Wow that is some detailed stuff!!

I looked it up before creating the Formula, and didnt find anything. Now I see that yes, it is pretty much trivial...

Steps:

1 - Look if its an arithmetic or a geometric sequance, if so - trivial

2 - If not, lets say we have 3 numbers (it will be much easier without a computer to process):

seq={26,25,192}

so A[n] = ((n-3)(n-2)26 + (n-1)(n-3)25 + (n-2)(n-1)192)/((n-1)!(3-n)!(-1)^(3-n))

And by wolfram, that is:

A[n] = (243n^2 - 806n + 615) / ((n-1)!(3-n)!(-1)^(3-n))

Pretty lame compared to what you just showed me... wolfram returns 84n^2 - 253n + 195 - much more simple..

• Which is much more simple than deviding by factorials, but I dont have a PRO subscription so I cant see step-by-step...

Thanks for enlightning me... And of course there are infinity options, because I always can add to the sequance another number that is different than the number it would have given me.