I can't put my hand on the book, but I think Vaughn has shown that every sufficiently large number is of the form a^2+b^3+c^5. The key is that the sum 1/2 + 1/3 + 1/5 = 31/30 > 1, so the expected number of representations for a number N is, on average, K * N^1/30. K is some mess of gamma(fractions). [It's also necessary to check that there's no modular exclusion.] I once tried to estimate how large an N was "sufficiently large", and IIRC, got around 10^60. The results for a^2+b^3+c^4+d^5 and a^2+b^3+c^4+d^5+e^6 would follow immediately, although proving an upper bound is a big question. I'm not even sure if Vaughn's proof is constructive. Chris Landauer worked on the 2...6 problem long long ago; I think his program got up to a million or so. There was a paper in Math Comp about five years ago, about the problem of 4 cubes. Beeler & I tried to find the empirical last- unrepresentable number, but couldn't reach it. The MC paper found the likely candidate. It was around 10^13 or 14. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of David Wilson Sent: Sun 10/16/2005 1:07 PM To: Math-Fun; Sequence Fans Subject: [math-fun] Sums of powers Under 10^8, it looks as if all n >= 0 not of the form a^2+b^3+c^4+d^5 for a,b,c,d >= 0 are 15 23 55 62 71 471 478 510 646 806 839 879 939 1023 1063 1287 2127 5135 6811 7499 9191 26471 which one would surmise is a finite list. all n <= 10^8 appear to be of the form a^2+b^3+c^4+d^5+e^6 -------------------------------- - David Wilson _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun