Persistent Tau Numbers

I got to thinking about tau numbers, those numbers that are divisible by their number of divisors, because it's relevant to my divisors sequence which arises when these rules are applied recursively to a positive integer. The rules for a number \(n\) are:

Let \(d\) be the number of divisors of \(n\)

• if \(d \,| \,n \) then \( n \rightarrow d \, | \,n \)
• if \( d \, \nmid \, n\) then \(n \rightarrow d \times n\)
• if \(d\) is prime then:
• if \(d\) is a \(4k+1 \) prime then \( d \rightarrow 2 \times d +1\)
• if \(d\) is a \(4k+3 \) prime then \( d \rightarrow \dfrac {d -1}{2}\)

I'll get back to that sequence later but what interested me were those tau numbers that remain tau numbers after being divided by their number of digits. I got Gemini to investigate and the results revealed that there are 160 such numbers in the range from 1 to 40000 (permalink) and they are:

1, 2, 8, 12, 80, 96, 108, 240, 384, 640, 720, 864, 896, 1200, 1408, 1440, 1664, 1920, 2016, 2176, 2240, 2432, 2560, 2688, 2944, 3168, 3456, 3600, 3712, 3744, 3968, 4032, 4224, 4736, 4896, 4992, 5040, 5248, 5472, 5504, 6016, 6528, 6624, 6784, 7296, 7552, 7808, 7920, 8100, 8352, 8576, 8832, 8928, 9088, 9344, 9360, 10112, 10624, 10656, 10800, 11136, 11392, 11808, 11904, 12240, 12384, 12416, 12928, 13184, 13536, 13680, 13696, 13952, 14208, 14464, 15264, 15744, 16256, 16512, 16560, 16768, 16992, 17536, 17568, 17792, 17920, 18048, 19072, 19296, 19328, 20096, 20160, 20352, 20448, 20864, 20880, 21024, 21376, 22144, 22320, 22656, 22752, 22912, 23040, 23168, 23424, 23904, 24300, 24448, 24704, 25216, 25472, 25632, 25728, 26244, 26640, 27008, 27264, 27936, 28032, 28224, 28544, 29056, 29088, 29312, 29520, 29664, 29824, 30000, 30336, 30592, 30720, 30816, 30848, 30960, 31392, 31872, 32128, 32544, 32896, 33664, 33840, 34176, 34432, 34560, 34688, 35456, 35968, 36224, 36576, 37248, 37504, 37728, 38160, 38400, 38784, 39296, 39456, 39552, 39808

Most survive only one division but some go further and these are (in the given range):

• \(80 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(96 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(240 \rightarrow 12 \rightarrow 2 \rightarrow 1 \)
• \(2240 \rightarrow 80 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(3456 \rightarrow 108 \rightarrow 9 \rightarrow 3 \)
• \(3600 \rightarrow 80 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(4032 \rightarrow 96 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(23040 \rightarrow 384 \rightarrow 24 \rightarrow 3 \)
• \(30720 \rightarrow 640 \rightarrow 40 \rightarrow 5 \)
• \(38400 \rightarrow 640 \rightarrow 40 \rightarrow 5 \)

I discussed the divisor sequence in my post titled Number's Divisors to Sequence Algorithm. Applied to 23040 (one of the numbers listed in the bullet points above), the sequence is very short indeed: 23040, 384, 24, 3, 1 Once the sequence reaches 3 then, because it is a \(4k+3\) prime, 3 becomes 1 and 1 has one divisor and so 1 times 1 or 1 divided by 1 is still 1.

Only 5.257% of the numbers up to 100,000 are tau numbers, so the divisors to sequence algorithm is strongly weighted toward multiplication. A composite number is about 19 times more likely to get multiplied than it is to get divided. The primes strike a balance because of the \(4k+1\) versus \(4k+3\) rule. The rule is actually not needed as primes always have two divisors. I introduced the prime rule because in the two factors to sequence algorithms (using prime factors with and without multiplicity), the primes have no prime factors.

