I do meth with an a

<3 number theory <3

general info

Every Sunday, I post a number and one or more mathematical facts about that number at 12:30 MST. All daily number posts are tagged with their number and #number, all other posts are tagged with with #no number. I will probably get things wrong so please inform me through the askbox when I do. If you have a question about a fact or number or whatever, don't be afraid to ask.

about me

This blog is not about me, it's about numbers so i'll keep it short.

I use he/him pronouns. I'm currently studying mathematics, so math is definitely a special interest of mine. My main blog is @z-score.

about submissions

There's a ton of fun facts about this number. 384 is the sum of a twin prime pair (191 + 193), as well as the sum of six consecutive primes (53 + 59 + 61 + 67 + 71 + 73)

Also, it is the value of the double factorial of 8:

8!! = 8 * 6 * 4 * 2 = 384

It is also an abundant number, meaning that the sum of its proper divisors is more than the number.

In recreational mathematics, a Harshad number in a given base is an integer that is divisible by the sum of its digits when written in that base. For example, in base 10, 18 is a Harshad number because 1 + 8 = 9, and 18 is divisible by 9. 384 is a Harshad number in bases 2, 3, 4, 5, 7, 8, 9, 13, 17 and 62 other bases.

Computing

384 also appears in computing quite a bit since it is a low multiple of a power of 2:

384 = 3 x 128 = 3 x 2^7

For example, the secure hash function SHA-384 has a 384-bit digest.

Also, the Virtual Boy's screen resolution is 384 x 224.

Thanks for the number suggestion!

While playing Sudoku Topology, one can calculate that for Sphere32, there are 2¹⁶ choices for swapping around pairs of cells that are in the exact same pair of regions, 4!² choices for permuting rows and columns, and 2 choices for switching the rows with the columns. that multiplies out to 2²³×3².

For the nested cuboctahedron grid, there are 48 ways of applying symmetries, 2¹⁸ choices for swapping pairs of cells that are in the exact same set of regions, and 3! ways of permuting the layers. This also multiplies out to 2²³×3². However, I think you need to divide by 2 because the first two lists of symmetries both include central inversion. This thus makes the number of symmetries equal to 2²²×3².

It is well known that this series:

1 + 1/2 + 1/3 + 1/4 + 1/5 + ...

(the harmonic series) diverges. In 1737, Euler showed that the sum of the reciprocals of the primes also diverges:

1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

But let's now consider the reciprocals of the twin primes, where a twin prime pair is a pair of primes (p,p+2):

(1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + (1/17 + 1/19) + ...

Each twin prime pair contributes both reciprocals, so primes that belong to two twin prime pairs (such as 5) are counted twice.

By Brun's theorem, this series converges to a finite number. Remarkably, this is true whether there are finitely or infinitely many twin primes (an open problem).

In 1976, Richard P. Brent calculated all twin primes up to 100 billion and obtained an estimate for what this series converges to, namely:

B ≈ 1.9021605823, with an error of about 8 x 10^-10.

Since then, the number of terms has been calculated using twin primes up to 10^16, giving the result

B ≈ 1.902160583104

This number is the largest known sublime number! But to understand what a sublime number is, we need to know what a perfect number is. A perfect number is a positive integer that is equal to the sum of its positive proper divisors (that is, divisors excluding the number itself). For example, 6 has proper divisors 1, 2, and 3, and

1 + 2 + 3 = 6. The next perfect number is 28, which has proper divisors 1, 2, 4, 7, and 14, and

1 + 2 + 4 + 7 + 14 = 28.

Sublime Numbers

A sublime number is a positive integer which has a perfect number of positive divisors (including itself), and whose positive divisors add up to another perfect number.

12 is an example of a sublime number. It has 6 positive divisors, and 6 is a perfect number:

1, 2, 3, 4, 6, and 12

and the sum of these is a perfect number:

1 + 2 + 3 + 4 + 6 + 12 = 28

As of July 2026, there are only two known sublime numbers, namely 12 and 6,086,555,670,238,378,989,670,371,734,243,169,622,657,830,773,351,885,970,528,324,860,512,791,691,264

Euclid-Euler Theorem

We can show that this number is a sublime number by the Euclid-Euler theorem. It states that a positive integer is an even perfect number if and only if it has the form 2^{p-1}(2^p - 1), where 2^p - 1 is a prime number.

In our case, that huge number has 8128 positive divisors, and

8128 = 2^6 * (2^7 -1),

where 2^7 - 1 is prime. By the Euclid-Euler theorem, 8128 is therefore a perfect number.

Even more remarkably, these 8128 divisors add up to

2^{126} * (2^{127} - 1),

which is the 12th even perfect number because 2^{127} - 1 is the 12th Mersenne prime.

This is one of those rare mathematical constants that appears almost everywhere, yet is surprisingly difficult to prove appears for any specific famous number.

A simple continued fraction is a continued fraction whose numerators are all equal to one, while the denominators are determined by a sequence of positive integers a_0, a_1, a_2, ... This is the general form:

Every real number has a simple continued fraction expansion. Rational numbers have finite continued fractions, while irrational numbers have unique infinite continued fraction expansions.

Below are some examples of the irrational kind:

Now that we have the basics out of the way, I can introduce one of my favorite constants. :3 Aleksandr Yakovlevich Khinchin proved something remarkable. For almost every real number x, if

x = [a_0, a_1, a_2, ...]

then the geometric mean of the coefficients converge to the same constant, regardless of which number you started with. That is: lim_{n -> ∞} (a_1 * a_2 * a_3 * ... * a_n)^{1/n} = K_0

The decimal value of Khinchin's constant is given by:

K_0 = 2.68545 20010 653006 44530 ...

Almost all real numbers satisfy this property, such as pi, cubic roots, quartic roots, natural logarithms, the Euler-Mascheroni constant, Apery's constant, and Khinchin's constant itself.

Here, "almost all" has a precise mathematical meaning: the set of exceptions has Lebesgue measure zero. So while there are infinitely many exceptions, they occupy essentially no space on the real number line. However, there are some real numbers that do NOT satisfy this property, such as irrational square roots of integers, the golden ratio, rational numbers (so any finite simple continued fraction), and Euler's number.

Curiously, no famous constant such as π or Khinchin's constant has actually been proven to satisfy this limit. Their convergence is supported only by extensive numerical computations.

Below is a plot of the geometric means of the continued fraction coefficients for pi (red), the Euler-Mascheroni constant (blue), and the cubic root of 2 (green). All three curves appear to approach the black line, which represents Khinchin's constant.

A holyhedron is a type of 3-dimensional shape. It is a polyhedron where every face has at least one hole. The full definition excludes faces where the hole touches the outer boundary of the face and where the boundaries of two holes intersect.

In other words, each face is a planar polygon containing at least one interior hole, with holes disjoint from each other and from the boundary in a precise topological sense.

So instead of faces being simple polygons, each face becomes something closer to a "Swiss chess polygon". This makes holyhedra very different from familiar polyhedra like cubes or dodecahedra. Even though the definition sounds like a small modification, it forces the geometry to become extremely complicated extremely quickly.

Essentially, a hole in a face requires additional edges. These edges must belong to adjacent faces. Adjacent faces now inherit the hole constraint that force further subdivisions. Subdivision creates more faces, which may require more holes. So instead of a single modification, you get a recursive geometric bookkeeping problem where every local fix propagates globally.

A Brief History

The idea of a holyhedron was introduced by John H. Conway (known famously as the creator of Conway's Game of Life) in 1997 as part of a prize problem. The challenge was to construct a finite holyhedron. At the time, it was not obvious that such an object could exist at all. The requirement that every face must contain a hole creates strong global constraints: holes force extra structure, which forces more faces, which forces more holes, and so on.

Conway's original expectation was that if a solution existed, it would likely have on the order of 100 faces or so. So he offered a prize of $10,000 USD divided by the number of faces, expecting to pay out around $1,000.

That intuition turned out to be dramatically optimistic

Various Constructions

When Jade P. Vinson first heard about this problem, he was a grad student at Princeton. He spent a great deal of time working on finding a solution. Eventually, it paid off with him being the first to discover a holyhedron, one with 78,585,627 faces in 1999.

This solution was worth about $0.00012724973 USD, but it's unknown if that money was ever paid out.

In 2003, Don Hatch presented a holyhedron with 492 faces, which was about $20.33 in prize money. And in 2026, a new construction was obtained by Geby Jaff using Archivara with 476 faces.

These shapes are (unsurprisingly) hard to visualize, but here are a few images showcasing the construction of Don Hatch's 492-face shape, which is built in 9 layers. It should give good intuition for what I mean by "holes", and should instill fear for the first holyhedron discovered with over 78 million faces.

To see the full construction (with .wrl files), below is a link:

And here's a cutaway of the full construction:

78,585,627 is remarkable because it is forced. It comes from a simple-sounding rule applied relentlessly across a geometric object until intuition completely breaks down.

That's one of the recurring surprises in mathematics: constraints that look harmless locally can become overwhelming globally, and the resulting objects often tell you more about structure than about shape.

A (0,1)-matrix is an integer matrix whose entries are only 0s and 1s. It's often called a logical matrix or a binary matrix.

Now let F(m, n) be the number of m x n binary matrices in which no two horizontally or vertically adjacent entries are both 1.

