## An oddity...

**Moderators:** gmalivuk, Moderators General, Prelates

### An oddity...

If you are familiar with narcissistic numbers, then this may be intriguing to you...

Let's say you take a block of text like this...

The term "narcissistic number" is sometimes used in a wider sense to mean a number that is equal to any mathematical manipulation of its own digits.

If you nix the symbols such as the spaces and the punctuation marks, you are left with plaintext. Counting the individual letters reveals that this sentence has 120 letters in it.

Write 120 in plaintext => one hundred twenty => 16 letters

Write 16 in plaintext => sixteen => 7 letters

Write 7 in plaintext => seven => 5 letters

Write 5 in plaintext => five => 4 letters

Write 4 in plaintext => four => 4 letters

Like the narcissistic number algorithm, this letter-counting algorithm always comes back to "FOUR" as the end result of the algorithm. I have mapped out paths with "FOUR" as the ending node and the next level nodes as being "FIVE" and "NINE". The rest can be extrapolated from there.

My main problem is how I would go about proving this.

I would say: Let N = plaintext, then f(N) = number of letter characters in plaintext

and Let P = new plaintext, then f(P) = number of letter characters in new plaintext

and through repitition of the algorithm: f(FOUR) = 4 ==> f(FOUR) = 4

What say you, fellow users?

Let's say you take a block of text like this...

The term "narcissistic number" is sometimes used in a wider sense to mean a number that is equal to any mathematical manipulation of its own digits.

If you nix the symbols such as the spaces and the punctuation marks, you are left with plaintext. Counting the individual letters reveals that this sentence has 120 letters in it.

Write 120 in plaintext => one hundred twenty => 16 letters

Write 16 in plaintext => sixteen => 7 letters

Write 7 in plaintext => seven => 5 letters

Write 5 in plaintext => five => 4 letters

Write 4 in plaintext => four => 4 letters

Like the narcissistic number algorithm, this letter-counting algorithm always comes back to "FOUR" as the end result of the algorithm. I have mapped out paths with "FOUR" as the ending node and the next level nodes as being "FIVE" and "NINE". The rest can be extrapolated from there.

My main problem is how I would go about proving this.

I would say: Let N = plaintext, then f(N) = number of letter characters in plaintext

and Let P = new plaintext, then f(P) = number of letter characters in new plaintext

and through repitition of the algorithm: f(FOUR) = 4 ==> f(FOUR) = 4

What say you, fellow users?

We all pay for life with death, so everything in between should be free.

-Bill Hicks

-Bill Hicks

- NathanielJ
**Posts:**882**Joined:**Sun Jan 13, 2008 9:04 pm UTC

### Re: An oddity...

I would first say something like "it can be proven by trial-and-error that for n <= 100, the conjecture that f^{k}(n) = 4 for some k >= 1 is true" (ie. just point out that the result is known to be true for small n).

Then the "proof" comes by noting that for any n sufficiently large (say, larger than 100), it is clearly the case that f(n) <= n, so it will eventually fall back into the "base case" above.

As far as how "rigorous" this proof is - as far as I can tell, it's essentially impossible to get more rigorous, because the assumption that f(n) <= n seems to be absolutely necessary (though of course quite believable). However, it could (however unrealistically) be the case that the word for a 1 followed by 361 zeroes takes more than that many letters to write out, in which case the sequence starting there could balloon out to infinity, rather than collapsing to four.

Then the "proof" comes by noting that for any n sufficiently large (say, larger than 100), it is clearly the case that f(n) <= n, so it will eventually fall back into the "base case" above.

As far as how "rigorous" this proof is - as far as I can tell, it's essentially impossible to get more rigorous, because the assumption that f(n) <= n seems to be absolutely necessary (though of course quite believable). However, it could (however unrealistically) be the case that the word for a 1 followed by 361 zeroes takes more than that many letters to write out, in which case the sequence starting there could balloon out to infinity, rather than collapsing to four.

### Re: An oddity...

I think you may be right, NathanielJ.

I know that just by mapping out possible solutions, several of them are already taken care of. So for n < 100, this would work. I know even if we were to take the most trivial case n = 0, then falls back to FOUR automatically. I think you may be onto something. I will look into it more.

Thank you very much.

I know that just by mapping out possible solutions, several of them are already taken care of. So for n < 100, this would work. I know even if we were to take the most trivial case n = 0, then falls back to FOUR automatically. I think you may be onto something. I will look into it more.

Thank you very much.

We all pay for life with death, so everything in between should be free.

