Testing AI

I thought I might challenge AI to guess the next number in a given sequence. This was my query:

Can you propose what the next number in this sequence of numbers might be and explain your reasoning$$1, 3, 7, 13, 7, 31, 43, 19, 73, 13, 37, \dots$$First up I asked Grok and got a rather curious response:

Uh-oh, too much information for me to digest all at once. You know, sometimes less is more!

That wasn't much help so I turned to Gemini and ChatGPT. They both accepted the challenge and tried their best but neither came up with anything convincing. Finally I gave the latter the following clue:

I'll give a clue. The numbers are generated by a simple quadratic function (using the integers 0, 1, 2, 3 , ... progressively as input) that collapses composite numbers when they appear according to a simple division rule. 

 Nice clue — that nails it.

The quadratic is \(f(n)=n^2+n+1\) for integer inputs \(n=0,1,2,\dots \). 

The collapse rule: whenever \(f(n) \) is composite, divide it by its smallest prime factor (and repeat if necessary) until the result is prime.

Checking the given terms (I'll show \(n\) \( \rightarrow \) raw \(f(n)\) \( \rightarrow \) collapsed result):

• \(n=0: f=0^2+0+1=1 \rightarrow 1\)

• \(n=1: f=1^2+1+1=3 \rightarrow 3\)

• \(n=2: f=4+2+1=7 \rightarrow 7 \)

• \(n=3: f=9+3+1=13 \rightarrow13 \)

• \(n=4: f=16+4+1=21 \rightarrow 21/3=7 \)

• \(n=5: f=25+5+1=31 \rightarrow 31\)

• \(n=6: f=36+6+1=43 \rightarrow 43 \)

• \(n=7: f=49+7+1=57 \rightarrow 57/3=19 \)

• \(n=8: f=64+8+1=73 \rightarrow 73\)

• \(n=9: f=81+9+1=91 \rightarrow 91/7=13 \)

• \(n=10: f=100+10+1=111 \rightarrow 111/3=37\)

All match your sequence.

Now the next input:$$n=11: f(11)=121+11+1=133 \rightarrow 133/7 = 19$$ So the next number in the sequence is 19.

So a little clue and ChatGPT, as it said, nailed it. This was just a little experiment to see what how AI would handle a problem that is actually quite challenging. The sequence is NOT listed in the OEIS by the way.

I challenged ChatGPT to come with a similar challenge and this is what it generated:

Could you propose a similar but different challenge. You should try to be creative and innovative but adhering to the general algorithm, namely generate a sequence of numbers according to a definite rule but modify certain numbers according to another definite rule whenever they satisfy a certain preestablished criterion. Go for it.

Nice — here’s a fresh puzzle in the same spirit as yours: a definite generating rule (a simple quadratic) plus a deterministic “collapse” rule that kicks in whenever the raw value meets a criterion.

I had to take a screenshot of the puzzle as copying and pasting introduces tedious formatting challenges.

The Universe of Catalan Numbers

 I uploaded two of my posts to NotebookLM. The posts were:

• Catalan Numbers from the 30th September 2015
• Motzkin Numbers from the 23rd March 2023

The former post was in fact only my third to Mathematical Meanderings and was made over ten years ago now (shortly after I retired at the age of 66). NotebookLM suggested "The Universe of Catalan Numbers" as a title so I've kept it. Here is the URL of the video.

NotebookLM again does a pretty good job at jazzing up the content and making it interesing. It does make an error early on as I explained in my comments to the video:

This video was generated by NotebookLM based on two posts from my mathematical blog, namely Catalan Numbers and Motzkin Numbers. These two posts are my own and AI was not used in their creation. The video doesn't quite get everything right. For example, it says at the outset that \(24285\) is a Catalan number. In fact \(24285\) is NOT a Catalan number but rather the residue after we divide the 15th Catalan number \(9694845\) by \(2^{15}\) as I explain my blog post. Apart from this, the video does a reasonable job of summarising the content of the two blog posts.

This is now the second video I've created from my blog posts and I'll sure I'll continue to generate more.

Claude to the Rescue

The number 27983, associated with my diurnal age today, has a property than confers it membership in OEIS A335752: the lexicographically earliest sequence of primes, each two of which sum to a triprime. The sequence begins: 3, 5, 47, 239, 3023, 27983, 53231, ... so let's confirm that these numbers satisfy using 3 as a starting point:

• \(3 + 5 = 8 =2^3\)
• \(3 + 47 = 50 = 2 \times 5^2\)
• \(3 + 239 = 242 = 2 \times 11^2 \)
• \(3 + 3023 = 3026 = 2 \times 17 \times 89 \)
• \(3+27983 = 27986 = 2 \times 7 \times 1999\)
  
  

The sums above are all triprimes and we will find that any number in the list paired with any other number will produce a triprime. The code for this was included in the OEIS comments but was in Mathematica. I got Gemini to convert it to Python code and then Grok but in both cases the primeomega function was misunderstood with both claiming that primeomega(12) =2 and not 3. 

I didn't fully understand the Python code and was looking to Gemini and then Grok to explain it to me. It was only Claude that succeeded to my satisfaction. Here is the code followed by Claude's explanation. I think Claude will be my go to resource for anything to do with coding.

from sympy import nextprime, primeomega

ps = [3]

for _ in range(10):

    p = nextprime(ps[-1]) 

    while not all(primeomega(x + p) == 3 for x in ps):

        p = nextprime(p)

    ps.append(p)

print(ps)

I'd be happy to explain this code line by line! This code generates a sequence of prime numbers with a special property.

Code Breakdown

from sympy import nextprime, primeomega

Imports two functions from SymPy:

• nextprime(\(n\)): Returns the next prime number greater than \(n\)
• primeomega(\(n\)): Returns the total number of prime factors of \(n\) (counting multiplicity). For example, primeomega(12) = 3 because 12 = 2² × 3 (two 2's and one 3)

ps = [3]

Initializes the sequence with the first prime number 3. This list will store our special sequence of primes.

for _ in range(10):

Loops 10 times to find 10 more primes (so we'll have 11 total including the initial 3). The underscore _ is used when we don't need the loop variable.

p = nextprime(ps[-1])

Gets the next prime after the last element in our list. 

• ps[-1] accesses the last element of the list
• Initially, this gives nextprime(3) = 5
• This is our candidate prime to add to the sequence

while not all(primeomega(x + p) == 3 for x in ps):

    p = nextprime(p)

This is the key part! It checks if the candidate prime \(p\) satisfies a special condition:

• For every number \(x\) already in our list ps
• The sum \(x + p\) must have exactly 3 prime factors (counting multiplicity)
• If this condition fails for any \(x\), we move to the next prime candidate
• This continues until we find a prime where all sums have exactly 3 prime factors

ps.append(p)

Adds the qualified prime to our sequence once it passes the test.

print(ps)

Prints the final sequence** of primes.

Example Walkthrough

Let's trace the first iteration:

1. Start with ps = [3]

2. Try p = 5: Check if 3 + 5 = 8 = 2³ has 3 prime factors ✓ (yes, three 2's)

3. Add 5 to list: ps = [3, 5]

4. Try p = 7: Check 3 + 7 = 10 = 2 × 5 (only 2 factors) ✗

5. Try p = 11: Check 3 + 11 = 14 = 2 × 7 (only 2 factors) ✗

6. Continue until finding a prime that works with both 3 and 5...

This creates a sequence where each new prime maintains the property that when added to any previous prime in the sequence, the result always has exactly 3 prime factors!

*************************

Now if we change the condition to primeomega = 2, we get the following sequence: 3, 7, 19, 139, 859, 8179, 173059, ... where it can be seen there is a big jump from 8179 to 173059. Testing this out using 3 again we see that:

• \(3+7=10=2 \times 5\)
• \(3+19=22=2 \times 11\)
• \(3+139=142=2 \times 71\)
• \(3+859=862=2 \times 431\)
• \(3+8179=8182=2 \times 4091\)
• \(3+173059=173062=2 \times 86531\)

The sequence is included in the OEIS as A113875: slowest growing sequence of primes having the prime-pairwise-average property: if i<j, (a(i)+a(j))/2 is prime. This makes sense in that 2 seems to be the common factor is all of the sums. 

The sequence with additional (very large) terms looks like this:

3, 7, 19, 139, 859, 8179, 173059, 1026199, 1827139, 15828679, 13187242759, 18732483199, 912492556939, 9130567625119

When we set primeomega = 4 we get the following sequence: 3, 13, 137, 1193, 4373, 26017, 33889, 495433, ... which is NOT listed in the OEIS. Again let's use 3 as our test number:

• \(3+13=16=2^4\)
• \(3+137=140=2^2 \times 5 \times 7\)
• \(3+1193=1196=2^2 \times 13 \times 23\)
• \(3+4373=4376=2^3 \times 547\)
• \(3+26017=26020=2^2 \times 5 \times 1301\)
• \(3+33889=33892=2^2 \times 37 \times 229\)
• \(3+495433=495436=2^2 \times 29 \times 4271\)

That's probably a good place to stop.

From Blog Post To YouTube Video

The idea occurred to me to save one of my blog posts as a PDF and then upload it to NotebookLM. I then generated a video overview which looked surprisingly good I thought. The blog post in question was Free Free Fibonacci Sequences from July 18th 2021. Here is the video:

I have hundreds of blog posts and I've always wanted to highlight some of them and this seems an excellent way of doing so. I was lazy to go to the trouble of creating a video but if it can be done for you then why not? I'd done all the hard work to create the post in the first place and NotebookLM is taking that content and presenting in a visually interesting way.

I'm looking forward to taking some of my other blog posts and converting them to video. I think I'll keep to the Fibonacci theme in the short term as I have several other videos on this subject that I've made over the years. In fact I have over ten years of blog posts so there's no shortage of material to draw on.

Passcodes and Repeated Digits

The content in this post is taken from this video on the YouTube channel MindYourDecisions. Figure 1 shows if the digits of a four digit code are known, there are 24 possibilites of the code contains four distinct digits but 36 possibilities if the code contains only three digits (with one digit repeated).

Figure 1

If no digits are repeated, then there are:$$4! = 4 \times 3 \times 2 \times 1 =24 \text{ possibilities}$$However if one digit is repeated then we have:$$ \frac{4!}{2!} \times 3 = 36 \text{ possibilities}$$What happens with a six digit passcode when we know the digits but not the order. If six digits are used, then we have:$$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1  = 720 \text{ possibilities}$$However, if one digit is repeated then we have:$$ \frac{6!}{2!} \times 5 =1800 \text{ possibilities: see Figure 2}$$

Figure 2

The formulae in general for distinct digits versus one repeated digit are:$$ n! \leftarrow \text{ versus } \rightarrow \frac{(n-1) \times n!}{2}$$Once we have \(n \geq 4\) then \( (n-1)/2 \) is greater than 1 and so the single repeated digit passcode will always yield more possibilities than the distinct digits. What about repeating more than one digit? 

Figure 3 shows the situation for six digit passcodes where it can be seen that one repeated digit yields the most possibilities:

Figure 3

However, for a seven digit posscode, using five distinct digits (with two digits repeated) yields the most possibilities. See Figure 4.

Figure 4

The video referenced earlier concludes with a mention of the general case and a rather formidable formula that I won't go into here.

Palindromic Day 27972

Today is palindromic day 27972 and the last palindromic day of the current millenium. The next palindromic day will occur in the new millenium and will 28082, one hundred and ten days from now.

My first observation is that the digits on either side of the central 9 also add to 9 to give a 9 - 9 - 9 pattern with 9 being the arithmetical digital root as well. This is the first and last time that such a triple pattern will occur in the current millenium. It cannot occur in the next two millenia (2 8 x 8 2 and 2 9 x 9 2) because the digits on either side of the central digit add to 10 and 11 respectively. So already 27972 is rather special. 

27972 is also a member of OEIS A344422: palindromes having more divisors than all smaller palindromes. The table below lists the initial members of this sequence and it can be seen that 27972 has a record 48 divisors.

  number   factorisation        divisors

  1        1                    1

  2        2                    2

  4        2^2                  3

  6        2 * 3                4

  44       2^2 * 11             6

  66       2 * 3 * 11           8

  252      2^2 * 3^2 * 7        18

  2112     2^6 * 3 * 11         28

  2772     2^2 * 3^2 * 7 * 11   36

  6336     2^6 * 3^2 * 11       42

  27972    2^2 * 3^3 * 7 * 37   48

  48384    2^8 * 3^3 * 7        72

27972 is also a member of OEIS A020485: least positive palindromic multiple of \(n\), or 0 if none exists. Here are the multiples for the initial values of \(n\).

  count   n       multiple

  1       1       1

  2       2       1

  3       3       1

  4       4       1

  5       5       1

  6       6       1

  7       7       1

  8       8       1

  9       9       1

  10      0       0

  11      11      1

  12      252     21

  13      494     38

  14      252     18

  15      525     35

  16      272     17

  17      272     16

  18      252     14

  19      171     9

  20      0       0

  21      252     12

  22      22      1

  23      161     7

  24      696     29

  25      525     21

  26      494     19

  27      999     37

  28      252     9

  29      232     8

  30      0       0

  31      434     14

  32      2112    66

  33      33      1

  34      272     8

  35      525     15

  36      252     7

  37      111     3

  38      494     13

  39      585     15

  40      0       0

  41      656     16

  42      252     6

  43      989     23

  44      44      1

  45      585     13

  46      414     9

  47      141     3

  48      2112    44

  49      343     7

  50      0       0

  51      969     19

  52      676     13

  53      212     4

  54      27972   518

  55      55      1

  56      616     11

  57      171     3

  58      232     4

  59      767     13

  60      0       0

  61      26962   442

27972 requires 13 steps to reach the palindrome 4964444694 under the Reverse and Add algorithm. 

27972 is a decagonal or ten-sided number and is the 84th decagonal number and the second non-trivial palindromic decagonal number after 232. See Figure 1 where 232 is shown but not 27972.

Figure 1: source

27972 is a member of OEIS A356854: palindromes that can be written in more than one way as the sum of two distinct palindromic primes. In the case of 27972 we have:

10501 + 17471 = 27972

11311 + 16661 = 27972

11411 + 16561 = 27972

12421 + 15551 = 27972

Prime Number Spiral

The number associated with my diurnal age today is a prime number: 27967 and it has the property that it is on the north spoke of a prime number spiral that begins as shown in Figure 1.

Figure 1: source

I thought it might be an interesting challenge for Gemini to write the Python code to generate this spiral up to the largest prime below 40000. To cut a long story short, Gemini failed miserably despite several attempts to improve the code. Grok on the other hand succeeded on its first try although it did time out several times and needed to be reconnected. However, it got the job done and the result is shown in Figure 2 (link to Grok).

Figure 2: code

The formatting is excellent as can be seen in the detail shown in Figure 3 that was taken from Figure 2:

Figure 3: detail from Figure 2

So Gemini proved to be not up to the task despite my persevering with additional prompts whereas Grok succeeded on its first attempt. A lesson learned.

Prime Factor Fibonacci

A recent number, 27960, associated with my diurnal age has an interesting property if you look at the prime factors more closely:$$27960=2^3 \times 3 \times 5 \times 233$$Let's ignore multiplicity and look at the sum of digits of each distinct prime factor. The 2, 3 and 5 remain the same but 233 becomes 8 and the sequence of prime factors then becomes:$$2, 3, 5, 8$$This is the Fibonacci sequence. How many numbers with four prime factors (ignoring multiplicity) have this property. Well, in the range up to 40000, it turns out that there are 115. These numbers are (permalink):

510, 1020, 1530, 1590, 2040, 2130, 2550, 3060, 3180, 3210, 4080, 4260, 4590, 4770, 5100, 6120, 6360, 6390, 6420, 6990, 7314, 7530, 7650, 7950, 8160, 8520, 8670, 9180, 9540, 9630, 9798, 10200, 10650, 12240, 12720, 12750, 12780, 12840, 12930, 13038, 13770, 13980, 14310, 14628, 14766, 15060, 15090, 15300, 15630, 15900, 16050, 16320, 17040, 17085, 17340, 17466, 18360, 19080, 19170, 19260, 19596, 20400, 20970, 21030, 21300, 21942, 22590, 22950, 23850, 24480, 25440, 25500, 25560, 25680, 25860, 26010, 26076, 26322, 27540, 27960, 28620, 28890, 29256, 29394, 29532, 30120, 30180, 30600, 31260, 31800, 31830, 31950, 32100, 32154, 32637, 32640, 34080, 34530, 34638, 34680, 34932, 34950, 35445, 35511, 36690, 36720, 37650, 38160, 38250, 38340, 38520, 38790, 39114, 39192, 39750

Lets consider the last number in this list: 39750. In this case we have:$$ \begin{align} 39750 &= 2 \times 3 \times 5^3 \times 53 \\ & \rightarrow 2, 3, 5, 8 \end{align}$$If we extend the number of prime factors to five, then only one number satisfies in the range up to 40000:$$ \begin{align}34170 &= 2 \times 3 \times 5 \times 17 \times 67 \\ &\rightarrow 2, 3, 5, 8, 13 \end{align}$$However, there are 160 numbers that satisfy in the range up to one million (permalink). The algorithm actually looks for generalised Fibonacci sequences:$$a,b,c,d,e, \dots \text{ such that } c = a+b, d=b+c, e=c+d \dots$$However, all of the 160 numbers begin with 2. For example, the last of the numbers is 9988520:$$ \begin{align} 998520 &= 2^3 \times 3 \times 5 \times 53 \times 157 \\ &\rightarrow 2, 3, 5, 8, 13 \end{align} $$I tried with six prime factors but without success up to 100 million. Finally I realised that any suitable number must add to 21 and thus be divisible by 3. Such a number can never be a prime factor and so the prime factor sequence must end at 13 and can never progress further.

A Practical Example of a Convolution

Over the past decade, for some reason, I've chosen to ignore \( \textbf{convolutions} \). Whenever they were mentioned in an OEIS sequence, I simply skipped over the reference. However, I'm now attempting to redress that neglect and to that end I was lucky to find two excellent YouTube videos about convolutions made by 3Blue1Brown (this guy has 7.76 million subscribers and for good reason). The two videos are:

• But what is a convolution?
• Convolutions | Why X+Y in probability is a beautiful mess

One example of the use of a convolution is that of a weighted die with probabilities of a particular face showing up being given by:

• p(1) = 0.1
• p(2) = 0.2
• p(3) = 0.3
• p(4) = 0.2
• p(5) = 0.1
• p(6) = 0.1

If this die twice is rolled twice, what are the probabilities of throwing a 2, 3, 4, ..., 10, 11, 12? Well, the convolution of the two sequences representing the probabilities of the two dice rolls we tell you. Let's call the sequence A = [0.1, 0.2, 0.3, 0.2, 0.1, 0.1] and the classic way to facilitate the convolution is the so-called "slide and roll". We'll flip A so that it becomes B = [0.1, 0.1, 0.2, 0.3, 0.2, 0.1] and slide B progressively over A. Figure 1 shows the situation for the initial moves.

Figure 1

Of course it would be tedious to have to construct this every time we needed to evaluate a convolution and so Python makes it easier by use of the following code (input in blue, output in red):

A=[0.1,0.2, 0.3, 0.2,0.1,0.1]

result=convolution(A,A)

L=[]

for n in result:

    L.append(numerical_approx(n,digits=2))

print(L)

[0.010, 0.040, 0.10, 0.16, 0.19, 0.18, 0.14, 0.10, 0.050, 0.020, 0.010] 

Reading the output we can see that the probabilities of the various sums are as follows:

• p(2) = 0.01
• p(3) = 0.04
• p(4) = 0.10
• p(5) = 0.16
• p(6) = 0.19
• p(7) = 0.18
• p(8) = 0,14
• p(9) = 0.10
• p(10) = 0.05
• p(11) = 0.02
• p(12) = 0.01

Figure 2 shows another view of what's going on. In this case, the various sum are calculated by adding up along the marked diagonals:

Figure 2

This post is simply the first in what I hope will be a series of posts relating to convolutions. As I've already discovered, convolutions linked to Fourier transforms and Laplace transformations so it's a big topic to investigate but at least I've finally made a start.

