This is a reponse to one of Matt Parker’s videos on his StandUpMaths YouTube channel at

https://www.youtube.com/watch?v=LYKn0yUTIU4
Matt talked about a function defined for non-negative integers wherein the result is the number of letters in the name of the number.

For example,

`F(23) = 11` (11 letters in “twenty three”)

`F(11) = 6` (6 letters in “eleven”)

`F(6) = 3` (3 letters in “six”)

`F(3) = 5` (5 letters in “three”)

`F(5) = 4` (4 letters in “five”)

And that’s the end of the Parker chain, because `F(4) = 4`.

In English, all starting points eventually lead to 4.

Matt showed us that 23 is the lowest number that starts a Parker chain with a length of 6.

I am defining `C_n` as the lowest number starting a Parker chain of length n using the Parker algorithm.

`C_6 = 23`

Matt challenged us to find `C_7`, and to find any other interesting information about this algorithm, and to post any such findings as a comment on his video. I got a little bit carried away, and wrote a little more than will fit comfortably in a comment.

Assuming use of the short scale (for now) and not using the word “and”.

The longest number name under `10^30` is 373,373,373,373,373,373,373,373,373,373, or three hundred seventy-three octillion three hundred seventy-three septillion three hundred seventy-three sextillion three hundred seventy-three quintillion three hundred seventy-three quadrillion three hundred seventy-three trillion three hundred seventy-three billion three hundred seventy-three million three hundred seventy-three thousand three hundred seventy-three, with 321 alphabetic characters.

Therefore, in the longest possible Parker chain whose first number is `10^30` or lower, the second number in the chain is 321 or lower. All Parker chains starting with 321 or lower have 6 or fewer numbers in the chain. Therefore, the longest possible Parker chain whose first number is `10^30` or lower has a maximum of 7 numbers.

The first Parker chain of 7 numbers is 323, 23, 11, 6, 3, 5, 4.

`C_7 = 323`

How strange is that? Out of all of the possible starting numbers 30 digits or less, you only have to try as high as 323 to find an example Parker chain of the maximum possible length in that range.

As others have noted, the first Parker chain of 8 numbers has a starting number of just a bit over `10^30`.

`C_8 ~~ 10^30`

That’s a rather large gap.

Which makes me wonder, how large is the next gap?

Within each power of a thousand, the longest number name is approximately 32 letters longer than the longest number name in the previous power of a thousand. That’s an additional 24 letters for “three hundred seventy-three” and an average of 8 letters for the appropriate “*illion” for that power.

(There are many other, possibly infinite, numbers with the same name length, but a number with “373” repeating will be the lowest number of that name length.)

From that we can derive a simple formula to find the approximate location of the start of the next longer Parker chain.

`C_(n+1)` has as least `C_n` letters in its name. The first number with `C_(n+1)` letters in its name will be approximately

`C_(n+1) ~~ 10^(3/32 C_n)`

We can see this works for `C_7 = 323`.

`C_8 ~~ 10^(3/32 323)`

`C_8 ~~ 10^30`

And we can use that to estimate the size of the next gap.

At this scale, when `C_n` in is the form `10^x`, we can simplify the approximation.

`C_(n+1) ~~ 10^(3/32 10^x)`

`C_(n+1) ~~ 10^(1/10 10^x)`

`C_(n+1) ~~ 10^(10^(x-1))`

`C_9 ~~ 10^(10^(30-1))`

`C_9 ~~ 10^(10^29)`

That’s quite a gap.

Now we’re at a scale where we can simplify the approximation even farther, as our x in `10^x` is so large that subtracting one from it is a trivial difference. So for very large numbers:

`C_(n+1) ~~ 10^(C_n)`

From here on out, the gaps are easy to predict.

`C10 ~~ 10^(10^(10^29))`

`C11 ~~ 10^(10^(10^(10^29)))`

`C12 ~~ 10^(10^(10^(10^(10^29))))`

`C13 ~~ 10^(10^(10^(10^(10^(10^29)))))`

Now you may have noticed I made an assumption early on that you might take issue with. I am assuming an average of 8 letters for the name of each group of three digits, the so-called illion word. But it is actually reasonable.

As going to `C_9` takes us well beyond the range of numbers in any existing number naming system, we need a hypothetical number naming system that can handle these ridiculously large numbers. A reasonable extension of existing systems is possible.

To get up to `10^(10^29)`, we can’t just keep making up new illion words. We’ll run out of possibilities long before we get there. So we need something different. We already have something that can work, the so-called Long System previously used in Britain and still used in the rest of Europe and parts of Canada. We just need to follow it to a different logical conclusion than was done in the past.

Here is the Short System, used in the US, Britain and parts of Canada.

| | | Short system |

| `1000^0` | `10^0` | |

`1000^(2^0)` | `1000^1` | `10^3` | thousand |

`1000^(2^1)` | `1000^2` | `10^6` | million |

| `1000^3` | `10^9` | billion |

`1000^(2^2)` | `1000^4` | `10^12` | trillion |

| `1000^5` | `10^15` | quadrillion |

| `1000^6` | `10^18` | quintillion |

| `1000^7` | `10^21` | sextillion |

`1000^(2^3)` | `1000^8` | `10^24` | septillion |

| `1000^9` | `10^27` | octillion |

| `1000^10` | `10^30` | nonillion |

Compare that to the Long System. I like the long system. Mathematically it is more logical, with major names based on powers of one million. And we can use it for much larger numbers without needing to add new words.

| | | Short system | Long system |

| `1000^0` | `10^0` | | |

`1000^(2^0)` | `1000^1` | `10^3` | thousand | thousand |

`1000^(2^1)` | `1000^2` | `10^6` | million | million |

| `1000^3` | `10^9` | billion | thousand million |

`1000^(2^2)` | `1000^4` | `10^12` | trillion | billion |

| `1000^5` | `10^15` | quadrillion | thousand billion |

| `1000^6` | `10^18` | quintillion | trillion |

| `1000^7` | `10^21` | sextillion | thousand trillion |

`1000^(2^3)` | `1000^8` | `10^24` | septillion | quadrillion |

| `1000^9` | `10^27` | octillion | thousand quadrillion |

| `1000^10` | `10^30` | nonillion | quintillion |

| `1000^11` | `10^33` | | thousand quintillion |

| `1000^12` | `10^36` | | sextillion |

| `1000^13` | `10^39` | | thousand sextillion |

| `1000^14` | `10^42` | | septillion |

| `1000^15` | `10^45` | | thousand septillion |

`1000^(2^4)` | `1000^16` | `10^48` | | octillion |

| `1000^17` | `10^51` | | thousand octillion |

| `1000^18` | `10^54` | | nonillion |

| `1000^19` | `10^57` | | thousand nonillion |

| `1000^20` | `10^60` | | |

But that only gets us twice as far, and we need to go much, much farther. Here is my hypothetical Really Long System. It starts the same as the long system, but more than one pattern can start from there, and I took the one less traveled by.

| | | Short system | Long system | Really long system |

| `1000^0` | `10^0` | | | |

`1000^(2^0)` | `1000^1` | `10^3` | thousand | thousand | thousand |

`1000^(2^1)` | `1000^2` | `10^6` | million | million | million |

| `1000^3` | `10^9` | billion | thousand million | thousand million |

`1000^(2^2)` | `1000^4` | `10^12` | trillion | billion | billion |

| `1000^5` | `10^15` | quadrillion | thousand billion | thousand billion |

| `1000^6` | `10^18` | quintillion | trillion | million billion |

| `1000^7` | `10^21` | sextillion | thousand trillion | thousand million billion |

`1000^(2^3)` | `1000^8` | `10^24` | septillion | quadrillion | trillion |

| `1000^9` | `10^27` | octillion | thousand quadrillion | million trillion |

| `1000^10` | `10^30` | nonillion | quintillion | thousand trillion |

| `1000^11` | `10^33` | | thousand quintillion | million trillion |

| `1000^12` | `10^36` | | sextillion | thousand million trillion |

| `1000^13` | `10^39` | | thousand sextillion | billion trillion |

| `1000^14` | `10^42` | | septillion | thousand billion trillion |

| `1000^15` | `10^45` | | thousand septillion | million billion trillion |

`1000^(2^4)` | `1000^16` | `10^48` | | octillion | thousand million billion trillion |

| `1000^17` | `10^51` | | thousand octillion | quadrillion |

| `1000^18` | `10^54` | | nonillion | thousand quadrillion |

| `1000^19` | `10^57` | | thousand nonillion | million quadrillion |

| `1000^20` | `10^60` | | | thousand million quadrillion |

| `1000^21` | `10^63` | | | billion quadrillion |

| `1000^22` | `10^66` | | | thousand billion quadrillion |

| `1000^23` | `10^69` | | | million billion quadrillion |

| `1000^24` | `10^72` | | | thousand million billion quadrillion |

| `1000^25` | `10^75` | | | thousand trillion quadrillion |

| `1000^26` | `10^78` | | | million trillion quadrillion |

| `1000^27` | `10^81` | | | thousand million trillion quadrillion |

| `1000^28` | `10^84` | | | billion trillion quadrillion |

| `1000^29` | `10^87` | | | thousand billion trillion quadrillion |

| `1000^30` | `10^90` | | | million billion trillion quadrillion |

| `1000^31` | `10^93` | | | thousand million billion trillion quadrillion |

`1000^(2^5)` | `1000^32` | `10^96` | | | quintillion |

| | | | | `vdots` |

`1000^(2^6)` | `1000^64` | `10^99` | | | sextillion |

`1000^(2^7)` | `1000^128` | `10^102` | | | septillion |

`1000^(2^8)` | `1000^256` | `10^768` | | | octillion |

`1000^(2^9)` | `1000^512` | `10^1536` | | | nonillion |

In the short system, with every new illion term added, we can count up through another power of a thousand.

In the long system, with every new illion term added, we can count up through another 2 powers of a thousand.

In the really long system, it’s an exponential growth rate. With every new illion term added, we can count up through the square of the previous word. That is, we double the number of powers of a thousand we can count through.

With this system, we only need 95 illion words (and a really long time) to count all the way up to `10^(10^29)`.

So this is a system that hypothetically would work.

In the table above, it looks like some of those names are quite a bit longer than 8 letters, but they are only used for a single group like that when talking about round numbers. For example, this number 1,000,000,000,000,000,000,000 would “one thousand million billion” but this number 1,002,003,004,005,006,007,008 would be “one thousand two million three thousand four billion five thousand six million seven thousand eight”

So with this system we do maintain an average of eight letters per power of a thousand group into very, very large numbers. Eventually we will be forced to start using longer illion words that will drive the average above 8 letters, but by then the approximate numbers we are describing will be so monstrous in size that the increase will be relatively trivial.