The Erdős primitive set conjecture is that the following summation:$$ \sum _{n \, \in A} \frac {1}{n\log{n}}$$where A is any primitive set (a set where no member of the set divides another member) attains its maximum at the set of primes numbers. It was proved by Jared Duker Lichtman (pictured above) in 2022. I was informed of this via a YouTube video first released in 2022. Here is a link to the academic paper by Lichtman. The constant turns out to be about1.6366 ... and summations of the form shown above can be no larger than this. Thus we have:$$ \sum _{p \, \in P} \frac {1}{p\log{p}} \approx 1.6366$$where P is the set of primes and \(p\) is any prime number.
Let's take the summation of the elements of the set S of semiprimes. These elements form a primitive set. Let's suppose each semiprime can be represented by its prime factors \(p\) and \(q\) where \(p \leq q\). We then have:$$ \sum _{pq \, \in S} \frac {1}{pq\log {pq}} \approx 1.1448 \dots$$We can continue this process and consider the primitive set containing all numbers with three not necessarily distinct prime factors and so on. In each case, the sum converges to a constant which can be designated as \(f_k\) where \(k\) represents the number of prime factors. Thus we've seen that \(f_1 \approx 1.6366\) and \(f_2 \approx 1.1448\). In general we write:$$f_k=\sum \frac {1}{n\log{n}}\\ \text{where } n \text{ has } k \text{ prime factors}$$The long term behaviour of \(f_k\) is shown in Figure 1 where the term "fingerprint numbers" is used to identify these types of numbers:
While the values of \(f_k\) initially decrease and drop below 1 for \(f_3\), it can be seen that the values bottom out around \(f_5\) and \(f_6\) and then increase slowly as they approach a value of 1 asymptotically from below. This behaviour can be seen more clearly in Figure 2.
The term "fingerprint numbers" is an informal term used in the referenced videos and is not used formally in mathematical circles. There are many different types of primitive sets but they all have the property that no member of the set divides another member.
On February 2nd 2018, I made a short post titled The Mathematics of Music and this current post builds on the content that I first introduced there.
The idea of doing a series of posts on Mathematics and Music occurred to me recently and one of the first and most basic topic in this regard were the ratios involved in the Western musical scales. I found a very good resource for this topic titled Why 12 notes to the Octave and the author begins with this statement:
The Greeks realised that sounds which have frequencies in rational proportion are perceived as harmonious. For example, a doubling of frequency gives an octave. A tripling of frequency gives a perfect fifth one octave higher. They didn't know this in terms of frequencies, but in terms of lengths of vibrating strings. Pythagoras, who experimented with a monochord, noticed that subdividing a vibrating string into rational proportions produces consonant sounds. This translates into frequencies when you know that the fundamental frequency of the string is inversely proportional to its length, and that its other frequencies are just whole number multiples of the fundamental.
The key point is that "sounds which have frequencies in rational proportion are perceived as harmonious" and the most important of these ratios is 3:2. The author continues:
The chromatic scale reflects this fact. In the 18th and 19th centuries, the chromatic scale was tuned using the idea of 3/2. In the most elegant of these, Thomas Young's tuning, several of the fifths were set exactly to 3/2, and the others were tempered slightly (to make octaves exact).
In the modern equal temperament (which came into practical use during the early part of the 20th century), all fifths are tuned to 2^(7/12)=1.49651..., slightly less than 3/2, and 12 repetitions of this ratio gets us back to where we started (after dropping down 7 octaves).
Of the various intervals, the only ones that are really well captured by tempered versions of the 3/2 scheme are: unison, 5th, major 2nd, and their reciprocals (octave, 4th, minor 7th).
The author then asks two key questions:
Why 3/2? The choice of 3/2 says that, next to the octave, it should be regarded as the most important interval.
Why do 12 steps work nicely? Interestingly, this can be explained in terms of simple number theory, namely continued fractions.
Ah, continued fractions! This is where the Mathematics comes in. The author remarks that it is necessary to understand when a power of 3/2 will be close to a power of 2 (because 2 represents an octave and we want a power of 2 that will be close to a power of 3/2). So we set an equation:$$\begin{align} \left ( \frac{3}{2}\right )^a &=2^b \text{ where }a \text{ and }b \text{ are natural numbers}\\\frac{3}{2} &=2^{\frac{b}{a}}\\&=2^x \text{ where }x \text{ is a real number}\\ x&=\frac{\log \left ( \frac{3}{2} \right )}{\log(2)}\\
&\approx 0.584962500721 \dots \end{align}$$There are no rational values of \(a\) and \(b\) that satisfy the equation which is why it is necessary to approximate with a real number \(x\). The continued fraction approximations to \(x\) are shown in the SageMath code in Figure 1 with permalink included:
We see that 7/12 gives a reasonable approximation (0.5833333... versus 0.5849625...). If we start with the octave between note A3 (220 Hz) and A4 (440 Hz) and divide it into 12 semitones according to \(220 \times 2^{k/12}\) where \(k=0 \dots 12\), we get what's shown in Figure 2.
As the author of Why 12 notes to the Octave remarks, there are other possible divisions and one of them is into 19 parts because 11/19 = 0.578947... is pretty close to 0.5849625... and this produces the situation shown in Figure 4 where octave is divided into 19 "semitones" according to \(220 \times 2^{k/19}\) where \(k=0 \dots 19\).
Let's remember that the above scales, and in fact nearly all modern scales, use equal temperament. As Wikipedia explains:
There are two main families of tuning systems: equal temperament and just tuning. Equal temperament scales are built by dividing an octave into intervals which are equal on a logarithmic scale, which results in perfectly evenly divided scales, but with ratios of frequencies which are irrational numbers. Just scales are built by multiplying frequencies by rational numbers, which results in simple ratios between frequencies, but with scale divisions that are uneven.
This is a big topic and I've only scratched the surface of it. More later.
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:$$\frac{3^{\overline{3}}}{3!}=\frac{3 \times 4 \times 5}{3 \times 2 \times 1}=\frac{60}{6}=10$$
