PhilX wrote:I do. In fact that pattern is one of the famous series in math. But before you can see it, factor out the 8. So now we have:

8•1 = 8

8•3 = 24

8•6 = 48

8•10 = 80...

Do you recognize the series 1, 3, 6, 10...? This is the famous triangular series and the process I used which led to those numbers I call cyclic multiplication.

This is merely scratching the surface. I've found out more with cyclic multiplication (e.g. other series) and it could lead towards other areas I haven't considered (it already has).

Here is what I think lies underneath the pattern you found.

Lets us take one of the sequence that you have provided, as an example:

1,2,3,4 => 24

Now, if you take the first member of this sequence (i.e. 1) and add it to the third member of the sequence (i.e. 3), you will get 4.

Lets us now take the second member of the sequence (i.e. 2) and add it to the fourth (i.e. 4), we get 6.

If we now multiply those two numbers, we get 24!

If we do this to the other sequences you have provided the answer is as you have provided in the OP.

0,1,2,3 => [(0+2)*(1+3)]= [2*4]= 8.

2,3,4,5 => [(2+4)*(3+5)]= [6*8]= 48.

3,4,5,6 => [(3+5)*(4+6)]= [8*10 = 80.

Now, how does all this relates to the triangular numbers? Well, here is the relation:

In effect, what we have been doing by this addition and multiplication scheme can be given a general representation as follows:

n, (n+1), (n+2), (n+3).

What we have been doing is then this: [n+ (n+2)] * [(n+1)+(n+3)]

The latter formula can be expanded into the quadratic expression 4*(n^2 + 3n + 2) and then factorized as 4*(n+1)(n+2)!

You will recall that (n+1)(n+2)/2 is the formula for evaluating the triangular numbers! (See Wikipedia for a refresher)

So, our formula, can be expressed as 4*2 [(n+1)(n+2)/2] = 8*[(n+1)(n+2)/2], for n ∈ +Z □