Stirling and Catalan Numbers

Solution [x]1 = x, and hence b1,1 = 1. [x]2 = x2-x, and hence x2 = [x]2 + x = [x]2 + [x]1. Therefore b2,2 = 1 and b2,1 = 1. [x]3 = x3-3x2 + 2x, and hence x3 = [x]3 + 3x2-2x = [x]3 + 3([x]2 + [x]1)–2[x]1 = [x]3 

+ 3[x]2 + [x]1. Therefore b3,3 = 1, b3,2 = 3, and b3,1 = 1. [x]4 = x4-6x3 + 11x2-6x, and hence x4 = [x]4 + 6x3-11x2 + 6x =      [x]4 + 6([x]3 + 3[x]2 + [x]1)–11([x]2 + [x]1) + 6[x]1 = [x]4 + 6[x]3 + 7[x]2 + [x]1.