What if instead of multiplying by the number of divisors, we simply added 1? It would be interesting to see what sort of sequence develops if we do this.

28241: Analysis of an Interesting Number

My diurnal age today, 28241, has a number of interesting properties and a couple of them I've not met before. Let's examine them one by one.

FIRST INTERESTING PROPERTY

28241 can be formed by a concatenation of powers of 2 since all its digits are powers of 2 but the digits of its prime factors are all powers of 3:$$ \begin{align} 28241 & \equiv 2^1 \, || \, 2^3 \, || \, 2^1 \, || \, 2^2 \, || \, 2^0 \\ 28241 &= 31 \times 911 \\ &\equiv 3^1 \, || \, 3^0 \times 3^2 \, || \, 3^0 \, || \, 3^0 \end{align}$$In the range up to 40000, this doesn't happen often (ignoring 1 and 121 whose factors contain only the digits 1):$$ \begin{align} 81 &= 3^4  \\ 1441 &= 11 \times 131 \\1881 &= 3^2 \times 11 \times 19 \\ 21241 &= 11 \times 1931 \\ 28241 &= 31 \times 911 \end{align}$$SECOND INTERESTING PROPERTY

28241 is a member of OEIS A227942:

A227942: semiprimes formed by inserting a semiprime between the semiprime's ordered factors.

In the range up to 40000, only the following numbers qualify:

393, 2105, 2147, 3155, 5255, 5357, 25829, 26231, 28241, 29447, 33913, 35719, 39331$$ \begin{align} 3 \textbf{9} 3 &= 3 \times 131 \\ 2 \textbf{10} 5 &= 5 \times 421 \\ 2 \textbf{14} 7 &= 19 \times 113 \\ 3 \textbf{15} 5 &= 5 \times 631 \\ 5 \textbf{25} 5 &= 5 \times 1051 \\ 5 \textbf{35} 7 &= 11 \times 487 \\ 2 \textbf{58} 29 &= 23 \times 1123 \\ 2 \textbf{62} 31 &= 17 \times 1543 \\ 2 \textbf{82} 41 &= 31 \times 911 \\ 2 \textbf{94} 47 &= 11 \times 2677 \\ 3 \textbf{39} 13 &= 11 \times 3083 \\ 3 \textbf{57} 19 &= 23 \times 1553 \\ 3 \textbf{93} 31 &= 37 \times 1063 \end{align}$$THIRD INTERESTING PROPERTY

28241 is a member of OEIS A367337:

A367337: member of the commas sequence  starting with 5.

I'm familiar with the commas sequences and have written about them in a post titled The Commas Sequence in December of 2023. I won't show the full sequence but this snippet gives an idea of what's going on:$$ \dots, 28077, 28149, 28241, 28253, 28285, \dots $$Up to 40000, the remaining members of the sequence are:

28253, 28285, 28337, 28409, 28501, 28513, 28545, 28597, 28669, 28761, 28773, 28805, 28857, 28929, 29021, 29033, 29065, 29117, 29189, 29281, 29293, 29325, 29377, 29449, 29541, 29553, 29585, 29637, 29709, 29801, 29813, 29845, 29897, 29969, 30062, 30085, 30138, 30221, 30234, 30277, 30350, 30353, 30386, 30449, 30542, 30565, 30618, 30701, 30714, 30757, 30830, 30833, 30866, 30929, 31022, 31045, 31098, 31181, 31194, 31237, 31310, 31313, 31346, 31409, 31502, 31525, 31578, 31661, 31674, 31717, 31790, 31793, 31826, 31889, 31982, 32005, 32058, 32141, 32154, 32197, 32270, 32273, 32306, 32369, 32462, 32485, 32538, 32621, 32634, 32677, 32750, 32753, 32786, 32849, 32942, 32965, 33018, 33101, 33114, 33157, 33230, 33233, 33266, 33329, 33422, 33445, 33498, 33581, 33594, 33637, 33710, 33713, 33746, 33809, 33902, 33925, 33978, 34061, 34074, 34117, 34190, 34193, 34226, 34289, 34382, 34405, 34458, 34541, 34554, 34597, 34670, 34673, 34706, 34769, 34862, 34885, 34938, 35021, 35034, 35077, 35150, 35153, 35186, 35249, 35342, 35365, 35418, 35501, 35514, 35557, 35630, 35633, 35666, 35729, 35822, 35845, 35898, 35981, 35994, 36037, 36110, 36113, 36146, 36209, 36302, 36325, 36378, 36461, 36474, 36517, 36590, 36593, 36626, 36689, 36782, 36805, 36858, 36941, 36954, 36997, 37070, 37073, 37106, 37169, 37262, 37285, 37338, 37421, 37434, 37477, 37550, 37553, 37586, 37649, 37742, 37765, 37818, 37901, 37914, 37957, 38030, 38033, 38066, 38129, 38222, 38245, 38298, 38381, 38394, 38437, 38510, 38513, 38546, 38609, 38702, 38725, 38778, 38861, 38874, 38917, 38990, 38993, 39026, 39089, 39182, 39205, 39258, 39341, 39354, 39397