Another Prime To Remember

In November of 2024, I created a post titled A Prime To Remember. The prime on that occasion was 27617 and you can read about its properties by following the link. I think it's time to celebrate another prime and that prime is \( \textbf{27947}\). I sometimes struggle to find a single interesting property for the number, on any given day, that is associated with my diurnal age. With 27947 I had no such problem.

First and foremost, it has the property that the sum of its digits, the sum of the squares of its digits and the sum of cubes of its digits are all prime. Thus we have:

• \(2+7+9+4+7 = 29\) 
• \(2^2+7^2+9^2+4^2+7^2 = 199\)
• \(2^3+7^3+9^3+4^3+7^3 = 1487\)

This property affords it membership in OEIS A176179 and there are 322 such numbers in the range up to 40,000. 27947 shares this property with 27617 and so it is included in my blog post A Prime To Remember. 

27947 also has the property that the absolute differences between successive pairs of digits, and also the first and last digits, are all prime. I discuss these sorts of primes in my blog post titled Fun With Primes and Digit Pairs. Thus we have:

• \( | 2 - 7 | = 5\)
• \(| 7 - 9 | = 2\)
• \(| 9 - 4 | = 5\)
• \(| 7 - 2 | = 5\)

This property affords it membership in OEIS A087593. This next property relates to the prime producing quadratic polynomial \( (4n-29)^2 + 58 \). This polynomial generates 28 distinct primes in succession from \(n=1\) to \(n=28\). When \(n=49\), the polynomial produces the prime 27947. This property affords it membership of OEIS A320772. See my blog post Another Prime Generating Polynomial.

Still on the subject of primes, 27947 has the property that it is a balanced prime of order 100 and thus a member of OEIS A363168. A prime \(p\) is in this sequence if the sum of the 100 consecutive primes just less than \(p\), plus \(p\), plus the sum of the 100 consecutive primes just greater than \(p\), divided by 201 equals \(p\). In the case of 27947, we have:$$ \begin{align} p_{3050} &= 27947 \\ p_{2950} &= 26891\\ p_{3150} &= 28933\\ \sum_{n=2950}^{3049} p_n &=2742922 \\ \sum_{n=3051}^{3150} p_n &= 2846478 \\ \text{average } &= \frac{ 2742922+27947 + 2846478}{201} \\ &= 27947 \end{align}$$See my blog post titled Varieties of Balanced Primes.

Extending Fibonacci-like Numbers

I noticed that in an earlier post titled Consolidating Fibonacci-like Numbers, I looked at numbers like 21347 where we have 2 + 1 = 3 and 3 + 4 = 7 from left to right and even 21101 where 1 + 0 = 1,  0 + 1 = 1 and 1 + 1 = 2 from right to left. In these I only considered additions that resulted in a sums that resulted in a single digit. In this system, a recent diurnal age number (27916) would be ignored and yet 2 + 7 = 9 and 9 + 7 = 16 really does qualify as a Fibonacci-like number.

For this reason I developed an algorithm (permalink) that returns all five digit numbers that follow a Fibonacci-like sequence that will include numbers like 21347 and 27916. Only 28 numbers qualify and these are:

10112, 11235, 12358, 15611, 16713, 17815, 18917, 20224, 21347, 24610, 25712, 26814, 27916, 30336, 31459, 34711, 35813, 36915, 40448, 43710, 44812, 45914, 53811, 54913, 62810, 63912, 72911, 81910

The sequence starts with 10112, 11235, 12358 but then jumps to 15611. What happened to the numbers beginning with 13 and 14? Let's investigate. 

1 + 3 --> 4 and so we have 134

3 + 4 --> 7 and so we have 1347

4 + 7 --> 11 and so we have 134711

However, this is a six digit number and only five digit numbers are being considered. The same holds for the number beginning with 15.

We can reverse the order and reckon from right to left instead of left to right. In this case, we get another 28 numbers. They are (permalink):

10642, 10734, 10826, 10918, 11651, 11743, 11835, 11927, 12752, 12844, 12936, 13761, 13853, 13945, 14862, 14954, 15871, 15963, 16972, 17981, 21101, 42202, 53211, 63303, 74312, 84404, 85321, 95413

If we multiply the digits from left to right (excluding any initial zeroes), we have 15 numbers that satisfy. These are (permalink):

11111, 12248, 14416, 15525, 16636, 17749, 18864, 19981, 21224, 23618, 24832, 31339, 32612, 33927, 42816

If we multiply the digits from right to left (excluding any initial zeroes), we also have 15 numbers that satisfy (permalink):

11111, 12623, 16441, 16824, 18632, 25551, 27933, 32842, 36661, 42212, 49771, 64881, 81991, 84221, 93313

Solve For X

Here's an interesting little problem. The first time I saw it, I immediately thought Lambert W function and indeed that will yield the two real solutions but there is a simpler way to find one of the solutions. The following is taken from the Mind Your Decisions YouTube channel and I'm including the content here to practise my LaTeX skills and help consolidate what I learnt from the video. Here it is:$$ \begin{align} 3^x &= x^9 \\ (3^{x})^{1/9x} &= (x^{9})^{1/9x} \\ 3^{1/9} &= x^{1/x} \\ ({3^3})^{1/27} &= x^{1/x} \\ 27^{1/27} &= x^{1/x} \\ x &=27 \end{align}$$Quite a neat little trick but there are two solutions so how to we find the other one? That's where the Lambert W function comes in handy.$$ \begin{align} 3^x &= x^9 \\ e^{\ln{3^x}} &= x^9 \\ e^{x \ln{3}} &= x^9 \\ 1 &= \frac{x^9}{e^{x \ln{3}}} \\ 1 &= x^9 e^{-x \ln{3}} \\   (1)^{1/9} &= (x^9 e^{-x \ln{3}})^{1/9} \\ 1 &=xe^{(-x \ln{3})/9} \\ \frac{-\ln{3}}{9} &= x  \Big (\frac{-\ln{3}}{9}\Big ) e^{(-x \ln{3})/9} \\  W \Big (\frac{-\ln{3}}{9} \Big)&= W \Big (x \Big (\frac{-\ln{3}}{9} \Big) e^{(-x \ln{3})/9} \Big ) \\  W \Big (\frac{-\ln{3}}{9} \Big ) &= -(x \ln{3})/9 \\  \frac{-9}{\ln{3} }W \Big (\frac{-\ln{3}}{9} \Big ) &=x  \\ x_1 &= \frac{-9}{\ln{3} }W_{-1} \Big (\frac{-\ln{3}}{9} \Big )  = 27 \\ x_2 &= \frac{-9}{\ln{3} }W_{0} \Big (\frac{-\ln{3}}{9} \Big ) \approx 1.15 \end{align}$$The syntax for evaluation of the above in WolframAlpha uses \( \textbf{productlog} \) instead of W and goes like this:

• -9/ln 3 * productlog(-1, -ln(3)/9)
• -9/ln 3 * productlog(0, -ln(3)/9)

Because \( -\ln(3)/9 \approx -0.122 \), this is why we get two solutions. See Figure 1.