Restricting our attention to square matrices, for n = 1, you have 2 ways (you can either have a 1 in the entry or a 0), and for n = 2, you have 7 ways, as shown below: | 0 0 | | 0 0 | | 0 0 | | 0 1 | | 1 0 | | 1 0 | | 0 1 | | 0 0 | , | 0 1 | , | 1 0 | , | 0 0 | , | 0 0 | , | 0 1 | , | 1 0 |

For n = 3, there are 63 matrices. For n = 4, there are 1234, and so on.

Then the hard square entropy constant is defined by

κ = lim_{n -> ∞} [F(n,n)]^{1 / n^2} ≈ 1.503048082...

As n grows, F(n,n) increases extremely rapidly. The hard square entropy constant is the exponential growth rate per cell. Roughly speaking,

F(n,n) ≈ κ^n^2

so each additional cell contributes a factor of about 1.503048082 to the total number of valid matrices.

This number is somehow prime. In particular, it is the largest known prime number with consecutively increasing digits.

But what makes this prime even more special is that it trivially answers Gelfand's fourth question.

For context, Israel M. Gelfand asked 4 questions about if the leading (most significant) digits of the powers of numbers (i.e. 2^n, 3^n, . . . , 9^n) ever form all possible rows of digits or follow a particular pattern over time.

The problem investigates the behavior of the most significant decimal digits of these numbers as n increases (n = 1, 2, 3, . . . ). The four specific questions are:

Will the digit 9 ever occur in the leftmost position for 2^n? As it turns out, yes. The first such occurrence is at 2^{53}, namely 9,007,199,254,740,992

Will the row 23456789 ever appear for n > 1? That is, for some n, is is true that the leading digit of 2^n is 2, leading digit of 3^n is 3, . . . , and the leading digit of 9^n is 9? As it turns out, the only integer n > 0 where this holds is n = 1 (a link to the paper proving this is below)

Will a sequence of all the same digits ever occur? By the same paper linked below, this does not hold for any n.

Will the decimal expansion of an 8-digit prime ever occur? That is, if you concatenate the leading digits of these powers, is the result ever prime? This is, in my opinion, the most interesting of the 4. Unlike the other 3 questions that took years to prove, the case n = 1 solves it. That is, 23456789 is prime. As a quick final note, there are precisely 17,596 integers in the set of decimal expansions, of which 1,127 are prime numbers.

Overall, I consider it to be an interesting number. 2 of Gelfand's questions were "trivial" in some sense, and the fact that the fourth problem is solved by a prime with consecutively increasing digits is fascinating to me.

Wow, two numbers in one post! Taken individually, neither 714 nor 715 is especially remarkable. Together, however, they are connected by one of the most delightful coincidences in recreational mathematics. On April 8, 1974, baseball player Hank Aaron hit his 715th major league home run. The prior record of 714 was held by Babe Ruth. Understandably, the event received considerable coverage in newspapers and on TV.

However, mathematicians Carol Nelson, David E. Penney, and Carl Pomerance, who all worked at the University of Georgia, found another fascination for these numbers.

Namely, they noted that 714 x 715 is the product of the first seven prime numbers!

714 = 2 x 3 x 7 x 17

715 = 5 x 11 x 13

Thus, 714 x 715 = 510,510 = 2 x 3 x 5 x 7 x 11 x 13 x 17

Pomerance and his colleagues wondered whether there were other pairs of consecutive numbers whose product is also the product of the first k primes.

The first few instances are easy enough to find:

1 x 2 = 2

2 x 3 = 2 x 3

5 x 6 = 2 x 3 x 5

14 x 15 = 2 x 3 x 5 x 7

714 x 715 = 2 x 3 x 5 x 7 x 9 x 11 x 13 x 17

The mathematicians then used a computer to search for such pairs, going as far as products of the first 3,049 primes (numbers up to 6,021 digits long). They didn't find any additional examples.

It is now conjectured that 714 and 715 are the last pair of consecutive integers whose product is the product of the first k primes for some k. To this day, the conjecture remained an open problem.

Another interesting fact they found is that the sum of the prime factors of 714 is 29, and 715 is the same:

2 + 3 + 7 + 17 = 29

5 + 11 + 13 = 29

Again, Pomerance and his colleagues wondered if there were other numbers that had this property and again a computer was used to search for such pairs up to 20,000. Here are the first few examples. Note that repeated prime factors are counted with multiplicity:

5 -> 5 = 5 and 6 -> 2 + 3 = 5

8 -> 2 + 2 + 2 = 6 and 9 -> 3 + 3 = 6

15 -> 3 + 5 = 8 and 16 -> 2 + 2 + 2 + 2 = 8

77 -> 7 + 11 and 78 -> 2 + 3 + 13 = 18

125 -> 5 + 5 + 5 = 15 and 126 -> 2 + 3 + 3 + 7 = 15

714 -> 2 + 3 + 7 + 17 = 29 and 715 -> 5 + 11 + 13 = 29

948 -> 86 and 949 -> 86

Such numbers are now known as Ruth-Aaron pairs after the two baseball players.

