Anyone seen this before? http://www.users.globalnet.co.uk/~perry/maths/morerecentstuff/morerecentstuf f.htm An invariant number, p ---------------------- If we define p as an invariant number, we can find solutions to equations. An invariant number is one that has the following properties: p=-1 p+k=p p^k=p for all k (real or complex or invariant) This, for example, provides roots to 1+x+x^2+x^3+..., and more generally for P(x), the generating function of the partition numbers. Let x=p/2^(1/k), so for k=1, x=p/2 and for k=2, x=p/sqrt(2), etc... Then 1+x+x^2+x^3 = 1 + p/2 + p^2/4 + p^3/8 + ... = 1-1/2-1/4-1/8-... = 0. And 1+x^2+x^4+x^6+... is zero for the k=2 case, and so on. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com