Figure 1: graph of \(x=y \, e^y\) and \(x=-\ln(3)/9\)
with intercepts A and B

Primes Within Transcendental Numbers

One of the properties of the number associated with my diurnal age today, 27943, is that it is a prime formed by merging five successive digits in the decimal expansion of \(e \). Primes with this property form OEIS A104846. This got me thinking about whether all possible primes could be expressed by merging successive digits in the decimal expansions of all transcendental numbers like \(e\) and \( \pi \).  

However, as Grok explained:

No, it cannot be proven that the decimal expansions of transcendental numbers like \( \pi \) or \(e\) contain representations of all prime numbers through progressive concatenations of consecutive digits. This interpretation hinges on whether every possible finite sequence of digits—including the decimal representations of all primes—appears as a substring somewhere in their infinite, non-repeating expansions. While this would be true if \( \pi \) and \(e\) are normal numbers (where every digit sequence of length \(k\) appears with equal frequency \(1/10^k\) in the limit), the normality of \( \pi \) and \(e\) remains an open problem in mathematics. 

If numbers like \( \pi \) and \(e\) could be proven to be normal then not only would every prime occur but it would occur an infinite number of times. It's interesting to look at 27943 and see how many times it appears in the first million digits of \(e\). Here are the initial positions for the "2" (permalink):

166, 46606, 217029, 234391, 248347, 258504, 483659, 832027, 941722

For \( \pi \), it takes a while before the first occurrence but afterwards the occurrences of 27943 are more frequent (permalink):

94943, 146738, 216793, 253032, 283291, 290571, 302660, 350882, 358637, 520562, 632389, 661691, 715324, 805161, 843112, 890626

I wondered if there was any difference between algebraic numbers like \( \phi \), the golden ratio, and transcendental numbers in regards to being able to represent all primes as concatenations of successive digits. Here's was Grok's response:

No, it cannot be proven that the decimal expansions of algebraic irrational numbers like the golden ratio (\( \phi  \approx 1.6180339887 \dots \)) contain representations of all prime numbers through progressive concatenations of consecutive digits, for essentially the same reasons as with transcendentals like \( \pi \) and \(e\). Their normality in base 10 remains unproven. 

Just as with π and e, the golden ratio's decimal expansion is infinite and non-repeating (since it's irrational), but this doesn't guarantee it contains all sequences. Empirical checks of millions of digits suggest randomness consistent with normality, but no proof exists—not even that all digits 0-9 appear infinitely often. It's widely conjectured that φ (a quadratic irrational algebraic number) is normal in base 10, mirroring beliefs about π and e. Proving this would be a major result in number theory.

There's a specific conjecture (dating back to Borel in 1909, refined later) that every irrational algebraic number is absolutely normal (normal in every integer base \(\geq 2)\). This includes \( \phi, \sqrt{2}, \sqrt{3} \), and others. No counterexamples exist, but crucially, no irrational algebraic has ever been proven normal in any base. In contrast, for transcendentals, normality isn't conjectured for all—only for certain "natural" ones like \( \phi \) and \(e\)—while many non-normal transcendentals are known (e.g., Liouville numbers designed to lack certain digits). 

Proven examples: We have constructed transcendentals proven to be normal, such as Champernowne's constant ( 0.123456789101112 ...)  or the Copeland–Erdős constant (formed by concatenating primes: 0.235711131719... ). These do contain every finite digit sequence infinitely often, including all primes. No such proven-normal algebraic irrationals exist, partly because algebraics satisfy polynomial equations, which might impose subtle constraints on their expansions (though the conjecture suggests not). 

In summary, algebraic irrationals like \( \phi \) aren't fundamentally "different" in the sense that the core question—whether all primes appear as substrings—remains unproven and conjectured true, just like for \( \phi\) and \(e\). The main distinction lies in the broader landscape: Algebraics have a stronger universal conjecture without exceptions or proven cases, while transcendentals show a mix of proven normal and non-normal examples. If the conjectures hold, then yes, \( \phi \)'s expansion would contain all primes, but we can't prove it yet.

So at what positions does the prime 27943 appear in the decimal expansion of \( \phi \). Here are the locations (permalink):

33280, 127164, 129935, 529433, 597796, 795399, 971620

Which is Larger?

The question was posed in a YouTube video as to which was larger: \(3.14^{\pi}\) or \( \pi^{3.14}\) ? The use of a calculator was not allowed. The solution provided in the video is interesting.$$ \begin{align} &3.14^{\pi} \text{ versus } \pi^{3.14} \\ &\text{raise each to the } 1/3.14 \text{ power} \\ &3.14^{\pi /3.14} \text{ versus } \pi^{3.14/3.14} \\ &3.14^{\pi/3.14} \text{ versus } \pi  \\ & \text{raise each to the } 1/ \pi \text{ power}  \\ &3.14^{\pi/(3.14 \pi)} \text{ versus } \pi ^{1/\pi} \\   &3.14^{1/3.14} \text{ versus } \pi ^{1/\pi} \end{align} $$Now consider the function \(y=x^{1/x}\) which we will modify as follows:$$ \begin{align} y &=x^{1/x} \\ &= e^{\ln{x^{1/x}}} \\ &=e^{1/x \ln{x}} \end{align}$$Let's find its first derivative in order to determine stationary values:$$ \begin{align} \frac{dy}{dx} &=e^{1/x \ln{x}} \frac{d}{dx} (1/x \ln{x})\\ &= \frac{e^{1/x \ln{x}}}{x^2}(1-\ln{x})  \end{align} $$the numerator and denominator on the left are always positive so it comes down to:$$ \begin{align} 1-\ln{x} = 0  \text{ when } x=e \\  \\ \text{if } x<e, dy/dx > 0 \text{ and if } x>e, dy/dx < 0 \end{align}$$Thus there is a maximum turning point at \(x=e\) and since, after this point, \(y\) decreases as \(x\) increases, we must have:$$3.14^{1/3.14} > \pi^{1/\pi} \text{ since } 3.14 < \pi$$This in turn means that by reversing our original transformation, we have:$$3.14^{\pi} > \pi^{3.14}$$Figure 1 shows what the graph of \(y=x^{1/x}\) look like.

