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%I #74 Aug 20 2023 10:49:22
%S 6,9,12,18,19,20,24,25,27,28,29,30,36,38,40,41,42,44,45,46,48,50,53,
%T 54,56,57,58,60,63,66,67,69,70,71,72,75,76,78,80,81,82,84,85,87,88,89,
%U 90,92,93,95,96,97,99,100,102,103,105,106,108,110,111,112,113
%N Numbers whose cube is the sum of three distinct nonnegative cubes.
%C Numbers w such that w^3 = x^3+y^3+z^3, x>y>z>=0, is soluble.
%C A226903(n) + 1 is an infinite subsequence parametrized by Shiraishi in 1826. - _Jonathan Sondow_, Jun 22 2013
%C Because of Fermat's Last Theorem, sequence lists numbers w such that w^3 = x^3+y^3+z^3, x>y>z>0, is soluble. In other words, z cannot be 0 because x and y are positive integers by definition of this sequence. - _Altug Alkan_, May 08 2016
%C This sequence is the same as numbers w such that w^3 = x^3+y^3+z^3, x>=y>=z>0, is soluble as Legendre showed that a^3+b^3=2*c^3 only has the trivial solutions a = b or a = -b (see Dickson's History of the Theory of Numbers, vol. II, p. 573). - _Chai Wah Wu_, May 13 2017
%D Ya. I. Perelman, Algebra can be fun, pp. 142-143.
%H Nathaniel Johnston and Chai Wah Wu, <a href="/A023042/b023042.txt">Table of n, a(n) for n = 1..10000</a> (1..1500 from Nathaniel Johnston)
%H A. Russell and C. E. Gwyther, <a href="http://www.jstor.org/stable/3605742">The Partition of Cubes</a>, The Mathematical Gazette, 21 (1937), pp. 33-35.
%H <a href="/index/Di#Diophantine">Index to sequences related to Diophantine equations</a> (3,1,3)
%e 20 belongs to the sequence as 20^3 = 7^3 + 14^3 + 17^3.
%p for w from 1 to 113 do for z from 0 to w-1 do bk:=0: for y from z+1 to w-1 do for x from y+((w+z) mod 2) to w-1 by 2 do if(x^3+y^3+z^3=w^3)then printf("%d, ",w); bk:=1: break: fi: od: if(bk=1)then break: fi: od: if(bk=1)then break: fi: od: od: # _Nathaniel Johnston_, Jun 22 2013
%t lst={};Do[Do[Do[Do[y=a^3+b^3+c^3; x=z^3; If[y<x,Break[], If[y==x, AppendTo[lst,z]]], {c,b-1,0,-1}], {b,a-1,0,-1}],{a,z-1,0,-1}],{z,2,3*5!}]; Union@lst (* _Vladimir Joseph Stephan Orlovsky_, Apr 11 2010 *)
%o (PARI) has(n)=my(L=sqrtnint(n-1,3)+1, U=sqrtnint(4*n,3)); fordiv(n,m, if(L<=m && m<=U, my(ell=(m^2-n/m)/3); if(denominator(ell)==1 && issquare(m^2-4*ell), return(1)))); 0
%o list(lim)=my(v=List(),a3,t); lim\=1; for(a=2,sqrtint(lim\3), a3=a^3; for(b=if(a3>lim, sqrtnint(a3-lim-1,3)+1,1), a-1, t=a3-b^3; if(has(t), listput(v,a)))); Set(v) \\ _Charles R Greathouse IV_, Jan 25 2018
%Y Cf. A001235, A003072, A114923, A225908, A226903.
%K nonn
%O 1,1
%A _N. J. A. Sloane_
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