In the book Those Fascinating Numbers by Jean-Marie De Konick they find interesting (or `interesting') things to say about many numbers. I reviewed the book in a SIGACT News book review column here. The entry for 13 is odd: 13 is the third Horse Number. The nth Horse number is the number of ways n horses can finish a race. You might think: OH, that's just n!. AH- horses can tie. So it's the number of ways to order n objects allowing ties.
Is there a closed form for H(n)? We will come back to that later.
0) The Wikipedia Entry on horse races that ended in a dead heat is here. They list 78 dead heats (two horses tie for first place) and 10 triple dead heats (three horses tie for first place). For the horse numbers we care if (say) two horses tie for 4th place. In reality nobody cares about that.
1) I have found nowhere else where these numbers are called The Horse Numbers.
2) They are called the Ordered Bell Numbers. The Wikipedia entry here has some applications.
3) They are also called the Fubini Numbers according to the Ordered Bell Number Wikipedia page.
4) I had not thought about the Horse Numbers for a long time when they came up while I was making slides for the proof that (Q,<) is decidable (the slides are here).
5) There is an OEIS page for the Horse Numbers, though they are called the Ordered Bell Numbers and the Fubini Numbers. It is here. That page says H(n) is asymptotically \(\frac{1}{2}n!(\log_2(e))^{n+1}\) which is approx \(\frac{1}{2}n!(1.44)^{n+1}\).
6) There is a recurrence for the Horse Numbers:
H(0)=1
H(1)=1
H(2)=3
For all \(n\ge 3\) we split H(n) into what happens if i horses are tied for last place (choose i out of n) and if the rest are ordered H(n-i) ways. Hence
\( H(n) = \binom{n}{1}H(n-1) + \binom{n}{2}H(n-2) + \cdots + \binom{n}{n}H(0) \)
Using \(\binom{n}{i} = \binom{n}{n-i}\) we get
\( H(n) = \binom{n}{0}H(0) + \binom{n}{1}H(1) + \cdots + \binom{n}{n-1}H(n-1) \)
STUDENT: Is there a closed form for H(n)?
BILL: Yes. Its H(n).
STUDENT: That's not closed form.
BILL: Is there a closed form for the number of ways to choose i items out of n?
STUDENT: Yes, \(\binom{n}{i}\) or \( \frac{n!}{i!(n-i)!}\)
BILL: Does that let you compute it easily? No. The way you compute \(\binom{n}{i}\) is with a recurrence. The way you compute H(n) is with a recurrence. Just having a nice notation for something does not mean you have a closed form for it.
STUDENT: I disagree! We know what n! is!
BILL: Do not be seduced by the familiarity of the notation.