Subfactorials, Semifactorials and Others

I've written about subfactorials before in a post titled Derangements on Sunday, 13th January 2019 where I explained that:

There are formulae and examples of subfactorials (derangements) aplenty there, so in this post I'm going to focus on what rekindled my interest. It was a YouTube video that I'll link to later and the problem posed was to connect the 3's shown in Figure 1 by mathematical operations so the LHS = RHS.

Figure 1

It wasn't made clear what constituted mathematical operations. Certainly the basic arithmetical operations of addition, subtraction, multiplication, division and exponentiation are sufficient, together with brackets, to balance most of the equations in Figure 1. For example:

• \( (3-3) \times 3 =0\)
• \(3^{3-3} = 1 \)
• \(3+3-3=3 \)
• \( \frac{3}{3}+3=4\)
• \(3 \times 3 - 3=6 \)
• \( 3 + 3 + 3 =9 \)

However, for 5, 7 and 8, the factorial is required as shown in Figure 2 where some alternative solutions for the numbers 0, 1, 2, 3, 4, 6 and 9 are shown:

Figure 2

The final problem of balancing the three 3's with 10 does not yield even to the factorial. It's here that the subfactorial is required. Firstly, let's list the subfactorials, up to ten using SageMathCell (permalink) as shown in Figure 3.

Figure 3

It can be seen that subfactorial 3, which we'll write as !3, is equal to 2 and that's what we need to balance the equation. Specifically:$$3! + !3 +!3=6+2+2=10$$If other mathematical operations such as log, square root and differentiation are allowed, then some other solutions are possible, as shown in Figure 4.

Figure 4

Clearly, the subfactorial offers the simplest solution. Here is the video from which the above screenshots were taken:

So much for subfactorials. What are semifactorials? Well, the semifactorial or double factorial of a number \(n\), denoted by \(n!! \) is the product of all the integers from \(1\) up to \(n\) that have the same parity (odd or even) as \(n\). Thus \(5!! = 5 \times 3 \times 1 \) and \( 4!! = 4 \times 2 \). The factorial function iterated twice should be written as \((5!)! = 6.689503 \times 10^{198}\), a very large number. However, the Google Calculator will return this same result when 5!! is entered and thus does not recognise the semifactorial.

The nomenclature associated with factorials can be quite confusing. For example, when the double factorial or semifactorial is odd, it's sometimes called the odd factorial. Thus the odd factorial of 5 is 5 x 3. To confuse matters even more, there's the falling factorial and the rising factorial and this pair goes by a variety of names as well. The former can be referred to as the descending factorial, the falling sequential product, or the lower factorial while the latter can be referred to as the Pochhammer function, the Pochhammer polynomial, the ascending factorial, the rising sequential product, or the upper factorial. 

Again, there are different ways of representing the falling and rising factorials. The respective different representations and definitions are shown below:$$(x)_n = x^{\underline{n}} = x(x-1)(x-2)\cdots(x-n+1) = \prod_{k=1}^n(x-k+1) = \prod_{k=0}^{n-1}(x-k)$$ $$x^{(n)} = x^{\overline{n}} = x(x+1)(x+2)\cdots(x+n-1) = \prod_{k=1}^n(x+k-1) = \prod_{k=0}^{n-1}(x+k)$$Clearly, there's some need for standardisation in the world of factorials. Anyway, let's apply these different sorts of factorials to the number 5, just to highlight the differences:

• factorial \(5\) is represented as \(5! =120\)
• factorial \(5\) iterated twice is represented as \((5!)! = 6.689503 \times 10^{198}\)
• double factorial \(5\) is represented as \(5!! = 5 \times 3 \times 1 = 15 \)
• \( 5^{\underline{3}}\) would be a falling factorial, equal to \(5 \times 4 \times 3 = 60\)
• \( 5^{\overline{3}}\) would be a rising factorial, equal to \(5 \times 6 \times 7 = 210\)

Returning to the 3 3 3 = 10 problem, the rising factorial can be used to provide another solution, namely:$$\frac{3^{\overline{3}}}{3!}=\frac{3 \times 4 \times 5}{3 \times 2 \times 1}=\frac{60}{6}=10$$