-Bill Hicks

-Bill Hicks

### Re: An oddity...

There's not going to be any mathematical proof (unless you make some assumptions about language), because "how many letters are in the word that describes this number" isn't a mathematical property at all, it's a linguistic one. The function isn't even well defined, because all numbers aren't named. I mean, there's not really a way to describe 10

^{51}, unless you say "ten to the fifty first", but then you have a problem because things like 1000 could be "one thousand" or "ten to the third".### Re: An oddity...

This sounds fun. But as cycle just pointed out, you will have to rigorously define what the English expression for large numbers is, or else state your conjecture (and possible future theorem) only for integers no larger than some large N. It seems that the usual naming system will work up to M=999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999 and then break down.

By the way, 10^51 is "one sexdecillion".

However you could select a different naming system for large numbers. There may be a standardized method for producing Latin prefixes so that you can continue the (Latin root)illion pattern a while longer, but you will still need to develop a recursive way of naming numbers. I would be in favor of just sticking to integers -M to M.

By the way, 10^51 is "one sexdecillion".

However you could select a different naming system for large numbers. There may be a standardized method for producing Latin prefixes so that you can continue the (Latin root)illion pattern a while longer, but you will still need to develop a recursive way of naming numbers. I would be in favor of just sticking to integers -M to M.

### Re: An oddity...

That's ok, since the largest number is around 45,000,000,000.

This is a placeholder until I think of something more creative to put here.

- BeetlesBane
**Posts:**138**Joined:**Sun Apr 27, 2008 2:32 pm UTC**Location:**Not Chicago

### Re: An oddity...

Robin S wrote:That's ok, since the largest number is around 45,000,000,000.

Well someone's going to have nitpick this, guess it might as well be me. A google is enough larger to obviate "around".

### Re: An oddity...

Don't be silly; a google isn't a number. It's a web search engine, and also a stuffed toy (I read about this in Dave Gorman's Googlewhack Adventure). A googol is a number, but it's obviously not real since it's much bigger than 45,000,000,000. And since non-real numbers can't be bigger than real numbers under the standard definition of "bigger", it is a nonsensical concept and doesn't exist.

This is a placeholder until I think of something more creative to put here.

### Re: An oddity...

NathanielJ wrote:I would first say something like "it can be proven by trial-and-error that for n <= 100, the conjecture that f^{k}(n) = 4 for some k >= 1 is true" (ie. just point out that the result is known to be true for small n).

Then the "proof" comes by noting that for any n sufficiently large (say, larger than 100), it is clearly the case that f(n) <= n, so it will eventually fall back into the "base case" above.

Of course this doesn't actually work, there are plenty of functions that always get less but don't get arbitrarily low. None of them are on the integers but that's difficult to prove, it's quicker and more rigorous to use Strong Induction.

- Indigo is a lie.

Which idiot decided that websites can't go within 4cm of the edge of the screen?

There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

### Re: An oddity...

When I read NathanielJ's post, I got the impression that

was intended to be proven by induction (if a rigorous proof was required).it will eventually fall back into the "base case" above

This is a placeholder until I think of something more creative to put here.

### Re: An oddity...

BeetlesBane wrote:Robin S wrote:That's ok, since the largest number is around 45,000,000,000.

Well someone's going to have nitpick this, guess it might as well be me. A google is enough larger to obviate "around".

Someone's going to have to nitpick a reference to a joke? A clearly linked reference? Why?

All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

- BeetlesBane
**Posts:**138**Joined:**Sun Apr 27, 2008 2:32 pm UTC**Location:**Not Chicago

### Re: An oddity...

Because it's there.

### Re: An oddity...

Because people are too scared of rickrolls to click on links any more?

- Indigo is a lie.

Which idiot decided that websites can't go within 4cm of the edge of the screen?

There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

- skeptical scientist
- closed-minded spiritualist
**Posts:**6142**Joined:**Tue Nov 28, 2006 6:09 am UTC**Location:**San Francisco

### Re: An oddity...

Macbi wrote:Because people are too scared of rickrolls to click on links any more?

Seriously? Which is worse - the occasional rickroll, or never being able to click on a YouTube link ever again?

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

### Re: An oddity...

skeptical scientist wrote:Macbi wrote:Because people are too scared of rickrolls to click on links any more?

Seriously? Which is worse - the occasional rickroll, or never being able to click on a YouTube link ever again?

You say "never being able to click on a YouTube link ever again" like it's some sort of bad thing.