FOURTH INTERESTING PROPERTY

28241 has the property that:

• sum of digits ( \(  \, \text{SOD}  \, \)  ) is prime: \(2 + 8 + 2 + 4 + 1 =17 \)
  
  
• sum of digits squared (  \( \, \text{SOD}^2 \, \) ) is prime: \(2^2 + 8^2 + 2^2 + 4^2 + 1^2 =89 \)
  
  
• sum of digits cubed ( \( \, \text{SOD}^3 \, \) ) is prime:  \(2^3 + 8^3 + 2^3 + 4^3 + 1^3 = 593 \)

This property is relatively common but nonetheless interesting.

Finding Numbers Within The Copeland-Erdos Constant

I watched this video recently about the Copeland-Erdos constant.

There following text was included below the video:

Write the primes in a row — 2, 3, 5, 7, 11, 13, and on forever — put a decimal point in front, and you get a single number: 0.235711131719232931… the Copeland–Erdős constant. In 1946, Copeland and Erdős proved it is normal in base ten: every finite string of digits appears in its expansion, and appears exactly as often as pure chance would predict — each digit one tenth of the time, each pair one hundredth, every block its fair share. So the opening of pi (314159) is in there, at digit 6,903. So is 42, at digit 214. So is every birthday, phone number, and message ever written.

What makes this remarkable: normality is the rule, not the exception — a randomly chosen real number is normal with probability one (Borel, 1909) — yet we still cannot prove that pi, e, or the square root of two are normal. The first explicit example was Champernowne's constant (0.123456789101112…), normal because it was built by pasting in every integer on purpose. The primes do no such favor: they thin out, follow no pattern, and were never designed to be tidy — and their concatenation is normal anyway. This video builds the number, defines normality, shows why the famous constants resist proof, and walks the intuition (via the prime number theorem) for why the patternless primes still even out in the limit.

I was interested in determining if and where a particular number, let's say my diurnal age today, might occur within this constant. Gemini created a program to do this, although I've written similar programs in the past myself for finding numbers within other constants like \( \pi\), \(e\) and \( \sqrt{2}\). Here is a permalink for the Gemini program. I've incorporated this program into my daily number analysis. Here is the output for the number associated with my diurnal age today: 28239.

The position in the Copeland-Erdos constant

Target '28239' first appears at position: 61014

What's of interest is whether a number can be equal to its index. I got Gemini to investigate and here is what it came up with:

While no fixed point where the 1-based index $P(N)$ equals $N$ itself has been found among smaller integers in the Copeland–Erdős constant, number-theoretic density principles dictate that such numbers must exist among larger composite numbers, even though they are mathematically impossible among prime numbers.

To see why this dichotomy exists, we have to look at the growth rate of the string indices for primes versus composite numbers.

