Special Numbers, Special Dates

On the 26th January 2023, I posted about Turning Dates Into Numbers and made the observation that the 10th February 2023 was a special day because the number associated with this date (20230210) had the interesting property that it was the only date in the year that produced a number that had seven distinct prime factors:

20230210 = 2 x 5 x 7 x 11 x 13 x 43 x 47

It turns out that this number has another interesting property in that the sum of its digits equals the last two digits (representing the day of the month). In fact, it begins a run of ten consecutive numbers that end with 20230219 (the 19th February 2023) ... permalink:

20230210, 20230211, 20230212, 20230213, 20230214, 20230215, 20230216, 20230217, 20230218, 20230219

Another run of ten similar numbers occurs in November, starting on the 10th and ending on the 19th.

20231110, 20231111, 20231112, 20231113, 20231114, 20231115, 20231116, 20231117, 20231118, 20231119

For many, many years to come, the sum of the digits of the numbers associated with the dates will always contain two digits. Today is the 15th February 2023 and so we are into the second half of the first-mentioned run.

This number can be expressed as a sum of three cubes
See Figure 2

Similar runs will occur every year but the dates will differ naturally. For example, in 2024, the first run of numbers will begin on 11th January 2024 with 20240111 and end on 19th January 2024 with 20240119. 

In my previous post Turning Dates Into Numbers I looked at what numbers were prime, semiprime and so on. In this post I'll turn my attention to cubes and see what numbers can be expressed as the sum of two cubes and as a sum of three cubes. 

It turns out that there is only one number in the entire year that can be expressed as a sum of two cubes and that is \( 20231020 = 69^3 + 271^3\). This corresponds to the 20th October 2023.

The sums of three cubes are more common. The complete list is shown in Figure 2, arranged in chronological order with 20th January being the first date and 26th December the last.

Figure 2: permalink

It can be noted that some dates produce numbers that can be expressed as sums of three cubes in more than one way. Specifically, 20230426 (two ways), 20230523 (two ways), 20230722 (two ways), 20230904 (two ways), 20230911 (three ways), 20231028 (two ways) and 20231226 (two ways). Thus the 11th September (the famous 911 in United States date format) produces the only number (20230911) that can be expressed as a sum of three cubes in three ways.

42

On April 1st 2019, my blog post was titled "42 is the new 33" and that time a solution to the problem of expressing 42 as a sum of three cubes had not been found. However, very recently, a solution was found as this Numberphile Video explains:

The solution is visible there in the thumbnail for the video but I'll repeat it below:$$X^3+Y^3+Z^3=42 \text{ when }$$ \(X = -80538738812075974\) 

\(Y = 80435758145817515\) 

\(Z = 12602123297335631\)

Certain numbers cannot be the sum of three cubes. The reason is that any cubic number must be equal to 0 or ±1 mod 9. If we add three cubic numbers together, the sum must therefore lie between -3 and +3 and so any number that is equal to 4 or 5 mod 9 (or equivalently ±4) cannot be expressed as a sum of three cubes. So between 1 and 99, the following numbers fall into this category: 4, 5, 13, 14, 22, 23, 31, 32, 40, 41, 49, 50, 58, 59, 67, 68, 76, 77, 85, 86, 94, 95.

It wasn't obvious to me why "any cubic number must be equal to 0 or ±1 mod 9". However, I finally realised why this was the case. Here is what I discovered:

Consider a number \(x\) that can be divided, by 9, \(n\) times leaving a remainder of \(r\). This means that \(x=9n +r\) and so \(x^3=(9n+r)^3 \). This latter expression becomes:$$(9n)^3+3 \times (9n)^2 r+ 3 \times (9n)r^2+r^3$$Clearly 9 divides each term of the above expression, except perhaps the \(r^3\) term. Because we are considering modulus 9, the remainder \(r\) must lie between -4 and +4 (-4 is the same as +5, -3 is the same as +6 etc.). Let's consider each possible remainder:

• \(±4^3=±64\) and \(±64 \equiv ±1 \bmod 9\)
• \(±3^3=±27\) and \(±27 \equiv 0 \bmod 9 \)
• \(±2^3=±8\) and \(±8 \equiv ±1 \bmod 9\)
• \(±1^3=±1\) and \(±1 \equiv ±1 \bmod 9 \)
• \(0^3=0\) and \(0 \equiv 0 \bmod 9 \)

So there you have it, any cubic number must be equal to ±1 or 0 and so the sum of three cubes must lie between ±3. This excludes any number that is ±4 or ±5 mod 9.

Here are the numbers from 1 to 99 expressed as a sum of three cubes (zeroes allowed). The table was taken from an old site and I had to manually enter the values for 33, 42 and 74 because they were still empty:

Figure 1: numbers between 1 and 29

Figure 2: numbers between 30 and 64

Figure 3: numbers between 65 and 99

As the YouTube video makes clear, the search is on for a solution to 114, the next "unknown" after 42, but also for a third solution to 3. The two current solutions are \(1^3+1^3+1^3=4^3+4^3+(-5)^3=3\) but is there another solution? As yet, we don't know.

