I think lgdegf means that f'(x) / f(x) is the rational function EGF for the given terms. So if f(x) given by superseeker is (a(n) - 3)^3 then the EGF is 3 / (a(n) - 3)??? hm, now I'm not making sense even to myself. Maybe you're supposed to find the function a(n) that makes this identically zero and then take a'(n) / a(n)? No, that doesn't make sense either, since a(n) = 3 is an obvious solution. Oh, but the higher-order terms probably give solutions like a(n) = n * 3^n or something? And then a'(n) / a(n) could be an EGF, and maybe it is then a rational function since the 3^n will divide out? As you can tell, it's too late at night for me to think this through, but maybe my muttering here will at least enable someone else to figure out what I should have said if only I knew what I was talking about. --Josh On Fri, Aug 20, 2010 at 12:10 AM, Joerg Arndt <arndt@jjj.de> wrote:
* Fred lunnon <fred.lunnon@gmail.com> [Aug 20. 2010 08:19]:
<< Superseeker finds ... 2 3 [-27 + 27 a(n) - 9 a(n) + a(n) , lgdegf]
2 3 4 5 [-243 + 405 a(n) - 270 a(n) + 90 a(n) - 15 a(n) + a(n) , lgdegf]
where "lgdegf" stands for "logarithmic derivative of exponential generating function" The polynomials are simply (a(n) - 3)^3 and (a(n) - 3)^5 respectively.