Why Primes Never Equal Their Index

For any prime number $p$, its "natural" appearance occurs when $p$ itself is concatenated into the decimal stream. The 1-based index where $p$ begins is equal to $1$ plus the total number of digits of all primes strictly smaller than $p$:

$$P_{\text{natural}}(p) = 1 + \sum_{q < p} \lfloor \log_{10}(q) + 1 \rfloor$$

Using Chebyshev's function $\theta(p) = \sum_{q \le p} \ln(q)$, which by the Prime Number Theorem is asymptotically equivalent to $p$, we can evaluate the sum of the base-10 logarithms of all primes up to $p$:

$$\sum_{q < p} \log_{10}(q) = \frac{1}{\ln 10} \sum_{q < p} \ln(q) \sim \frac{p}{\ln 10} \approx 0.4343 \times p$$

Because $\frac{1}{\ln 10} \approx 0.4343 < 1$, the cumulative number of digits written into the stream grows at less than half the rate of the prime numbers themselves. By the time we concatenate a large prime $p$, its starting position is only around the $0.4343p$ mark. If $p$ happens to appear even earlier as an accidental substring across prime boundaries, its index is smaller still.

Therefore, for all primes, $P(p)$ is strictly less than $p$, making a fixed point $P(p) = p$ impossible.

Why Composites Can Form Fixed Points

Composite numbers behave completely differently because they never get a "natural" turn in the prime concatenation. They can only appear in two ways:

1. As an accidental boundary string formed across the junction of two or more adjacent primes (for example, 4 first appears at index 21, spanning across the end of $3\mathbf{7}$ and the start of $\mathbf{4}1$).
2. As a substring entirely within a much larger prime $q > N$.

Because the Copeland–Erdős constant is normal in base 10 (proven by Copeland and Erdős in 1946), every finite decimal string is guaranteed to appear eventually. However, the expected waiting time to see a specific $k$-digit composite string by chance in a normal sequence is roughly $10^k \ln 10 \approx 2.3 \times 10^k$.

Because $N$ itself sits between $10^{k-1}$ and $10^k$, a typical composite number that does not get an early boundary match will appear at an index much larger than $N$.

The Oscillation of $P(N) - N$

We can see this sharp split by examining the first few integers in the constant (OEIS A229190):

Integer $N$	Index $P(N)$	Type	Behavior
2	1	Prime	$P(N) < N$
3	2	Prime	$P(N) < N$
4	21	Composite	$P(N) > N$
5	3	Prime	$P(N) < N$
6	31	Composite	$P(N) > N$
7	4	Prime	$P(N) < N$
8	41	Composite	$P(N) > N$
9	12	Composite	$P(N) > N$
10	47	Composite	$P(N) > N$

Notice how consistently the primes lag behind their own values ($P(N) < N$), while the composite numbers overshoot ($P(N) > N$).

As we examine larger numbers, "lucky" composite numbers will occasionally appear unusually early due to boundary concatenations, while "unlucky" ones will appear late. Because the difference function $f(N) = P(N) - N$ oscillates wildly between negative values (for primes and early-matching composites) and large positive values (for late-matching composites), the graph of $P(N)$ repeatedly jumps across the line $Y = N$.

While finding a specific integer where $P(N) = N$ requires extensive string-searching through millions of digits, this continuous crossing confirms that fixed points exist among composite numbers.

Circles Inscribed In Triangles

I watched an interesting video that shows the area of a circle inscribed in a 3 : 4 : 5 right-angled triangle is equal to \( \pi \).

This got me thinking about right-angled triangle in general and I noted in the comments to the video that for a right-angled triangle with sides of \(a, b, c\) where \(c\) is the hypotenuse, the radius \(r\) of the inscribed circle is given by:$$r=\frac{a+b-c}{2}$$One can see that when \(a=3\), \(b=4\) and \(c=5\), we get:$$ r =\frac{3+4-5}{2} = \frac{2}{2}=1$$The area of the inscribed circle is thus \( \pi\) square units. So I investigated the general case, for any triangle with sides of \(a\), \(b\) and \(c\), and this is the explanation and formula that Gemini came up with:

Inscribed Circles in Any General Triangle

The area of the inscribed circle (the incircle) can be expressed purely in terms of the side lengths $a$, $b$, and $c$ for any triangle, whether it is right-angled, acute, or obtuse.