Figure 1: note the gradual descent as \(x\) gets larger

The approximate values of our two expressions are as follows:$$ \begin{align} 3.14^{\pi}&\approx 36.4041195357888 \\ \pi^{3.14} &\approx 36.3957438848941 \end{align}$$The solution to this problem also settles the more general issues surrounding when \(a^b\) is larger or smaller than \(b^a\) with \(a,b>1\). Let's consider the problem of:$$a^b \text{ versus } b^a \text{ for the particular case of }2024^{2025} \text{ versus } 2025^{2004}$$We know from earlier that this comparison reduces to:$$ \begin{align} &a^{1/a} \text{ versus } b^{1/b} \text{ or } 2024^{1/2024} \text{ versus } 2025^{1/2005} \\  &\text{and if }a>e, b>e \text{ and } a<b \text{ then:} \\ &\text{because } 2024 < 2025 \text{ we have:}\\ &2024^{2025} > 2025^{2004} \end{align} $$ However, what about:$$ \phi^{1.618} \text{ versus } 1.618^{\phi}$$Here the problem reduces to:$$\phi^{1/\phi} \text{ versus } 1.618^{1/1.618}$$Here \(\phi<e\) and \(1.618<e\) and \(1.618 < \phi\), so:$$ 1.618^{1/1.618} < \phi ^{1/\phi} \text{ and } 1.618^{\phi}<\phi^{1.618}$$Double checking we find that:$$ \begin{align} 1.618^{\phi} \approx 2.17838352536880\\ \phi^{1.618} \approx 2.17842193783803 \end{align}$$If the values of \(a\) and \(b\) fall on either side of \(e\) then the previous methods don't work. I'll leave this problem for another post possibly as I've spent enough time on this.

Obscure Difference of Two Squares

 Here's an interesting problem that appeared in this YouTube video:

Find the values of \(a, b \in \mathbb{N}\) such \(a^b-b=77\).

With two unknowns and only one equation, the problem seems daunting but the factorisation of 77 and the fact that the values of \(a\) and \(b\) are restricted to natural numbers will be enough to solve the problem. However, the solution that the video describes hinges on the use of a difference of two squares and this is not at all obvious. Let's begin:$$ \begin{align} a^b-b &= 77 \\ (a^{b/2})^2-(\sqrt{b})^2 &= 77 \\ (a^{b/2}-\sqrt{b}) \times  (a^{b/2}+\sqrt{b}) &= 7 \times 11 \end{align} $$We know that \(a^{b/2}-\sqrt{b} < a^{b/2}+\sqrt{b}\) and so there are only two possible factorisations of 77. These are \(1 \times 77\) and \(7 \times 11 \). Not only is the second option more likely but the first option in fact does not yield a solution, as we'll show later. Let's proceed using the \(7 \times 11\) factorisation.$$ \begin{align} a^{b/2}-\sqrt{b} &=7 \dots 1\\ a^{b/2}+\sqrt{b} &=11 \dots 2 \end{align}$$Adding equations 1 and 2 together gives:$$ \begin{align} 2 a^{b/2} &=18 \\ a^{b/2} &=9 \\ \end{align}$$Let's substitute \(a^{b/2}=9\) back in the equation 1:$$ \begin{align} 9 - \sqrt{b} &= 7 \\ \sqrt{b} &= 2 \\ b &=4 \\ \because a^{b/2} &= 9 \\ a^2 &= 9 \\ a&=3 \end{align}$$So we have a solution with \(a = 3\) and \(b=4\). Substituting these values back into the original equation should confirm the solution:$$ \begin{align} a^b -b &= 77 \\ 3^4-4 &=77\\77 &= 77 \end{align}$$But what of the \(1 \times 77\) factorisation? Let's confirm that this does not yield a solution:$$ \begin{align} a^{b/2}-\sqrt{b} &=1 \dots 1\\ a^{b/2}+\sqrt{b} &=77 \dots 2 \end{align}$$Adding equations 1 and 2 together gives$$ \begin{align} 2a^{b/2} &=78\\ a^{b/2} &= 39 \end{align}$$Substitute \(a^{b/2}=39\) back in the equation 1:$$ \begin{align} 39 - \sqrt{b} &=77 \\ \sqrt{b}=-38 \end{align}$$There is no value of \(b\) that satisfies this equation so we can confirm that the \(1 \times 77\) factorisation does not yield a solution. I did not think up the method of solution described here and all credit is due the author of the linked video. It did occur to me to generalise the problem and consider what other biprimes (or semiprimes) yield a solution. In other words, what prime values of \(x\) and \(y\), with \(x<y\), allow for natural number solutions to:$$a^b-b=xy$$The two equations we obtain from this relationship are:$$ \begin{align} a^{b/2}-\sqrt{b} &=x \dots 1\\ a^{b/2}+\sqrt{b} &=y \dots 2 \end{align}$$Adding equations 1 and 2 gives:$$ \begin{align} 2a^{b/2} &= x+y \\ a^{b/2} &= \frac{x+y}{2} \end{align}$$Substituting into equation 1 gives:$$ \begin{align} \frac{x+y}{2}-\sqrt{b} &= x \\ \sqrt{b} &= \frac{x+y}{2}-x \\ \sqrt{b} &= \frac{x+y-2x}{2} \\ b &= \frac{(y-x)^2}{4}  \end{align}$$Testing this equation with \(x=7\) and \(y=11\) we see that \(b=4\) and so all is well. So what other prime values of \(x\) and \(y\) will yield solutions? Testing we find that \(x=3\) and \(y=5\) yield a value of \(b=1\). This leads to:$$ \begin{align} a^{b/2} &=\frac{x+y}{2}\\ \sqrt{a} &= 4 \\ a &=16 \end{align}$$Checking, we see that:$$ \begin{align} a^b-b &=15 \\ 16^1-1 &=15\\15 &= 15 \end{align}$$Rather than randomly checking prime values of \(x\) and \(y\), the design of a search algorithm is a more organised approach (permalink). Let's exclude those cases where either \(a=1\) or \(b=1\) as these are rather trivial. For semiprimes with prime factors less than 300, I found the following additional semiprimes (the calculations are very processor intenstive). The first is 6557:$$ \begin{align} 6557 &=79 \times 83 \\ \text{with } a &= 9 \text{ and }b= 4 \end{align}$$Checking, we find that:$$ \begin{align} a^b-b &= 6557 \\9^4 - 4 &= 6557 \\ 6561-4  &= 6557\\ 6557&=6557 \end{align}$$The second is 50621:$$ \begin{align} 50621 &=223 \times 227\\ \text{with } a &= 15 \text{ and }b= 4 \end{align}$$Checking, we find that:$$ \begin{align} a^b-b &= 50621 \\ 15^4 - 4 &= 50621\\50625-4 &= 50621\\50621 &= 50621 \end{align}$$The third is 194477:$$ \begin{align} 194477 &=439 \times 443 \\ \text{with } a &= 21 \text{ and } b=4 \end{align}$$Checking, we find that:$$ \begin{align} a^b-b &= 194477 \\ 21^4-4 &= 194477 \\ 194481-4 &= 194477 \\ 194477 &= 194477 \end{align}$$That's probably enough. As can be seen, the semiprimes that satisfy are few and far between but it was an interesting exercise. Notice that the prime factors in each case are adjacent primes.