Pomerance and his colleagues published their results in the Journal of Recreational Mathematics where they conjectured that Ruth-Aaron pairs become increasingly sparse as numbers grow larger.

After publication, Pomerance received a letter from the legendary mathematician Paul Erdös, who offered to show him how to prove the conjecture. Erdös was invited to Georgia, and the meeting resulted in a joint paper giving the proof, which was published in 1978. It was the first of more than 40 papers the two would write together.

During his lifetime, Erdös collaborated with so many mathematicians that these efforts have been captured in something called the Erdös number. Erdös is assigned the number 0. People who have coauthored a paper with him have an Erdös number of 1. People who have coauthored a paper not with Erdös but with someone who coauthored a paper with Erdös get the number 2, and so on.

Somehow, I've done posts for 144 and 146, but not 145 . . .

In number theory, a happy number is a number that eventually reaches 1 when the number is replaced by the sum of the square of each digit.

For example, 13 is a happy number because

13 -> 1^2 + 3^2

= 10 -> 1^2 + 0^2

= 1

1 is itself a happy number because 1^2 = 1.

But there are numbers that are not happy. For example, 4 is an unhappy number because it gets stuck in an infinite loop (known as the Melancoil, whose name was popularized by Matt Parker):

4 -> 4^2

= 16 -> 1^2 + 6^2

= 37 -> 3^2 + 7^2

= 58 -> 5^2 + 8^2

= 89 -> 8^2 + 9^2

= 145 -> 1^2 + 4^2 + 5^2

= 42 -> 4^2 + 2^2

= 20 -> 2^2 + 0^2

= 4

145 is an interesting number because it is the only 3 digit number in the Melancoil.

All Numbers are Happy or Unhappy

I figured it was worth talking about how all numbers are happy or unhappy, and thus all numbers will eventually reach 1 or 145.

First note that large numbers always decrease. Suppose a number n has d digits. The largest possible value of f(n) occurs when every digit is 9, so f(n) <= 81d.

You can then compare 81d with the smallest d-digit number, 10^{d-1}.

For d >= 4, 81d <= 10^{d-1}

For example,

d = 4 : 324 < 1000 d = 5 : 405 < 10000

and it only gets more true as d grows.

Therefore, whenever n has at least 4 digits, f(n) < n. Thus, every sufficiently large number with at least four digits strictly decreases after one step.

If n has at most 3 digits, then

f(n) <= 3 * 81 = 243. Thus after at most a few iterations, every orbit enters the set

{1, 2, . . . , 243}

Once you're inside this set, every future value stays inside it because the sum of squares of three digits is still at most 243. Then you can use the fact that finite sets force cycles. If you keep iterating forever, eventually some value must repeat (this is essentially the pigeonhole principle).

Now the problem is finite: check the numbers 1, . . . , 243

A computer search shows there are exactly two cycles:

1 -> 1

4 -> 16 -> 37 -> 58 -> 89 -> 145 -> 42 -> 20 -> 4

Since every orbit must eventually enter a cycle, and these are the only two cycles, every positive integer eventually reaches either 1 or the Melancoil.

Factorion

As a final interesting fact, 145 is also a factorion, meaning that

145 = 1! + 4! + 5! = 1 + 24 + 120

It's one of only 4 factorions in base 10 (the others being 1, 2, and 40585).

Mathematicians have proven mathematically that no factorion can exist beyond seven digits. Any 8-digit number is at least 10,000,000, but the largest possible sum of the factorials of eight digits is only

8 * 9! = 2,903,040

Therefore, no 8-digit number can be a factorion. The same argument works even more strongly for numbers with more digits, so every factorion must have at most 7 digits.

Conclusion

142857 is one of the most famous "curious numbers" in mathematics. It appears as the repeating digits of 1/7, and it has several surprising properties that make it worth remembering.

Cyclic Number

This number is the best known cyclic number in base 10. It is the six repeating digits of 1/7 (0.\overline{142857}) If you multiply 142857 by 2, 3, 4, 5, or 6, the answer will be a cyclic permutation of itself, and will correspond to the repeating digits of 2/7, 3/7, 4/7, 5/7, and 6/7 respectively:

1 * 142,857 = 142,857 -> 1/7 = 0.\overline{142857}

2 * 142,857 = 285,714 -> 2/7 = 0.\overline{285714}

3 * 142,857 = 428,571 -> 3/7 = 0.\overline{428571}

4 * 142,857 = 571,428 -> 4/7 = 0.\overline{571428}

5 * 142,857 = 714,285 -> 5/7 = 0.\overline{714285}

6 * 142,857 = 857,142 -> 6/7 = 0.\overline{857142}

7 * 142,857 = 999,999 -> 7/7 = 0.\overline{999999} = 1

If multiplying by an integer greater than 7, there's a simple process to get to a cyclic permutation of 142857. By adding the rightmost six digits to the remaining digits and repeating this process until only 6 digits are left, it will result in a cyclic permutation of 142857. For example, 815 * 142,857 = 116428455

Take the last 6 digits (428,455), and add them to the remaining digits:

116 + 428,455 = 428,571, which is a cyclic permutation.

