I saw someone point out recently that

10! = 7! × 5! × 3! × 1!

Are there more examples like this?

What would you call the pattern on the right? I don’t think there’s a standard name, but here’s why I think it should be called double super factorial or super double factorial.

## Super factorial

The factorial of a positive number *n* is the product of the positive numbers up to and including *n*. The super factorial of *n* is the product of the *factorials* of the positive numbers up to and including *n*. So, for example, 7 super factorial would be

7! × 6! × 5! × 4! × 3! × 2! × 1!

## Double factorial

The double factorial of a positive number *n* is the product of all the positive numbers up to *n* with the same parity of *n*. So, for example, the double factorial of 7 would be

7!! = 7 × 5 × 3 × 1.

## Double superfactorial

The pattern at the top of the post is like super factorial, but it only includes odd terms, so it’s like a cross between super factorial and double factorial, hence double super factorial.

Denote the double super factorial of *n* as dsf(*n*), the product of the factorials of all numbers up to *n* with the same parity as *n*. That is,

dsf(*n*) = *n*! × (*n* − 2)! × (*n* − 4)! × … × 1

where the 1 at the end is 1! if *n* is odd and 0! if *n* is even. In this notation, the observation at the top of the post is

10! = dsf(7).

## Super double factorial

We can see by re-arranging terms that a double super factorial is also a super double factorial. For example, look at

dsf(7) = 7! × 5! × 3! × 1!

If we separate out the first term in each factorial we have

(7 × 5 × 3 × 1)(6! × 4! × 2!) = 7!! dsf(6)

We can keep going and show in general that

dsf(*n*) = *n*!! × (*n *− 1)!! × (*n *− 2)!! … × 1

We could call the right hand side super double factorial, sdf(*n*). Just as a super factorial is a product of factorials, a super double factorials is a product of double factorials. Therefore

dsf(*n*) = sdf(*n*).

## Factorials that equal double super factorials

Are there more solutions to

*n*! = dsf(*m*).

besides *n* = 10 and *m* = 7? Yes, here are some.

0! = dsf(0)

1! = dsf(1)

2! = dsf(2)

3! = dsf(3)

6! = dsf(5)

There are no solutions to

*n*! = dsf(*m*)

if *n* > 10. Here’s a sketch of a proof.

Bertrand’s postulate says that for *n* > 1 there is always a prime *p* between *n* and 2*n*. Now *p* divides (2*n*)! but *p* cannot divide dsf(*n*) because dsf(*n*) only has factors less than or equal to *n*.

If we can show that for some *N*, *n* > *N* implies (2*n*)! < dsf(*n*) then there are no solutions to

*n*! = dsf(*m*)

for *n* > 2*N* because there is a prime *p* between *N* and 2*N* that divides the left side but not the right. In fact *N* = 12. We can show empirically there are no solutions for *n* = 11 up to 24, and the proof shows there are no solutions for *n* > 24.