Pick a number. Any integer! Because four is the cosmic number.
Seven is five, five is four, four is the cosmic number!
Twenty-seven is twelve, twelve is six, six is three, three is five, five is four, four is the cosmic number!
What is the trick? (scroll down)
Give up? Okay, you're going to hit yourself once I tell you: count the lengths of the number. Not the number itself, just the English.
I can just hear you. Don't feel too bad: it's a hard trick; I'm making you think of math when you need to be thinking of words. Now you've learned a neat party trick, but what gives? Of course, I did link to you a working example (that works with any integer between ±undecillion, I think), but that's not what's interesting.
Here's my thought process. First, I wonder if all numbers tend to four. Obviously, four tends to four, it being the only number whose value numbers its letters; thus, it is a natural tendency. Is it, however, a hard-and-fast rule?
I conjecture that this is indeed a rule. Now, a proof is needed. So let us begin!
First, let us describe the process.
- Start with a number, call it N0.
- Create a marker, call it k. Set k initially to 0.
- Let E(x) be the English representation of a number x; for example, E(20) is twenty, E(25) is twenty-five (notice the hyphen), and E(123) is one hundred twenty-three (notice the lack of the improper “and” and punctuation other than the dash).
- Let length(S) be the number of characters in string S. For example, length("Hello!") is 6, and length("O'Brien") is 7.
- Set Nk+1 = length(E(Nk))
- Does Nk+1 = 4?
- If so, terminate the process, returning k.
- Otherwise, increment k and go to step 5.
Conjecture: All integers, when the above process is repeatedly applied, tend to four, i.e., have “fourness”; equivalently, there is no integer for which the above process does not halt.
My idea at first was to find the name of the longest number within the context (call it M), calculate length(E(M)), and prove that every number from 1 to M tends to four. However, this is [a] a laborious task (also consider that length(E(10M±1)) > length(E(M) ) and [b] is unsatisfactory for all numbers. But what then shall be the argument form?
As it stands, this problem seems to be in, or at the very least related to, a currently unsolvable class of problems. A famous example would be the Collatz Conjecture. As Paul Erdős said,
Mathematics may not be ready for such problems.
So perhaps the proof is unattainable; the Cosmic Problem, like the Collatz Conjecture, involves in a case function's n-ness (i.e. that functions values tending towards n unconditionally). They both have common “seed” strings that are found (e.g. three to five to four). I, however, like to think that, since the Cosmic Problem is different in that it requires a non-mathematical conversion, that a proof can indeed be placed.
As of now, I have not found such a proof. But I'm still looking... ;)