Recent content by kpkkpk

Many thanks for you Opalg for detailed answer. I also found this same series of squares from Online Ensyclopedia of Integer Sequences (OEIS) with code A038202.

1 = 1^2 1 = 1^2 9 = 3^2 1 = 1^2 81 = 9^2 729 = 27^2 225 = 15^2 324 = 18^2 X 82944 = 288^2 176400 = 420^2 215296 = 464^2 3444736 = 1856^2 So, I am trying to find short method to find factorials. In order to achieve this, I imagined factorials as squares, one edge of which corresponding square...

Yes, indeed! It took me some time before I understood what you meant, but now things are much clearer. If I understood correctly, from your triangle I pick two consecutive integers (their absolute values) from the same horizontal line, then I multiply the leftmost of them with vertical row...

x = 1x x(x-1) = 1x^2 - 1x x(x-1)(x-2) = 1x^3 - 3x^2 + 2x x(x-1)(x-2)(x-3) = 1x^4 - 6x^3 + 11x^2 - 6x x(x-1)(x-2)(x-3)(x-4) = 1x^5 - 10x^4 + 35x^3 - 50x^2 + 24x ... So, I am looking for a short method how to find these coefficients ahead of each raisings of x: 1 1,-1 1,-3,2 1,-6,11,-6...