(n)=2^F(n);a(n+1)=2^F(n+1) , where F(n+1) ,F(n) are Fibonacci numbers and F(n+1)=F(n)+F(n-1) , then a(n+1)/a(n)=2^[F(n+1)-F(n)]=2^F(n-1)
Fibonacci series S=[n_F(n)]=(0_0),(1_1),(2_1),(3_2),(4_3),(5_5),(6_8),(7_13),(8_21),(9_34)...
Product {a(n+1)/a(n)}{a(n)/a(n-1)}...{a(3)/a(2)}{a(2)/a(1)}=a(n+1)/a(1)
       =2^{F(n-1)+F(n-2)+...+F(1)+F(0)} =2^{F(n+1)-1}=2^F(n+1)/2
Example=>
(2^21/2^13)(2^13/2^8)(2^8/2^5)(2^5/2^3)(2^3/2^2)(2^2/2^1)(2^1/2^1)=2^21/2
 =(2^8)(2^5)(2^3)(2^2)(2^1)(2^1)=2^(8+5+3+2+1+1)=2^20
I hope this  example   suits to your needs.