This happens because 10^6 is congruent to 1 (mod 142857). Because of that, replacing a block of digits by the sum of its six-digit chunks preserves the remainder modulo 142857.

Mental Division

I've actually used this trick before when I needed to divide by 7 quickly. If you have some natural number, it's fairly trivial to find the integer part. Once you have it, you can then divide the remainder by 7 and consider 142857. If your remainder is n, you look at n/7 in the table above.

For example, to compute 103 / 7, first note that 14 * 7 is 98, leaving a remainder of 5. Then note that 5 / 7 = 0.\overline{714285}.

So the 103 / 7 = 14.\overline{714285}.

Kaprekar Number

142857 also appears in a completely different corner of recreational mathematics.

A p-Kaprekar number is a number whose square can be split into two parts (whose second part is p-digits long) and add to the original number. For example, 45 is a 2-Kaprekar number because 45^2 = 2025, and 20 + 25 = 45.

142857 is a 6-Kaprekar number since 142857^2 = 20,408,122,449, and 20,408 + 122,449 = 142,857.

Conclusion

1023 is interesting because it shows up everywhere from floating point arithmetic to error-correcting codes to satellites!

1023 is a sum of 5 consecutive primes (193, 197, 199, 211, and 223)

It is a sub-perfect power (a perfect power minus 1) that can be expressed in two different ways, namely 4^5 - 1 and 2^10 - 1. The latter form makes this a Mersenne number.

1023 is a palindromic number in base 2 and base 4. In base 2, it is a string of 1's, and in base 4 it is a string of 3's.

Polycubes are the 3-dimensional analogue of planar polyominoes (like Tetris pieces). Below is a picture of the tetracubes (4 blocks that are connected by one or more faces). If you consider the number of ways you can arrange 7 blocks to be distinct (polycubes with 7 blocks), then there are 1023 distinct combinations.

It is a Stirling number of the second kind, specifically S(11, 2). This means that if you have 11 objects, and you want to group them into 2 unlabeled nonempty subsets, then there are 1023 ways to do it.

In computer science, floating point numbers are often represented using something known as the IEEE-754 double-precision floating point standard. This is the standard used for representing doubles in a ton of different programming languages. It uses 1 sign bit, 11 exponent bits, and 52 fraction (mantissa) bits. To represent a number, you consider the 52 fraction bits to be 0.1101101011101... (52 binary bits beyond the decimal point), and then you multiply by the exponent bit (specifically, multiply by 2 ^ (exponent number - 1023)). For example, 3/4 can be represented as 0 01111111110 1000000000000000000000000000000000000000000000000000. The 0 in the sign bit means this is positive. The 01111111110 represents 1022 in binary, and with the bias, this is equal to 1022 - 1023 = -1. Then the mantissa is 1.1000... (there's a hidden 1 in front for more precision). So this number is 2^-1 * 1.1 = 0.11, which in binary is 1/2 + 1/4 = 3/4. The reason 1023 appears as the bias is because with 11 exponent bits, there are 2^11 = 2048 possible exponent values. IEEE-753 chooses the bias to be 2^{11-1} - 1 = 1023 which centers the representable exponents roughly around 0. So this makes the smallest normal exponent 1 - 1023 = -1022, and the largest normal exponent to be 2046 - 1023 = 1023 (Values 0 and 2047 in the exponent field are reserved for subnormals, infinity, and NaN). And as a quick note, IEEE-754 uses a bias instead of two's complement because comparing floating point numbers becomes easier in hardware. Positive exponents sort correctly as unsigned integers, and the exponent field can be compared lexicographically. It also avoids needing a separate sign bit for the exponent.

