From: Christian.Radoux@skynet.be Newsgroups: sci.math Subject: Re: Waring's problem, and why is G(4) so large? Date: Sun, 07 Jun 1998 12:29:48 GMT In article <35799D39.6BDC@idt.net>, tjr19@idt.net wrote: > > I have done some computer calculations on Waring's problem (express > all sufficiently-large numbers as sum of G(n) N-th powers). These > values are not proved, but they are strongly suggested by the computer > analysis. I have tested powers 3-6 to 10^9. > > G(3) probably is 5. Last number that requires 6 cubes is 1290740. > I found 4060 numbers that require 6 or more cubes. Calculations > suggest that G(5) is probably not 4. > > G(4) probably is 16. I think this has been proved (?). Last number > that requires 17 fourth powers is 13792. I found 96 numbers that > require 17 or more fourth powers. I found a large number of > values that require 16 fourth powers all the way out to a billion. > > G(5) probably is 8 or 9. Last number that requires 10 fifth powers > is 617597724. Calculations strongly suggest that G(5) > 6. > > G(6) probably is <= 15. Numbers requiring 16 sixth powers are spaced > many millions apart around 10^9. Probably all numbers above about > 1.5x10^9 are sums of 15 sixth powers. > > Question: Why are 16 fourth powers needed? This is very strange, as > it is more than the number of fifth or probably even sixth powers > that are needed. It is contrary to intuition. > > Other question: What are the latest proven results on this problem? > > --Tenie Remmel (tjr19@idt.net) > Indeed the "all sufficiently-large numbers" makes the problem of computing G(k) more difficult than computing Waring's g(k). Here is a small list of results : G(k) < k*(3*ln(k)+5.2) (Chen, 1958) G(k) < k*(2*ln(k)+4*ln(k)+2*ln(ln(k))+13) (Vinogradov) (Vinogradov's result is better than Chen's theorem for n > 6 103 975 350) G(k) < (ln(108)+3*ln(k))/(ln(k/(k-1))) - 4 (Balasubramanian and Mozzochi, 1984) I think there are still better and more recent results by Mozzochi. With best regards ! e-mail : Christian.Radoux@skynet.be URL : http://users.skynet.be/radoux -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading ============================================================================== From: bobs@rsa.com Newsgroups: sci.math Subject: Re: Waring's problem, and why is G(4) so large? Date: Mon, 08 Jun 1998 21:25:13 GMT In article <35799D39.6BDC@idt.net>, tjr19@idt.net wrote: > > I have done some computer calculations on Waring's problem (express > all sufficiently-large numbers as sum of G(n) N-th powers). These > values are not proved, but they are strongly suggested by the computer > analysis. I have tested powers 3-6 to 10^9. > > G(3) probably is 5. Last number that requires 6 cubes is 1290740. > I found 4060 numbers that require 6 or more cubes. Calculations > suggest that G(5) is probably not 4. On the contrary, calculations done by Bohman & Froberg (1981) and Romani (1982) strongly suggest G(3) = 4. > > G(4) probably is 16. I think this has been proved (?). Yes. By Balasubramanian et.al. G(4) = 16 > Last number > that requires 17 fourth powers is 13792. I found 96 numbers that > require 17 or more fourth powers. I found a large number of > values that require 16 fourth powers all the way out to a billion. > > G(5) probably is 8 or 9. Last number that requires 10 fifth powers > is 617597724. Calculations strongly suggest that G(5) > 6. I mean no insult, but you are probably wrong (from the work of others) about G(3). Please explain why we should believe your assertions about G(5) and higher? Note that taking calculations out to 10^9 is far short of what needs to be done in order to even guess what might be the true value. > > G(6) probably is <= 15. Numbers requiring 16 sixth powers are spaced > many millions apart around 10^9. Probably all numbers above about > 1.5x10^9 are sums of 15 sixth powers. > > Question: Why are 16 fourth powers needed? This is very strange, as > it is more than the number of fifth or probably even sixth powers > that are needed. It is contrary to intuition. Actually it is not mysterious. Consider the d'th powers of an integer taken modulo a prime p. Suppose for the moment d is prime. Now if p = 1 mod d then (p-1)/d integers will be dth powers mod p. But if p = -1 mod d, then alll integers will be dth powers mod p. On the other hand, 4th powers modulo a prime are relatively rare. And it can never be the case that 4th powers mod p consist of all the integers. Thus, 4th powers are not distributed in the same way as 5th or 7th or ....... etc. They are much rarer. Thus, when adding up 4th powers, since they fall into fewer congruence classes one needs 'more of them' to cover all the integers. The same will be true for 8th, 16'th etc. powers -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading ============================================================================== From: gerry@mpce.mq.edu.au (Gerry Myerson) Newsgroups: sci.math Subject: Re: Waring's problem, and why is G(4) so large? Date: Tue, 09 Jun 1998 11:44:07 +1100 In article <6lhkrp$18s$1@nnrp1.dejanews.com>, bobs@rsa.com wrote: => In article <35799D39.6BDC@idt.net>, => tjr19@idt.net wrote: => > => > Question: Why are 16 fourth powers needed? This is very strange, as => > it is more than the number of fifth or probably even sixth powers => > that are needed. It is contrary to intuition. => => => Actually it is not mysterious. Consider the d'th powers of an integer taken => modulo a prime p... It may be a little simpler than that. The fourth powers mod 16 are 0 and 1. If n is 15 mod 16 then it needs 15 fourth powers. There may be some simple twist on this reasoning that gets you the 16th. Gerry Myerson (gerry@mpce.mq.edu.au) ============================================================================== From: John Rickard Newsgroups: sci.math Subject: Re: Waring's problem, and why is G(4) so large? Date: 09 Jun 1998 13:12:32 +0100 (BST) Gerry Myerson wrote: : It may be a little simpler than that. The fourth powers mod 16 are 0 and 1. : If n is 15 mod 16 then it needs 15 fourth powers. There may be some simple : twist on this reasoning that gets you the 16th. Yes; if n is 0 mod 16 and is the sum of 15 fourth powers, then each power must be 0 mod 16, so n/16 is also the sum of 15 fourth powers. So if m is not the sum of 15 fourth powers (e.g. m = 31), then neither is (16^k)m for any k. Considering the density of nth powers suggests that every sufficiently large number M should be the sum of n+1 nth powers (in at least c*M^(1/n) ways for some c depending on n), unless there is some reason based on congruences why it can't be so. So I'd guess G(3) = 4, G(5) = 6, and G(6) = 9 (because 6th powers are 0 or 1 mod 9). I don't have any hope of proving these, though! Does this agree with the expectations of those who have studied the problem? -- John Rickard