To do this, we use two foundational concepts in geometry: Heron's Formula and the relationship between a triangle's area and its inradius.

Step 1: Define the Semi-perimeter

For any triangle with sides $a$, $b$, and $c$, we first calculate the semi-perimeter (half of the perimeter), usually denoted as $s$:

$$s = \frac{a + b + c}{2}$$

Step 2: The Radius Formula

The radius $r$ of an inscribed circle in any triangle is equal to the area of the triangle divided by its semi-perimeter. Using Heron's Formula for the area of the triangle, this gives us:

$$r = \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$$

Which simplifies algebraically to:

$$r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$

Step 3: Calculate the Area of the Incircle

Now that we have the radius purely in terms of the sides (via $s$), we just plug it into the circle area formula ($A = \pi \, r^2 \,$). The square root neatly disappears:

$$A = \pi \left( \frac{(s-a)(s-b)(s-c)}{s} \right)$$

Summary of the General Formula

If you want to write the area of the inscribed circle entirely in a single expression using just $a$, $b$, and $c$ (without using the $s$ substitution), it looks like this:

$$A = \pi  \left( \frac{(b+c-a)(a+c-b)(a+b-c)}{4(a+b+c)} \right)$$

Both the formula using $s$ and the expanded algebraic formula above will work perfectly for any valid triangle.

Golomb's Tromino Theorem

When reading this Quanta article about the Fields Medal recipient Yu Deng, I came across this paragraph:

On some of those hikes, his father gave him math problems to solve. He still remembers one: You have a square grid divided into a certain number of cells. Prove that if you remove any one cell from the grid — it doesn’t matter which — you can always perfectly divide the rest of the grid into L-shaped pieces made from three cells each. The solution, which Deng figured out, required a concept known as induction, usually taught in high school. Deng was 7 years old.

I asked Gemini about its solution and this was its response:

This classic tiling puzzle is formally known as Golomb’s Tromino Theorem. For this division to be guaranteed regardless of which cell you remove, the grid must specifically be a square of dimensions \(2^n \times 2^n\), where \(n\) is any positive integer (\(2 \times 2\), \(4 \times 4\), \(8 \times 8\), and so on).

Before diving into the mechanics, notice the foundational arithmetic that makes this possible: a \(2^n \times 2^n\) grid contains \((2^n)^2 = 4^n\) total cells. When you remove any one cell, you are left with \(4^n - 1\) cells. Because \(4^n \equiv 1 \pmod 3\) for all positive integers \(n\), the quantity \(4^n - 1\) is always perfectly divisible by \(3\). This guarantees that the remaining area at least has the potential to be covered by \(3\)-cell L-shaped tiles (trominoes) without any left-over space.

The Mathematical Induction Proof

The beauty of the induction approach lies in a clever divide-and-conquer strategy that reduces any large grid down to simple \(4\)-cell squares.

Step 1: The Base Case (\(n = 1\))

Consider the smallest possible power-of-two grid: a \(2^1 \times 2^1 = 2 \times 2\) square containing \(4\) cells. If you remove any one cell from a \(2 \times 2\) grid, the remaining \(3\) cells naturally form an L-shape. A single L-tromino covers this remaining area perfectly. The base case holds.

Step 2: The Inductive Hypothesis

Assume that for some arbitrary positive integer \(k\), any \(2^k \times 2^k\) grid with a single cell removed can be tiled completely by L-trominoes without overlapping or hanging off the edges.

Step 3: The Inductive Step (\(n = k + 1\))

Now consider a larger grid of size \(2^{k+1} \times 2^{k+1}\). We must prove it can also be tiled if any single cell is removed.

1. Divide into Quadrants: Slice the \(2^{k+1} \times 2^{k+1}\) grid horizontally and vertically through its center. This divides the board into four equal quadrants, each of size \(2^k \times 2^k\).
   