ADDENDUM: LINK

Well actually now we do know:$$ 3=569936821221962380720^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}$$The only remaining unsolved cases up to 1,000 are 114, 165, 390, 579, 627, 633, 732, 906, 921, and 975.

The Connectivity of Numbers

Figure 1

I thought it time to summarise what I've learned thus far about the number properties that I'll term intrinsic. By this term, I mean number properties that are not dependent on the number base used to represent the number. For example, 25652 is a palindromic number in base 10 but Figure 1 shows that this is not the case for the other bases from 2 to 16. Examples of intrinsic number properties would be the number of prime factors and the number of divisors. For example, 25652 factors to \(2^2 ⋅ 11^2 ⋅ 53\) and it has 18 divisors (1, 2, 4, 11, 22, 44, 53, 106, 121, 212, 242, 484, 583, 1166, 2332, 6413, 12826 and 25652). These factors and divisors are the same numbers in any number base.

The sum of the digits of a number is another example of a number property that is not intrinsic. In the case of 25652, the digit sum is 20. However, in base 8 the number is 62064 and the digit sum is \(22_8\), which is 18 in base 10. Contrast this with the sum of the prime factors of 25652 in bases 10 and 8 (ignoring multiplicity). In base 10, the sum is 66. In base 8, the factors are \(2_8\), \(13_8\) and \(65_8\) and so the sum is \(102_8\) or 66 in base 10. So having clarified the difference between intrinsic and base-dependent number properties, what are some of the most important properties of the former type of numbers?

I'll continue using 25652 as an example. To begin the divisors, and particularly the prime divisors, are all important. For 25652, as already noted, the divisors are:$$1, 2, 4, 11, 22, 44, 53, 106, 121, 212, 242, 484, 583, 1166, 2332, 6413, 12826, 25652$$The prime divisors are \( 2, 2, 11, 11, 53 \) and these uniquely define the number. 25652 has an interesting property that does not relate to its own divisors but instead relates to the proper divisors of other numbers. 25652 is an untouchable number, because it is not equal to the sum of proper divisors of any number. The proper divisors of 25652 are 1, 2, 4, 11, 22, 44, 53, 106, 121, 212, 242, 484, 583, 1166, 2332, 6413, 12826 and these add to 25137. This means that 25137 is not an untouchable number because it can be formed from the addition of the proper divisors of 25652. This talk of adding up divisors leads us on to the sigma function.

The sigma function can be used to find the number of divisors, the sum of these divisors, the sum of the squares of these divisors, the sum of the cubes of these divisors and so on. So-called sigma zero, written as \( \sigma_0\), returns the number of divisors or the sum of the divisors raised to the zero power. \( \sigma_1\) returns the sum of the divisors raised to the first power and so on. This is the result:

\( \sigma_0(25652)=18\)
\( \sigma_1(25652)=50274\)
\( \sigma_2(25652)=871164630\)
\( \sigma_3(25652)=19267967775942\)

Also of interest is the count of numbers that are relatively prime to 25652 where 1 is regarded as being relatively prime to all numbers. This is known as the totient of the number and can be written as \( \phi=11440 \). All multiples of 2, 11 and 53 (and 25652 itself) will be excluded from this count. For prime numbers, the totient will always be one less than the number itself. For example, 25667 is prime and its totient is 25666.

Moving along, we note that some numbers are so-called square numbers. An example of such a number is 25600 which is equal to the square of 160. Visually, this number could be represented as shown in Figure 2.

Figure 2

In the case of 25652, it is not a square number but it can be represented a sum of two squares, since$$25652=23716 + 1936 = 154^2 + 44^2 $$So, visually, the number could be represented as shown in Figure 3:

Figure 3

Only certain numbers can be represented as a sum of two squares. If a number has a 4k+3 prime factor that is raised to an odd number (1, 3, 5, .. ), then it cannot be represented as a sum of two squares. Most numbers can be represented as sum of three squares, provided that there is not a remainder of 7 when the number is divided by 8. 25652 gives a remainder of 4 when divided by 8 and it can be represented as a sum of three squares in 20 different ways as shown in Figure 4:

Figure 4

Figure 5

Visually, the nineteen triplets in Figure 4 that contain no zeroes, could be used as the sides of three squares, to represent 25652. The same approach can be used for cubes. For example:$$25665=11^3+23^3+23^3$$Rather less frequently, a number can be represented as a sum of two cubes. For example:$$25720 = 11^3 + 29^3 $$Such numbers can be envisaged as comprising three cubes (in the case of 25665) or two cubes (in the case of 25720). See Figure 5 for a not-to-scale representation of 25720.

In my previous post, I considered the seed numbers that would be needed to make a number a part of a Fibonacci sequence. In the case of 25652, these numbers would be 52 and 146, producing the sequence:$$25652,15854,9798,6056,3742,2314,1428,886,542,344,198,146,52$$Similarly, the three seed numbers for 25652 to be part of a tribonacci sequence are 10, 30 and 63, producing the sequence:$$25652,13947,7583,4122,2242,1219,661,362,196,103,63,30,10$$So I could go on but I'll leave off there and try to summarise things via Figure 6.

Figure 6