Divisible numbers

From the link :)
These are some formula`s I came across doing clock branching, but may serve many needs outside clocking.
Unlike a prime number, which is divisible by one number only. You can create numbers that are divisible by any integer up to a given point. The simplest form of making such numbers is by multiplication into a series that is constructed by multiplying integers like so.

12 = 3*4 (divisible by 2, 3 and 4)
30 = 3*4*5 (divisible by 2, 3, 4 and 5)
210 = 3*4*5*7 (divisible by 2, 3, 4, 5, 6 and 7)
1890 = 3*4*5*7*9 (divisible by 2, 3, 4, 5, 6, 7 and 9) and not yet by 8, but you can multiply it by 4 and be done.

Since even numbers are much more friendly than the odd ones for division, you can compress these series to fit your needs at any given point. By multiplying it by 2 or 4.

This works with the Fibonacci series as well.
To express a series rather than building it from the start.

2*3*5*8*13*21*34*55 = 122522400 which is compressed and divisible by any number up to 18, also it can be multiplied as a result by any fraction like 0.1 up to a given point and still produce an integer outcome.
Click to expand...
 
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I'm stopped at the very first clause, "unlike the square root of a number which is divisible by one number only"! What about "12" which is the square root of 144 and is divisible by 1, 2, 3, 4, and 6? The examples given don't seem to have anything to do with square roots.

And then the author says "even numbers are much more friendly than the odd ones for division". Even numbers are divisible by 2 and odd numbers aren't. That's not such a big deal. What about numbers that are divisible by 3 versus numbers that aren't? We don't have a special name for those but are numbers divisible by 3 "much more friendly for division" than those that aren't?

"You can compress these series to fit your needs at any given point. By multiplying it by 2 or 4." What series? And how does multiplying by 2 or 4 "compress" them. I really don't understand what he is trying to say!
 
OK. I get that n! is not the smallest integer evenly divisible by every postive integer less than or equal to n.

6! = 720, but 4 does not divide evenly into 30 so the second example in your list is wrong. The correct number is 60.

Until you specify the algorithm or formula for your series that generates the smallest number number evenly divisible by every integer from 1 through n and prove it, this is just silliness
 
Think that is a very big number from 1 to n. In order to calculate some form of convergence. But youre right, I stand corrected. For the first example I know it cant be new, it has to be done before. The latter is kind of cool to have it show up there, amazing feat of intelligent design.
 
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