   
2. Locate the Missing Cell: The single removed cell must lie inside exactly one of these four quadrants. We will call this the deficient quadrant. Because this deficient quadrant is a \(2^k \times 2^k\) grid missing exactly one cell, our inductive hypothesis guarantees that we can tile it completely with L-trominoes!
   
   
3. The Center Trick: We are now left with three intact \(2^k \times 2^k\) quadrants that have no missing cells. To apply our hypothesis to them, take one single L-tromino and place it squarely at the exact center intersection of the overall board. Orient this tile so that it covers exactly one corner cell of each of the three non-deficient quadrants—specifically, the three innermost corners that meet at the center of the board.
   
   
4. Recurse: By placing that one central tromino, we have effectively "removed" one cell from each of the three remaining quadrants. Now, all four quadrants are independent \(2^k \times 2^k\) grids that are each missing exactly one cell!
   
   
5. By our inductive hypothesis, all four quadrants can now be tiled perfectly by L-trominoes. Therefore, the entire \(2^{k+1} \times 2^{k+1}\) grid minus one cell is successfully tiled.

Figure 1 shows the 8 x 8 case:

Figure 1

Pronic Determinants of Circulant Matrices

Consider the number 28235 that is my diurnal age today. It has a circulant matrix with a determinant 10100 that is a pronic number since 10100 = 100 x 101.$$\begin{bmatrix}

2 & 8 & 2 & 3 & 5 \\

8 & 2 & 3 & 5 & 2 \\

2 & 3 & 5 & 2 & 8 \\

3 & 5 & 2 & 8 & 2 \\

5 & 2 & 8 & 2 & 3

\end{bmatrix}$$What's interesting is that most of the permutations of the digits of 28235 have determinants of their circulant matrices that are also pronic (permalink).

Number       | Determinant     | Factorisation  
------------------------------------------------
22358        | 15500           | 124 x 125      
22385        | 10100           | 100 x 101      
22538        | 19100           |                
22583        | 10100           | 100 x 101      
22835        | 19100           |                
22853        | 15500           | 124 x 125      
23258        | 19100           |                
23285        | 19100           |                
23528        | 10100           | 100 x 101      
23582        | 15500           | 124 x 125      
23825        | 15500           | 124 x 125      
23852        | 10100           | 100 x 101      
25238        | 15500           | 124 x 125      
25283        | 15500           | 124 x 125      
25328        | 10100           | 100 x 101      
25382        | 19100           |                
25823        | 19100           |                
25832        | 10100           | 100 x 101      
28235        | 10100           | 100 x 101      
28253        | 10100           | 100 x 101      
28325        | 15500           | 124 x 125      
28352        | 19100           |                
28523        | 19100           |                
28532        | 15500           | 124 x 125      
32258        | 10100           | 100 x 101      
32285        | 15500           | 124 x 125      
32528        | 15500           | 124 x 125      
32582        | 19100           |                
32825        | 10100           | 100 x 101      
32852        | 19100           |                
35228        | 19100           |                
35282        | 10100           | 100 x 101      
35822        | 15500           | 124 x 125      
38225        | 19100           |                
38252        | 15500           | 124 x 125      
38522        | 10100           | 100 x 101      
52238        | 10100           | 100 x 101      
52283        | 19100           |                
52328        | 19100           |                
52382        | 15500           | 124 x 125      
52823        | 10100           | 100 x 101      
52832        | 15500           | 124 x 125      
53228        | 15500           | 124 x 125      
53282        | 10100           | 100 x 101      
53822        | 19100           |                
58223        | 15500           | 124 x 125      
58232        | 19100           |                
58322        | 10100           | 100 x 101      
82235        | 15500           | 124 x 125      
82253        | 19100           |                
82325        | 19100           |                
82352        | 10100           | 100 x 101      
82523        | 15500           | 124 x 125      
82532        | 10100           | 100 x 101      
83225        | 10100           | 100 x 101      
83252        | 15500           | 124 x 125      
83522        | 19100           |                
85223        | 10100           | 100 x 101      
85232        | 19100           |                
85322        | 15500           | 124 x 125  