Truncated Pyramid

I was surprised to discover that the number associated with my diurnal age today (27930) has a connection to the volume of a truncated square pyramids. Firstly, let's recall the formula for the volume of such figure:$$V=\frac{1}{3} (a^2+ab+b^2) \, h$$where \(a\) is the side length of the square base, \(b\) is the side length of the top square and \(h\) is the height of the truncated pyramid. The formula is easily confirmed by integration from first principles without relying on the formula for the volume of a pyramid. 

Now 27930 is a member of OEIS A027444: \( \text{a}(n) = n^3 + n^2 + n\) with the following interesting comment attached to the entry:

For \(n>1\), a(\(n\)) is the volume of a truncated square pyramid with height \(n\) and base  lengths \(n+2\) and \(n-1\). See Figure 1.

Figure 1

Now if we substitute \(a=n+2\), \(b=n-1\) and \(n=h\) into our earlier formula we find the volume \(V\) of the truncated pyramid becomes:$$V=n^3+n^2+n$$When \(n=30\), the volume is 27930 cubic units. Here are the volumes for various values of \(n>1\):

  n    n-1   n+2   V 

  2    1     4     14

  3    2     5     39

  4    3     6     84

  5    4     7     155

  6    5     8     258

  7    6     9     399

  8    7     10    584

  9    8     11    819

  10   9     12    1110

  11   10    13    1463

  12   11    14    1884

  13   12    15    2379

  14   13    16    2954

  15   14    17    3615

  16   15    18    4368

  17   16    19    5219

  18   17    20    6174

  19   18    21    7239

  20   19    22    8420

  21   20    23    9723

  22   21    24    11154

  23   22    25    12719

  24   23    26    14424

  25   24    27    16275

  26   25    28    18278

  27   26    29    20439

  28   27    30    22764

  29   28    31    25259

  30   29    32    27930

  31   30    33    30783

  32   31    34    33824

  33   32    35    37059

Seven Eleven Rules

I was struggling to find something that caught my fancy regarding the number associated with my diurnal age today: 27929. I thought I'd look at its reverse, 92972, and compare their factorisations. The results were:$$ \begin{align} 27929 &= 11 \times 2539 \\ 92972 &=2^2 \times 11 \times 2113 \end{align} $$Clearly, the number and its reverse share a common prime factor of 11. I then realised that 27929 has a digit sum of 29 and the two digits, when added together, give 11. So I then decided to look for numbers with the following properties:

• number is divisible by 11
• its reverse is also divisible by 11
• its sum of digits gives a number whose digits sum to 11

It turns out that there are only 30 numbers that satisfy these criteria in the range up to 40000. They are:

20999, 21989, 22979, 23969, 24959, 25949, 26939, 27929, 28919, 29909, 30899, 30998, 31889, 31988, 32879, 32978, 33869, 33968, 34859, 34958, 35849, 35948, 36839, 36938, 37829, 37928, 38819, 38918, 39809, 39908

The details are (permalink):

  number   factors              reverse   factors              digit sum   sum

  20999    11 * 23 * 83         99902     2 * 11 * 19 * 239    29          11
  21989    11 * 1999            98912     2^5 * 11 * 281       29          11
  22979    11 * 2089            97922     2 * 11 * 4451        29          11
  23969    11 * 2179            96932     2^2 * 11 * 2203      29          11
  24959    11 * 2269            95942     2 * 7^2 * 11 * 89    29          11
  25949    7 * 11 * 337         94952     2^3 * 11 * 13 * 83   29          11
  26939    11 * 31 * 79         93962     2 * 11 * 4271        29          11
  27929    11 * 2539            92972     2^2 * 11 * 2113      29          11
  28919    11^2 * 239           91982     2 * 11 * 37 * 113    29          11
  29909    11 * 2719            90992     2^4 * 11^2 * 47      29          11
  30899    11 * 53^2            99803     11 * 43 * 211        29          11
  30998    2 * 11 * 1409        89903     11^2 * 743           29          11
  31889    11 * 13 * 223        98813     11 * 13 * 691        29          11
  31988    2^2 * 11 * 727       88913     11 * 59 * 137        29          11
  32879    7^2 * 11 * 61        97823     11 * 8893            29          11
  32978    2 * 11 * 1499        87923     11 * 7993            29          11
  33869    11 * 3079            96833     11 * 8803            29          11
  33968    2^4 * 11 * 193       86933     7 * 11 * 1129        29          11
  34859    11 * 3169            95843     11 * 8713            29          11
  34958    2 * 7 * 11 * 227     85943     11 * 13 * 601        29          11
  35849    11 * 3259            94853     11 * 8623            29          11
  35948    2^2 * 11 * 19 * 43   84953     11 * 7723            29          11
  36839    11 * 17 * 197        93863     7 * 11 * 23 * 53     29          11
  36938    2 * 11 * 23 * 73     83963     11 * 17 * 449        29          11
  37829    11 * 19 * 181        92873     11 * 8443            29          11
  37928    2^3 * 11 * 431       82973     11 * 19 * 397        29          11
  38819    11 * 3529            91883     11 * 8353            29          11
  38918    2 * 11 * 29 * 61     81983     11 * 29 * 257        29          11
  39809    7 * 11^2 * 47        90893     11 * 8263            29          11
  39908    2^2 * 11 * 907       80993     11 * 37 * 199        29          11

The algorithm can be modified to search for prime numbers other than 11. For example, there are 80 numbers in the range up to 40000 that satisfy these criteria:

• number is divisible by 7
• its reverse is also divisible by 7
• its sum of digits gives a number whose digits sum to 7

These numbers are (permalink):

259, 952, 1078, 1708, 2527, 2779, 3346, 3598, 4165, 5614, 5866, 6433, 6685, 7252, 8071, 8701, 8953, 9079, 9709, 9772, 10087, 10717, 10969, 11536, 11788, 12103, 12355, 13174, 13804, 14623, 14875, 15442, 15694, 17017, 17269, 17962, 18088, 18718, 19537, 19789, 20545, 20797, 21364, 22183, 22813, 23884, 24451, 25207, 25459, 26026, 26278, 26908, 26971, 27097, 27727, 27979, 28546, 28798, 29113, 29365, 30121, 30373, 31129, 31192, 31822, 32641, 32893, 33649, 34216, 34468, 35035, 35287, 35917, 36736, 36988, 37303, 37555, 38122, 38374, 39823