A Reed-Muller code is a type of error-correcting code where you turn a polynomial into a long pattern of bits, and the redundancy lets you recover from errors (it's one of the ways a CD can still play music even if it's scratched a little bit). A Boolean polynomial uses variables that are only 0 or 1. For example, f(x, y, z) = x + yz where arithmetic is mod 2, multiplication is AND and addition is XOR is an example of a boolean polynomial since yz = 1 only if both are 1 and x + yz means XOR. You then evaluate it on every possible input. x | y | z | f(x, y, z) 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 This gives the bit string 00011110. That bit string is the codeword. This helps with errors since not every 8-bit string is allowed. Only strings that come from evaluating low-degree polynomials are valid codewords. So if noise flips a few bits during transmission, the receiver can ask which valid polynomial-generated codeword is closest and recover the original message. Generally, Reed-Muller codes are written RM(r,m) where r = maximum degree allowed and m = number of variables. For example, RM(1,3) means you can have a max degree of 1 with 3 variables. So allowed polynomials are of the form a + bx + cy + dz. There is 1 constant terms and 3 linear terms, which gives 4 coefficients and thus a code dimension of 4. The code length is 2^3 = 8 because we evaluate on all 8 inputs. Note that the dimensions of Reed-Muller codes comes from this formula: dim(RM(r,m)) = \Sum_{i=0}^r (m, i) (add up the binomials from m choose 1 to m choose r). 1023 also appears in Reed-Muller codes. For the code RM(9,10), the dimension is \Sum_{i=1}^9 (10, i) = 1023 while the code length is 1024. So RM(9,10) encodes 1023 its of information into 1024 codeworks with enough redundancy for error correction.

This number also appears in technology. A GPS (Global Positioning System) satellite continuously broadcasts its current position, the current time, and a special pseudorandom bit pattern (the PRN code). Your phone receives signals from multiple satellites and computes distance = c * time delay (where c is speed of light and time delay is how long the signal took to arrive). With distances to several satellites, your phone triangulates your location. GPS time is represented as a week number plus the seconds within that week. Because GPS started on January 6th, 1980, you can store time as week 2417, second 123456 of that week instead of storing a full date like May 24th, 2026. GPS works on a 10-digit binary counter that runs for 1024 weeks, 0 to 1023, at which point an integer overflow causes its internal value to roll over to 0 again. 1024 weeks is around 19.7 years, which is why GPS rollovers happened in 1999, 2019, and will happen again around 2038.

But there's another place where 1023 is found with GPS. Each satellite sends a radio signal that looks roughly like a very accurate carrier wave, multiplied by a rapidly changing binary code, and then navigation data. That rapidly changing binary code is the famous C/A code which you can think of as an "identifier" for a given satellite. The C/A code exists because if every satellite transmitted just a plain sine wave, you would have satellites that interfere with each other which would making timing ambiguous. So each satellite transmits a unique pseudorandom sequence like 11011010011100110011101010111000111100 . . . These sequences look random, but they are actually deterministic and repeat every 1023 bits ("chips"). Your receiver (phone) generates the same code internally and slides it back and forth until it matches. The amount of shift tells you the travel time and which satellite gave the transmission. The civilian GPS C/A code is generated using a 10-stage linear feedback shift register (LFSR), which is a simple way to generate a long sequence of bits that looks random but is actually completely deterministic. It's basically a row of bits, a shift operation, and some XOR logic feeding back into the front. An n-stage maximal LFSR produces a sequence length of 2^n - 1 before repeating. So for n = 10, we have 2^10 - 1 = 1023. Thus the GPS C/A code has length 1023 chips. This also means that a GPS's radio signals at certain levels operate with a period of 1023, and the frequencies used are multiples of 1023 kHz. This makes it so that one full 1023-chip code period lasts exactly 1 millisecond. Since 1023 chips / ms = 1.023 * 10^6 chips / s, we get the famous 1.023 MHz chip clock. This makes timing and ranging calculations extremely convenient. Hence the system clock architecture was built around this base frequency.

A Smith number is a composite number which has the sum of its digits equal to the sum of the digits in its prime factorization.

For example, 85 has the prime factorization 5 * 17. Adding up the digits of 85, we get 8 + 5 = 13. Adding up the digits of the prime factorization, we get 5 + 1 + 7 = 13.

Today's number, 4,937,775, is also a Smith number and is the reason these numbers are called Smith numbers in the first place.

They were named by Albert Wilansky of Lehigh University as he noticed the property in the phone number (493-7775) of his brother-in-law Harold Smith.

The prime factorization of 4937775 = 3 * 5 * 5 * 65837.

Adding up the digits of the number gives 4 + 9 + 3 + 7 + 7 + 7 + 5 = 42

Most famous constants announce themselves immediately. For example, π appears wherever circles show up, e emerges from growth and calculus, and so on. But there's a stranger hiding deep within this infinite series:

ζ(3) = 1 + 1 / 2^3 + 1 / 3^3 + 1 / 4^3 + . . .

The number this sum converges to is called Apéry's constant, and despite looking innocent, it resisted proof for centuries. Mathematicians strongly suspected that it was irrational but nobody could prove it until 1978.

The Zeta Function Apéry's constant comes from one of the most important objects in mathematics: the Riemann zeta function.