Note that it is only when the determinant is 19100 that it is not pronic since 19100 = 100 x 191. In my post titled Determinants of Circulant Matrices, I listed all numbers up to 40000 with the property that the determinants of their circulant matrices were pronic. The numbers between 28000 and 40000 are:

28235, 28253, 28325, 28327, 28453, 28479, 28532, 28543, 28574, 28619, 28732, 28776, 29054, 29168, 29245, 29254, 29555, 29700, 29748, 30171, 30179, 30566, 30575, 30665, 30900, 31100, 31107, 31134, 31233, 31323, 31332, 31355, 31358, 31385, 31400, 31413, 31422, 31440, 31510, 31637, 31646, 31684, 31763, 31907, 32124, 32133, 32223, 32232, 32241, 32258, 32285, 32287, 32313, 32322, 32331, 32528, 32728, 32825, 32845, 32854, 32960, 33123, 33132, 33141, 33176, 33213, 33222, 33231, 33312, 33321, 33335, 33353, 33515, 33518, 33533, 33569, 33671, 33789, 33815, 33965, 33987, 34100, 34166, 34212, 34258, 34311, 34410, 34582, 34599, 34700, 34861, 35153, 35183, 35248, 35282, 35333, 35482, 35507, 35531, 35606, 35693, 35822, 35831, 35936, 35949, 35996, 35999, 36056, 36065, 36137, 36359, 36395, 36418, 36461, 36506, 36614, 36713, 36920, 36995, 37011, 37055, 37091, 37316, 37361, 37400, 37700, 37799, 37822, 37893, 37938, 37979, 38146, 38153, 38252, 38272, 38379, 38397, 38425, 38522, 38524, 38531, 39495, 39536, 39569, 39599, 39653, 39659, 39738, 39797, 39873, 39900, 39954, 39959, 39977, 39995

Note that with 28235 and its digit permutations there are TWO determinants that satisfy. These are:

• \(10100 = 100 \times 101\)
• \(15500 = 124 \times 125\)

This is generally not the case. Consider 28327 and its digit permutations where only the determinant 14762 = 121 x 122 satisfies (permalink).

Number       | Determinant     | Factorisation  
------------------------------------------------
22378        | 14762           | 121 x 122      
22387        | 11462           |                
22738        | 27962           |                
22783        | 11462           |                
22837        | 27962           |                
22873        | 14762           | 121 x 122      
23278        | 27962           |                
23287        | 27962           |                
23728        | 11462           |                
23782        | 14762           | 121 x 122      
23827        | 14762           | 121 x 122      
23872        | 11462           |                
27238        | 14762           | 121 x 122      
27283        | 14762           | 121 x 122      
27328        | 11462           |                
27382        | 27962           |                
27823        | 27962           |                
27832        | 11462           |                
28237        | 11462           |                
28273        | 11462           |                
28327        | 14762           | 121 x 122      
28372        | 27962           |                
28723        | 27962           |                
28732        | 14762           | 121 x 122      
32278        | 11462           |                
32287        | 14762           | 121 x 122      
32728        | 14762           | 121 x 122      
32782        | 27962           |                
32827        | 11462           |                
32872        | 27962           |                
37228        | 27962           |                
37282        | 11462           |                
37822        | 14762           | 121 x 122      
38227        | 27962           |                
38272        | 14762           | 121 x 122      
38722        | 11462           |                
72238        | 11462           |                
72283        | 27962           |                
72328        | 27962           |                
72382        | 14762           | 121 x 122      
72823        | 11462           |                
72832        | 14762           | 121 x 122      
73228        | 14762           | 121 x 122      
73282        | 11462           |                
73822        | 27962           |                
78223        | 14762           | 121 x 122      
78232        | 27962           |                
78322        | 11462           |                
82237        | 14762           | 121 x 122      
82273        | 27962           |                
82327        | 27962           |                
82372        | 11462           |                
82723        | 14762           | 121 x 122      
82732        | 11462           |                
83227        | 11462           |                
83272        | 14762           | 121 x 122      
83722        | 27962           |                
87223        | 11462           |                
87232        | 27962           |                
87322        | 14762           | 121 x 122   

Artin's Constant

My diurnal age today is 28229, a prime, with a reciprocal that has a period of 28228. It is thus one of those primes \(p\) with a period that is given by \( \dfrac{1}{p-1}\).

