From: Jan Kristian Haugland Subject: Re: Numbers as sums of powers Date: Mon, 31 Jul 2000 19:41:28 -0400 Newsgroups: sci.math Summary: [missing] On Tue, 1 Aug 2000, Kjell Fredrik Pettersen wrote: > It is known that any natural number is expressible as a sum of 4 (or less) > squares, of at most 9 cubes, etc. In general for any natural N there is a > P(N) such that any natural number is a sum of (at most) P(N) N-powers. > > I've been thinking more about the least number of N-powers needed for a > number, and how this is distributed all over the naturals. To be more > precise, this is the setting: > > Let N>=1 and m>=0 be natural numbers. For any x>0 define g(m,N,x) to be the > portion of natural numbers <=x that are expressible as a sum of m N-powers, > but not as a sum of m-1 N-powers. Let g(m,N) be the limit of g(m,N,x) as x > approaches infinity. Questions: > > 1) For which m,N does g(m,N) exist? All? > 2) If it exists, then what is g(m,N) ? > > Some obvious results: > > * g(m,N)=0 for any m N^(1/x), which is not enough). > * g(N,N), if it exists, is not more than 1/N. (Use similar argument) > * g(4,2)=1/6. This follows from the fact that the numbers that are possible > to write as sums of three squares are the numbers on the form (8k+7)*4^n. > > I don't know about g(2,2) or g(3,2). Not even if a closed formula exists for > any of them. For g(m,3) I have no idea when 3 <= m <= 9. > > > Kjell Fredrik Pettersen g(2,2)=0, g(3,2)=1. It is known that g(2,2,x) is asymptotically A x/sqrt(log x) for some constant A. g(9,3)=0: The only integers that need 9 cubes are 23 and 239. g(8,3)=0: ------------"-------------- 8 cubes are 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, 454. It is not known whether there are infinitely many that need 7 cubes, but it is believed that 8042 is the largest one, and that 7373170279850 is the largest one that needs 5 cubes: See Math. Comput. 69(1999) pp. 421-439. This would imply g(5,3)=g(6,3)=g(7,3)=0. I found all this stuff about cubes on the internet a little while ago and took some notes...I think maybe it was on Eric Weisstein's (sp?) math pages. ============================================================================== From: jk_haugland@my-deja.com Subject: Re: Numbers as sums of powers Date: Tue, 01 Aug 2000 01:35:50 GMT Newsgroups: sci.math In article , Jan Kristian Haugland wrote: > > On Tue, 1 Aug 2000, Kjell Fredrik Pettersen wrote: > > > It is known that any natural number is expressible as a sum of 4 (or less) > > squares, of at most 9 cubes, etc. In general for any natural N there is a > > P(N) such that any natural number is a sum of (at most) P(N) N-powers. > > > > I've been thinking more about the least number of N-powers needed for a > > number, and how this is distributed all over the naturals. To be more > > precise, this is the setting: > > > > Let N>=1 and m>=0 be natural numbers. For any x>0 define g(m,N,x) to be the > > portion of natural numbers <=x that are expressible as a sum of m N-powers, > > but not as a sum of m-1 N-powers. Let g(m,N) be the limit of g(m,N,x) as x > > approaches infinity. Questions: > > > > 1) For which m,N does g(m,N) exist? All? > > 2) If it exists, then what is g(m,N) ? > > > > Some obvious results: > > > > * g(m,N)=0 for any m > N^(1/x), which is not enough). > > * g(N,N), if it exists, is not more than 1/N. (Use similar argument) > > * g(4,2)=1/6. This follows from the fact that the numbers that are possible > > to write as sums of three squares are the numbers on the form (8k+7)*4^n. > > > > I don't know about g(2,2) or g(3,2). Not even if a closed formula exists for > > any of them. For g(m,3) I have no idea when 3 <= m <= 9. > > > > > > Kjell Fredrik Pettersen > > g(2,2)=0, g(3,2)=1. It is known that g(2,2,x) is asymptotically > A x/sqrt(log x) for some constant A. Sorry, that should be g(3,2)=5/6 of course. Also, if the values for N=3 are as boring as suggested below, N=4 might still be interesting since there are infinitely many integers that are the sum of 16 4th powers but not 15. > > g(9,3)=0: The only integers that need 9 cubes are 23 and 239. > g(8,3)=0: ------------"-------------- 8 cubes are 15, 22, 50, 114, > 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, 454. It is not > known whether there are infinitely many that need 7 cubes, but it > is believed that 8042 is the largest one, and that 7373170279850 > is the largest one that needs 5 cubes: See Math. Comput. 69(1999) > pp. 421-439. This would imply g(5,3)=g(6,3)=g(7,3)=0. > > I found all this stuff about cubes on the internet a little while > ago and took some notes...I think maybe it was on Eric Weisstein's > (sp?) math pages. > > Sent via Deja.com http://www.deja.com/ Before you buy.