A strict total order is a set $A$ with a binary relation $<$ on it, satisfying irreflexivity, transitivity and totality.
Could you give an example of a strict total order, please?
[Math] Example of a strict total order
order-theory
Related Solutions
Let $a_n$ denote the answer. I claim that
$$\sum_{n \ge 0} \frac{a_n}{n!} z^n = \frac{1}{2 - e^z}.$$
To see this, write the RHS as $\frac{1}{1 - (e^z - 1)}$, or
$$\sum_{k \ge 0} (e^z - 1)^k.$$
Then I claim that the coefficient of $\frac{z^n}{n!}$ in $(e^z - 1)^k$ is precisely the number of ways to weakly order $n$ objects such that there are $k$ equivalence classes. This is equivalent to the number of ways to partition $n$ objects into $k$ non-empty subsets in order, so you can prove this by using identities for Stirling numbers, but it actually follows directly from basic properties of exponential generating functions: $e^z - 1$ is the generating function for non-empty sets, so $(e^z - 1)^k$ is the generating function for $k$-tuples of non-empty sets.
The generating function above shows that one can't expect, say, a formula in terms of a bounded number of binomial coefficients or anything especially nice like that. However, $\frac{1}{2 - e^z}$ is a meromorphic function, and the pole closest to the origin is at $z = \log 2$. This turns out to imply, after some computations, that
$$\frac{a_n}{n!} \sim \frac{1}{2 (\log 2)^{n+1}}$$
and one can even precisely write down the rest of the terms in the asymptotic expansion. For example, the next poles are at $z = \log 2 \pm 2 \pi i$, so the next term in the asymptotic expansion is $$\frac{a_n}{n!} \sim \frac{1}{2 (\log 2)^{n+1}} + \frac{\cos (n+1) \theta}{r^{n+1}}$$
where $r = \sqrt{(\log 2)^2 + 4 \pi^2}, \theta = \arctan \frac{2\pi}{\log 2}$. It actually follows from the fact that all of the other terms are much smaller than this that the first term in the above estimate has exponentially small error. For example, the largest value in the OEIS is $$\frac{a_{18}}{18!} = 528.793...$$
whereas $$\frac{1}{2 (\log 2)^{19}} = 528.794...$$
For more on these kinds of techniques, see Flajolet and Sedgewick's Analytic Combinatorics.
Two distinct elements are called "comparable" when one of them is greater than the other. This is the definition of "comparable". When you have a partially ordered set, some pairs of elements can be not comparable. i.e. you can have two elements $x$ and $y$ such that $x\leqslant y$ is false and $y \leqslant x$ is also false.
For example, consider the set $\mathbb{R}^2$ and a partial order defined like this: $$ (x_1,x_2) \leqslant (y_1,y_2) \quad\textrm{iff}\quad x_1\leqslant y_1 \,\textrm{and}\,x_2 \leqslant y_2. $$ With this partial order, elements $(0,0)$ and $(1,2)$ of $\mathbb{R^2}$ are comparable, because $(0,0)\leqslant(1,2)$. But elements $(0,1)$ and $(1,0)$ are not comparable, because both statements "$(0,1)\leqslant(1,0)$" and "$(1,0)\leqslant(0,1)$" are false.
Best Answer
Well, here's a simple example. Consider the set $A=\{1,2,3\}.$ We'd like to put a strict order of some kind on this set--the natural choice (pun intended, for those who see it) is to make $1$ the least element and $3$ the greatest. Let's call our relation $\prec$, so we want $1\prec 2,$ $2\prec 3,$ and $1\prec 3$. That is, $$\prec\::=\bigl\{\langle 1,2\rangle,\langle 2,3\rangle,\langle 1,3\rangle\bigr\}.$$ This can readily be shown to be irreflexive, transitive, and total on $A$. We can also take the inverse relation $$\succ\::=\:\prec^{-1}\:=\bigl\{\langle 2,1\rangle,\langle 3,2\rangle,\langle 3,1\rangle\bigr\}$$ to get another total order on $A$, and there are $4$ other total orders on $A$, too. (Can you find them all?)