I got to wondering what proportion of primes have reciprocals that are one less than the prime itself. I asked Gemini. It's response follows and comprises the rest of this post.

Approximately 37.3956% of all prime numbers have reciprocals with a repeating decimal period of maximum length, \(p - 1\). In number theory, primes with this property are known as full reptend primes (or long primes) in base 10.

The Number Theory Behind Maximal Periods

The number of digits in the repeating period of the reciprocal \(1/p\) is determined by the multiplicative order of \(10\) modulo \(p\), denoted as \(\text{ord}_p(10)\).

By Fermat's Little Theorem, for any prime \(p \neq 2, 5\), the period length \(k\) is the smallest positive integer such that:

\[10^k \equiv 1 \pmod{p}\]

Since the multiplicative group \((\mathbb{Z}/p\mathbb{Z})^\times\) has order \(\phi(p) = p - 1\), Lagrange's theorem dictates that the period length \(k\) must always divide \(p - 1\).

For the period to achieve the absolute maximum length of \(p - 1\), the integer \(10\) must be a primitive root modulo \(p\). When \(10\) is a primitive root, the successive powers of \(10 \pmod{p}\) cycle through every possible non-zero remainder from \(1\) to \(p - 1\) before repeating, generating a cyclic period of length \(p - 1\).

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Artin's Constant and Asymptotic Density

The exact asymptotic density of primes for which a given integer is a primitive root is governed by Artin's conjecture on primitive roots.

Because the base \(a = 10\) is neither a perfect square nor \(-1\), Artin's conjecture predicts that the proportion of full reptend primes among all primes up to \(N\) converges to a universal mathematical constant as \(N \to \infty\). This value is known as Artin's constant (\(C_{\text{Artin}}\)), defined by the infinite product over all prime numbers \(q\):

\[C_{\text{Artin}} = \prod_{q \text{ prime}} \left(1 - \frac{1}{q(q-1)}\right) \approx 0.3739558136\dots\]

While Emil Artin's conjecture remains unconditionally unproven for specific bases like 10, Christopher Hooley proved in 1967 that the asymptotic density is exactly \(C_{\text{Artin}}\) under the assumption of the Generalized Riemann Hypothesis (GRH). Extensive computational verification confirms that the distribution stabilizes at this ~37.3956% threshold (cataloged as decimal expansion A005596 in the OEIS).

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Period Lengths of Small Primes

Comparing the first few primes highlights how the multiplicative order behaves when \(10\) is (and is not) a primitive root:

Prime (\(p\))	Reciprocal (\(1/p\))	Period Length (\(\text{ord}_p(10)\))	Full Reptend? (\(k = p - 1\))
3	\(0.\overline{3}\)	\(1\)	No
7	\(0.\overline{142857}\)	\(6\)	Yes
11	\(0.\overline{09}\)	\(2\)	No
13	\(0.\overline{076923}\)	\(6\)	No
17	\(0.\overline{0588235294117647}\)	\(16\)	Yes
19	\(0.\overline{052631578947368421}\)	\(18\)	Yes
23	\(0.\overline{0434782608695652173913}\)	\(22\)	Yes

The sequence of base-10 full reptend primes continues: \(7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, \dots\) (OEIS sequence A001122). Noticeably, primes like \(13\) (where \(10^6 \equiv 1 \pmod{13}\)) and \(37\) (where \(10^3 \equiv 1 \pmod{37}\)) fall into sub-cyclic periods because their multiplicative orders are proper divisors of \(p - 1\).