Binomial transform

First, the finite difference method, as shown in several websites; but I prefer not to use sequences starting with 0.By way of example, taking finite differences of (1, 4, 9, 16, 25,...) we get:1,.....4,.....9,.....16,.....25,.....36,.........3......5.......7.......9.......11...................2.......2......2.......2...................(when bottom row has all k,k,k,k,...the operation ends.). This implies that the Binomial transform of [1, 3, 2, 0, 0, 0,...] = (1, 4, 9, 16, 25, 36,...); and that theinverse Binomial transform of [1, 4, 9, 16, 25, 36,...] = (1, 3, 2, 0, 0, 0,...)....Pascal's triangle method. (C.f. A007318 in OEIS, http://www.tinyurl.com/4zq4qUsing a pocket calculator or online calculator we plug in rows of Pascal's triangle:11, 11, 2, 11, 3, 3, 11, 4, 6, 4, 1...preferably as many rows as possible. I use 17 rows. Then in the VECTOR section, plug in [1, 3, 2, 0, 0, 0,...], 17 terms. Let the triangle = P and the vector = V. Then simply multiply, P * V, getting (1, 4, 9, 16, 25, 36, 49,...); i.e. the vector in brackets, and (1, 4, 9, 16,...) in parentheses meaning: vector considered as a sequence; whereas Pascal's triangle A007318 is used as an infinite lower triangular matrix.Now for the inverse b.t.: Given the vector [1, 4, 9, 16,...] (17 terms); we have our Pascal's triangle already stored, so first invert it, getting:1-1, 11, -2, 1-1, 3, -3, 1...So, this matrix = M, with [1, 4, 9, 16,...] = V. Then take M * V, getting [1, 3, 2, 0, 0, 0,...] meaning that the inverse b.t. of [1, 4, 9, 16, 25,...] = [1, 3, 2, 0, 0, 0,...]....QUIZ: what's the b.t. of [1, 2, 3,...]. We plug in [1, 2, 3, 4, 5, 6,...] (17 terms) into the vector section and multiply it by our stored Pascal's triangle getting(1, 3, 8, 20, 48, 112, 256,...). Let's go in reverse, using the finite difference method:1,...3,...8,...20,...48,...112,.....2,....5,...12,...28,.....................3,....7,....16,............................4,....9,....................................5,............................But our bottom rows don't terminate in a row with k,k,k,k,...; but no matter.We can see by inspection that given enough rows, the inverse binomial transform of [1, 3, 8, 20, 48,...] is shaping up to be (1, 2, 3, 4, 5,...). But naturally, the more rows we have to deal with, the more certain is our conclusion..
Posted by Yifu Xero at 8:38 PM