The details are (permalink):

  number   factors             reverse   factors                digit sum   sum

  259      7 * 37              952       2^3 * 7 * 17           16          7
  952      2^3 * 7 * 17        259       7 * 37                 16          7
  1078     2 * 7^2 * 11        8701      7 * 11 * 113           16          7
  1708     2^2 * 7 * 61        8071      7 * 1153               16          7
  2527     7 * 19^2            7252      2^2 * 7^2 * 37         16          7
  2779     7 * 397             9772      2^2 * 7 * 349          25          7
  3346     2 * 7 * 239         6433      7 * 919                16          7
  3598     2 * 7 * 257         8953      7 * 1279               25          7
  4165     5 * 7^2 * 17        5614      2 * 7 * 401            16          7
  5614     2 * 7 * 401         4165      5 * 7^2 * 17           16          7
  5866     2 * 7 * 419         6685      5 * 7 * 191            25          7
  6433     7 * 919             3346      2 * 7 * 239            16          7
  6685     5 * 7 * 191         5866      2 * 7 * 419            25          7
  7252     2^2 * 7^2 * 37      2527      7 * 19^2               16          7
  8071     7 * 1153            1708      2^2 * 7 * 61           16          7
  8701     7 * 11 * 113        1078      2 * 7^2 * 11           16          7
  8953     7 * 1279            3598      2 * 7 * 257            25          7
  9079     7 * 1297            9709      7 * 19 * 73            25          7
  9709     7 * 19 * 73         9079      7 * 1297               25          7
  9772     2^2 * 7 * 349       2779      7 * 397                25          7
  10087    7 * 11 * 131        78001     7 * 11 * 1013          16          7
  10717    7 * 1531            71701     7 * 10243              16          7
  10969    7 * 1567            96901     7 * 109 * 127          25          7
  11536    2^4 * 7 * 103       63511     7 * 43 * 211           16          7
  11788    2^2 * 7 * 421       88711     7 * 19 * 23 * 29       25          7
  12103    7^2 * 13 * 19       30121     7 * 13 * 331           7           7
  12355    5 * 7 * 353         55321     7^2 * 1129             16          7
  13174    2 * 7 * 941         47131     7 * 6733               16          7
  13804    2^2 * 7 * 17 * 29   40831     7 * 19 * 307           16          7
  14623    7 * 2089            32641     7 * 4663               16          7
  14875    5^3 * 7 * 17        57841     7 * 8263               25          7
  15442    2 * 7 * 1103        24451     7^2 * 499              16          7
  15694    2 * 7 * 19 * 59     49651     7 * 41 * 173           25          7
  17017    7 * 11 * 13 * 17    71071     7 * 11 * 13 * 71       16          7
  17269    7 * 2467            96271     7 * 17 * 809           25          7
  17962    2 * 7 * 1283        26971     7 * 3853               25          7
  18088    2^3 * 7 * 17 * 19   88081     7 * 12583              25          7
  18718    2 * 7^2 * 191       81781     7^2 * 1669             25          7
  19537    7 * 2791            73591     7 * 10513              25          7
  19789    7 * 11 * 257        98791     7 * 11 * 1283          34          7
  20545    5 * 7 * 587         54502     2 * 7 * 17 * 229       16          7
  20797    7 * 2971            79702     2 * 7 * 5693           25          7
  21364    2^2 * 7^2 * 109     46312     2^3 * 7 * 827          16          7
  22183    7 * 3169            38122     2 * 7^2 * 389          16          7
  22813    7 * 3259            31822     2 * 7 * 2273           16          7
  23884    2^2 * 7 * 853       48832     2^6 * 7 * 109          25          7
  24451    7^2 * 499           15442     2 * 7 * 1103           16          7
  25207    7 * 13 * 277        70252     2^2 * 7 * 13 * 193     16          7
  25459    7 * 3637            95452     2^2 * 7^2 * 487        25          7
  26026    2 * 7 * 11 * 13^2   62062     2 * 7 * 11 * 13 * 31   16          7
  26278    2 * 7 * 1877        87262     2 * 7 * 23 * 271       25          7
  26908    2^2 * 7 * 31^2      80962     2 * 7 * 5783           25          7
  26971    7 * 3853            17962     2 * 7 * 1283           25          7
  27097    7^3 * 79            79072     2^5 * 7 * 353          25          7
  27727    7 * 17 * 233        72772     2^2 * 7 * 23 * 113     25          7
  27979    7^2 * 571           97972     2^2 * 7 * 3499         34          7
  28546    2 * 7 * 2039        64582     2 * 7^2 * 659          25          7
  28798    2 * 7 * 11^2 * 17   89782     2 * 7 * 11^2 * 53      34          7
  29113    7 * 4159            31192     2^3 * 7 * 557          16          7
  29365    5 * 7 * 839         56392     2^3 * 7 * 19 * 53      25          7
  30121    7 * 13 * 331        12103     7^2 * 13 * 19          7           7
  30373    7 * 4339            37303     7 * 73^2               16          7
  31129    7 * 4447            92113     7 * 13159              16          7
  31192    2^3 * 7 * 557       29113     7 * 4159               16          7
  31822    2 * 7 * 2273        22813     7 * 3259               16          7
  32641    7 * 4663            14623     7 * 2089               16          7
  32893    7 * 37 * 127        39823     7 * 5689               25          7
  33649    7 * 11 * 19 * 23    94633     7 * 11 * 1229          25          7
  34216    2^3 * 7 * 13 * 47   61243     7 * 13 * 673           16          7
  34468    2^2 * 7 * 1231      86443     7 * 53 * 233           25          7
  35035    5 * 7^2 * 11 * 13   53053     7 * 11 * 13 * 53       16          7
  35287    7 * 71^2            78253     7^2 * 1597             25          7
  35917    7^2 * 733           71953     7 * 19 * 541           25          7
  36736    2^7 * 7 * 41        63763     7 * 9109               25          7
  36988    2^2 * 7 * 1321      88963     7 * 71 * 179           34          7
  37303    7 * 73^2            30373     7 * 4339               16          7
  37555    5 * 7 * 29 * 37     55573     7 * 17 * 467           25          7
  38122    2 * 7^2 * 389       22183     7 * 3169               16          7
  38374    2 * 7 * 2741        47383     7^2 * 967              25          7
  39823    7 * 5689            32893     7 * 37 * 127           25          7