For real numbers s > 1,

ζ(s) = ∑_{n=1}^∞ 1 / n^s

(the infinite series of 1 / 1 ^ s + 1 / 2 ^ s + 1 / 3 ^ s + . . . )

At first glance, this is just an infinite sum. But the zeta function secretly connects prime numbers, complex analysis, quantum physics, probability, cryptography, and the distribution of the primes themselves.

Some values are beautifully understood. For example,

ζ(2) = π^2 / 6

ζ(4) = π^4 / 90

In fact, every even positive integer produces a formula involving powers of π. But the odd inputs are another story.

Nobody knows a comparably elegant formula for

ζ(3), ζ(5), ζ(7), . . .

These numbers are mysterious, and ζ(3) became the first "battleground".

Roger Apéry's Bombshell

Numerically, Apéry's constant equals approximately

1.202056903159694...

The question sounds deceptively simple: Is this number rational? For over 200 years, nobody knew. That's remarkable because Euler had solved the analogous problem for ζ(2) in the 1700s.

In 1978, French Mathematician Roger Apéry announced that ζ(3) was irrational in a lecture. The announcement was met with criticism in part because Apéry was relatively unknown at the time. He also gave the lecture in French, made jokes throughout, and omitted several key explanations needed to follow the proof..

For example, there was an equation at the beginning of his lecture that no one knew but formed the core of his proof. When asked where this equation came from, Apéry is said to have answered "They grow in my garden," which was said to have caused many in the audience to stand up and leave the room.

However, someone in attendance had an electronic calculator (uncommon at the time) and with a short program, checked Apéry's equation and found it correct.

The equation in question is below, which was an unknown series representation of ζ(3) at the time:

With this expression, he was able to use a condition for irrationality that German mathematician Gustav Lejeune Dirichlet had derived in the 19th century. It states that a number χ is irrational if there are an infinite number of integers p and q, so that the following inequality is satisfied:

Here, c and δ denote constant values. Although the formula looks complicated, it basically means that χ can be approximated by fractions, but there is no fractional number that corresponds to χ exactly. Apéry succeeded in deriving this inequality for ζ(3), and thus the number is irrational.

In simpler terms, Apéry's proof constructed two sequences of integers:

a_n and b_n

such that

a_n / b_n

approximate ζ(3) far too well for a rational number.

This is the key philosophical idea. If a number is rational, there are limits to how accurately fractions can approximate it without eventually becoming exact. Apéry built approximations that violated those limits.

The machinery involved strange recursive sequences and combinatorial identities that seemed to come out of nowhere.

Even now, many mathematicians describe the proof as "magical".

To honor his work, the value of ζ(3) now bears his name and is known as Apéry's constant. This doesn't answer all of the questions associated with the number, however. We are still looking for a clear numerical value for ζ(3) that can be expressed with known constants, much in the same way ζ(2) is.

But regardless, Apéry's constant feels like an accident. It emerges from a simple series, has no known simple closed form, and required centuries to understand even partially. And yet it keeps appearing across math and physics like a recurring character in a story nobody fully understands.

the First Inaccessible Cardinal

(this number was submitted by @hyperoperationfractallisation)

If you've spent any amount of time around set theory, you've probably seen increasingly large infinities, ℵ0, ℵ1, and so on, marching upwards in a tidy sequence. It's tempting to think that all infinite sizes behave roughly the same way: you take a set, build something bigger from it, and keep going forever.

But at some point, something strange happens.

Eventually, you encounter cardinals that cannot be obtained from smaller cardinals using the usual operations of taking successors, unions, and power sets. This is where inaccessible cardinals enter the picture.

Building Cardinals

I suppose it's best to start simple.

We begin with ℵ0, the size of the natural numbers. From there, we can build larger and larger infinities in a few standard ways.

1. You can take the next size up (successor cardinals like ℵ1, ℵ2, ...)

ℵ1 is the smallest cardinal strictly bigger than ℵ0. Then ℵ2 is the smallest cardinal strictly bigger than ℵ1. And so on.

2. Take unions of smaller sets

Suppose you have a whole collection of sets, A0, A1, A2, and so on, where each is "small" (say, countable). If you take the union of all of these sets, you get a countable union of countable sets, which is still countable. But if you take ℵ1 many sets, each of size less than ℵ1, then their union can reach size ℵ1.

More generally, by taking unions of many smaller sets, you can sometimes build larger cardinals.

3. Take power sets (which are always strictly bigger).

Given a set A, we write P(A) for the set of all subsets of A. For example, if A = {1,2}, then P(A) = {∅, {1}, {2}, {1,2}}. So the size doubles (from 2 to 4). For finite sets, this doubles the exponent. A set of size n has a power set of size 2^n.

