The problem with these bounds is that they're tricky to check. A computer can be programmed to do the computations, but they're awkward with just a hand-held calculator.

The critical computation involves computing some combinations. For example, if you have five numbered balls, one can ask how many ways there are to pick two of them. You could pick balls 1 and 2, or 1 and 3, or 1 and 4, or 1 and 5, or 2 and 3, or 2 and 4, or 2 and 5, or 3 and 4, or 3 and 5, or 4 and 5. That makes 10 possible ways to pick two balls out of five. If you pick ball 2 first and then ball 5, we count that as being the same as if you picked ball 5 first, and then ball 2, as you end up with the same two balls either way.

More generally, if you have n balls and want to pick k of them, there are n ways to pick the first ball. There are then n-1 ways to pick the second ball, as you can't pick the first one again, but can pick any of the other n-1 balls. There are n-2 ways to pick the third, n-3 ways to pick the fourth, and so forth. In general, there are n(n-1)(n-2)...(n-k+1) ways to pick the k balls in order.

However, this counts picking the same balls in different orders as separate. If you pick some particular set of k balls, then there are k! = k(k-1)(k-2)...(2)(1) different orders in which you could have picked them. We thus divide by k! to get that there are n(n-1)(n-2)...(n-k+1)/(k!) = n!/(k!(n-k)!) ways to pick the k balls out of n. Let's call this number C(n, k) for convenience.

Now suppose that you have some balls, and you want to pick four of them. You want for there to be at most 1000 ways to pick four balls out of what you have. What is the most balls that you could possibly have, without there being more than 1000 ways to pick four of them?

One could start by guessing, and then moving up or down. If we have eight balls, we can compute that there are C(8, 4) = 70 ways to pick four of them. That's far under 1000, so maybe we could have more. If we have twelve balls, there are C(12, 4) = 495 ways to pick four of them. That's still well under 1000, so let's try more balls. We can compute C(16, 4) = 1820, so 16 balls is too many. Thus, we can have as many as 12, but can't have 16, so the maximum must be between 12 and 15. We can try C(14, 4) = 1001, which is barely too many. We try C(13, 4) = 715, which is well under 1000. Thus, we can have 13 balls, but not 14, so the number we're looking for is 13.

As you can see, that's a bit of a pain to compute, but it's doable. The problem is that to check the bounds of the Kruskal-Katona theorem, we don't have to find the answer to this sort of problem just once. We have to do it a bunch of times. If there are k numbers in our list, then just to compute the bounds for the last pair of integers, we have to do the above problem once with k-1 balls, once with k-2 balls, once with k-3 balls, and so forth. This is the sort of thing that computers can readily handle, but it's a pain to do by hand.

Lovasz gave a pretty good approximation to the Kruskal-Katona theorem, which says that we only have to compute one constant as above. His bounds are much nicer to state than the full Kruskal-Katona theorem, though they're not quite as strong, so they only say, in effect, "the right bound is close to this, but a little bit bigger".

The problem with Lovasz's bounds is that they're still not that easy to compute. Instead of having discrete computations as above, we have to find the largest real root of a polynomial of degree k. If k = 2, then this is easy by the quadratic formula. For example, if we want to find the largest real root of x(x-1) = 10, we can use the quadratic formula to get that the roots are 1/2 + sqrt(41)/2 and 1/2 - sqrt(41)/2. The former is larger, so that's what we're looking for.

But k can be arbitrarily large. For a polynomial of degree 3 or 4, there are much messier formulas to find the exact roots. For polynomials of higher degree, no such formulas exist. We can't, for example, find an exact formula for the largest real root of x(x-1)(x-2)(x-3)(x-4)(x-5) = 267840. We can use computers to solve numerically and get a very close approximation to the largest real root, as the function will always be nicely behaved near the value that we're looking for (i.e., no double roots or complex roots with large real part). But if you're going to have a computer do the computations, then you might as well just get the exact numbers. Furthermore, "the computer said such and such" doesn't give very good intuition about how big the numbers ought to be.

I improve on this in two ways. First, I give a bound that doesn't require any messy computations. It's just a simple plug-and-chug formula, like plugging numbers into the quadratic formula to see what it says. This bound isn't quite as good as Lovasz's bound, but Lovasz's bound is usually closer to my bound than to the sharp bounds of the Kruskal-Katona theorem, so my bound is still pretty good.

Second, I give a bound that requires computing only one constant as in the sharp bounds of the Kruskal-Katona theorem, rather than arbitrarily many of them. This avoids having to find roots of polynomials, but works exactly the same as above. This second bound is usually better than Lovasz's bound, while also being much easier to compute. It still isn't as good as the sharp bounds of the Kruskal-Katona theorem, of course, since those are the exact numbers that we're trying to approximate.