### Re: An oddity...

This kind of thing is hard to prove, I think, because it depends on the use of the English language, which is not maths. For instance, it could work up to some big number, then I could simply call some number X a name which has X letters. But I think if we take the assumption that, say f(n)<1000000000000 or something (fairly reasonable, I think, since noone wants to call a number something that takes an approximation of infinite time to say).

Of course, in different languages the function will have different (or possibly no) fixed points. Maybe it will have cycles, and some fixed points. There is no requirement on the function to be continuous, although in (a sensible) naming convention there are likely to be patterns. Discuss.

Of course, in different languages the function will have different (or possibly no) fixed points. Maybe it will have cycles, and some fixed points. There is no requirement on the function to be continuous, although in (a sensible) naming convention there are likely to be patterns. Discuss.

- NathanielJ
**Posts:**882**Joined:**Sun Jan 13, 2008 9:04 pm UTC

### Re: An oddity...

pemcat wrote:This kind of thing is hard to prove, I think, because it depends on the use of the English language, which is not maths. For instance, it could work up to some big number, then I could simply call some number X a name which has X letters. But I think if we take the assumption that, say f(n)<1000000000000 or something (fairly reasonable, I think, since noone wants to call a number something that takes an approximation of infinite time to say).

One problem with assuming that f(n)<(some number) for all n is that it's necessarily impossible for that to be the case, because (some number) digits is only enough space to represent 26^(some number) different numbers, which is finite. This is why I proposed the assumption that f(n) < n for all n sufficiently large instead.

- imatrendytotebag
**Posts:**152**Joined:**Thu Nov 29, 2007 1:16 am UTC

### Re: An oddity...

Its not hard to show f(n) < n for all n greater than 999. Suppose 1000^k <= n < 1000^(k+1) for some integer k. Then n has k+1 distinct blocks of three digits (the last block may have less than 3 digits). For example, let n = 56,723,456. Then 1000^2 < n < 1000^3, and n has 3 "blocks" of 3 digits, 56, 723, 456. (53, being the last block, is okay having less than 3 digits).

Each block of three is written as "blah hundred an blah-dy blah blah-illion". For example, the "723" is read as "seven hundred and twenty three thousand" (yes, an exception to the "-illion" rule) and the "56" is read as "fifty six million").

By checking digits, we see the "blah hundred and blah-dy blah" part is maximized at "seven-hundred and seventy seven" (or other numbers, like 378, 877, etc.) which has 27 digits. Now check the "illion" part. Let N be the average of the letters all the "illions" we have to use (in our example we have nothing, "thousand" and "million", giving us 15 letters over three blocks, so N = 15/3 = 5). THEN the number of letters in the representation of our number is less than (k+1)*(27 + N).

If we were to increase, we would need (k+1)*(27 + N) > 1000^k or 27 + N > (1000^k)/(k+1). Now it is easy to show for k>=1, k+1 <= 2^k. So making this replacement we get 27 + N > (1000^k)/(2^k) = 500^k.

Looking at this, it is absurd. Even though we don't ACTUALLY have words for more than 10^51, for this to be feasible we would need to start needing representations with more than 500^k letters. This is, frankly, not going to happen. Even naive notation like million -> million million -> million million million -> etc. gives us N < 7*k, which is vastly less than 500^k.

Now we need to check 1 through 999, which can be done through similar arguments or brute force.

Each block of three is written as "blah hundred an blah-dy blah blah-illion". For example, the "723" is read as "seven hundred and twenty three thousand" (yes, an exception to the "-illion" rule) and the "56" is read as "fifty six million").

By checking digits, we see the "blah hundred and blah-dy blah" part is maximized at "seven-hundred and seventy seven" (or other numbers, like 378, 877, etc.) which has 27 digits. Now check the "illion" part. Let N be the average of the letters all the "illions" we have to use (in our example we have nothing, "thousand" and "million", giving us 15 letters over three blocks, so N = 15/3 = 5). THEN the number of letters in the representation of our number is less than (k+1)*(27 + N).

If we were to increase, we would need (k+1)*(27 + N) > 1000^k or 27 + N > (1000^k)/(k+1). Now it is easy to show for k>=1, k+1 <= 2^k. So making this replacement we get 27 + N > (1000^k)/(2^k) = 500^k.

Looking at this, it is absurd. Even though we don't ACTUALLY have words for more than 10^51, for this to be feasible we would need to start needing representations with more than 500^k letters. This is, frankly, not going to happen. Even naive notation like million -> million million -> million million million -> etc. gives us N < 7*k, which is vastly less than 500^k.

Now we need to check 1 through 999, which can be done through similar arguments or brute force.

Hey baby, I'm proving love at nth sight by induction and you're my base case.

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