If |A| = ℵ0, then |P(A)| = 2^ℵ0, which is the size of the real numbers. And crucially, 2^ℵ0 > ℵ0. So power sets always give you a strictly larger infinity.

Using these operations, we can climb higher and higher through the infinite landscape.

So a natural question is to ask if every infinite size is reachable this way.

A Cardinal You Can't Reach

An inaccessible cardinal is, roughly speaking, a size so large it cannot be obtained from smaller cardinals using these operations applied only below κ.

More precisely, an inaccessible cardinal κ has 3 properties:

Uncountable: which means that it is bigger than ℵ0

Regular: κ cannot be written as the union of fewer than κ sets, each of size less than κ

Strong limit: This means that even taking power sets of smaller cardinals never catches up to it. More formally, for every λ < κ, we have 2^λ < κ

Taken together, this means that κ is out of reach of all the standard ways we construct larger infinities.

Why is it called "Inaccessible"?

The name is quite literal. If you imagine building cardinals step-by-step from below, an inaccessible cardinal is one you simply can't arrive at using those steps. No construction using fewer than κ many stages, each involving only smaller cardinals, can reach κ.

It's like climbing a staircase where, at some point, there's a platform floating above you with no steps leading to it.

Do They Actually Exist?

Here's a fun fact. You cannot prove that inaccessible cardinals exist using the standard axioms of set theory (ZFC).

To assume one exists, you have to add a new axiom. That's why inaccessible cardinals are called the first "large cardinal" concept.

Inaccessible cardinals are the entry point into a whole hierarchy of stronger and stronger assumptions about infinity.

Rubik's Cubes are fun. Those of you who are interested in solving these may be well aware of God's Number, which is defined as the maximum number of moves needed to optimally solve any scrambled cube.

For the standard 3 x 3 x 3 Rubik's Cube, this number has been proven to be 20.

But there's the opposite idea as well. A Devil's Algorithm is defined as a single fixed sequence A such that for any starting state g, some power A^k(g) is solved. In other words, it's an algorithm that you can perform on any starting state, and eventually you will get to a solved state.

Then the Devil's Number is the corresponding number of moves in the shortest Devil's Algorithm.

An upper bound for this number is 43,252,003,274,489,856,000, which is the total number of permutations for a 3 x 3 x 3 Rubik's Cube. To see why, note that the Rubik's Cube has 8 corners and 12 edges. There are 8! ways to arrange the corner cubes. Each corner has 3 possible orientations, although only 7 of them can be oriented independently, so we have 3^7 (2,187) possibilities.

Then for the edges, there are 12! / 2 (239,500,800) ways to arrange them, restricted from 12! because edges must be in an even permutation when the corners are.

Eleven edges can be flipped independently, with the flip of the twelfth depending on the preceding ones (much like the corners), which gives 2^11 (2,048) possibilities. Multiplying all of those out gives:

8! * 3^7 * 12! / 2 * 2^11 = 43,252,003,274,489,856,000.

But this is a bit underwhelming. Of course, if your algorithm cycles through all group elements, then eventually it must hit the identity. Or in other words, if you go through every single permutation, then you will solve the cube. So can we do better?

Unfortunately, I don't believe we can. The upper bound we found is enormous because it comes from a very inefficient idea: blindly cycling through all possible cube states. To make progress, it helps to rephrase the problem in the language of group theory. The set of all Rubik's Cube configurations forms a group G, where each element represents a cube state, and each move sequence is also an element of this group.

A Devil's Algorithm, then would correspond to a single element A in G such that for every configuration g in G, some power of A sends g to the solved state. In other words, repeated application of A eventually reaches the identity, no matter where you start.

This means that the sequence

g, Ag, A^2g, A^3g, ...

must eventually hit the identity for every g in G. Equivalently, for every g in G, some power of A must equal g^-1.

In particular, this implies that the cyclic subgroup generated by A, denoted <A>, must be "large enough" to reach every part of the group. The size of this subgroup is exactly the order of A, say m, meaning that A^m = e.

Now here's the key constraint: If A has order m, then repeated application of A can only produce at most m distinct states before cycling. So for a single algorithm to eventually reach all |G| possible configurations (from appropriate starting points), we must have

m ≥ |G| / k

for some integer k that accounts for how many different starting positions can share the same cycle.

While making this precise requires more careful group-theoretic analysis, it suggests an important takeaway: any Devil's algorithm must have very large order comparable to the size of the Cube group itself.

So although the trivial upper bound is about 4.3 x 10^19, even optimistic lower bounds remain astronomically large. If a Devil's Algorithm exists at all, it cannot be short.

So to restate more clearly, in group terms, you're asking for a single element whose cyclic subgroup intersects every group element's orbit at the identity. This is a strong condition, and it's not obvious that such an element exists. And thus, it's not even obvious that such an algorithm exists for the cube.