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%I #82 Oct 16 2015 09:49:28
%S 1,30,71,100,104,485,1169,705,887,1727,421,775,4499,4260,3180,5748,
%T 9719,307,6092,18521,20304,18825,32255,6174,33082,7601,29400,17607,
%U 13457,52487,1727,44794,41772,76328,1801,29707,80999,119789,111226,132105,122730,171071,123117,237275
%N For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.
%C It seems to me the sequence can never drop all the way to 1 again.
%C From _Robert Israel_, Oct 13 2015: (Start)
%C There are only finitely many n with a(n) = 1.
%C Such n correspond to solutions of the Diophantine equation x (2 x^2 + 3 x + 3) = z^3.
%C Since gcd(x, 2 x^2 + 3 x + 3) = 1 or 3, we get two cases:
%C if x is not divisible by 3, x = s^3, z = s^3 t^3 where 2 s^6 + 3 s^3 + 3 = t^3,
%C otherwise x = 9 s^3, z = 3 s t, where 54 s^6 + 9 s^3 + 1 = t^3.
%C The algebraic curves 2 s^6 + 3 s^3 + 3 = t^3 and 54 s^6 + 9 s^3 + 1 = t^3 both have genus 4, so by Faltings's theorem they have only finitely many rational solutions.
%C (End)
%H Robert G. Wilson v, <a href="/A259753/b259753.txt">Table of n, a(n) for n = 1..368</a>
%F a(n) = A050793(n) - A050792(n). - _Robert G. Wilson v_, Jul 21 2015
%e 10 - 9 = 1 is the first number in the sequence because 10^3 + 9^3 = 12^3 + 1^3 and no other lower z produces a result.
%p Cubes:= {seq(x^3, x=2..10^4)}:
%p count:= 0:
%p for z from 1 to 10^4 do
%p s:= z^3+1;
%p M:= map(t -> s-t, select(`<`,Cubes,floor(s/2))) intersect Cubes;
%p for m in M do
%p count:= count+1;
%p y:= simplify(m^(1/3));
%p x:= simplify((s-m)^(1/3));
%p A[count]:= y-x;
%p od
%p od:
%p seq(A[i],i=1..count); # _Robert Israel_, Oct 13 2015
%t y = 3; lst = {}; While[y < 100001, x = 2; While[x < y, z = (y^3 + x^3 - 1)^(1/3); If[IntegerQ[z], AppendTo[lst, {z, y, x, y - x}]; Print[{z, y, x, y - x}]]; x++]; y++]; Last@ Transpose@ Sort@ lst (* _Robert G. Wilson v_, Jul 21 2015 and modified Oct 14 2015 *)
%o (Python 3.x)
%o start = 12
%o end = 100000
%o print(start,end)
%o cubes = []
%o for i in range(0,end):
%o cubes.append(i**3)
%o for z in range(start,end):
%o target = cubes[z]+1
%o for x in range(2,z):
%o t = target-cubes[x]
%o y = int(t**(1/3)+.5)
%o if cubes[y] == t:
%o print(target,y,x,z,1,y-x,'\a')
%o break
%Y Cf. A001235.
%Y Cf. x = A050792, y = A050793, z = A050791 , x^3+y^3 = A050794.
%K nonn
%O 1,2
%A _David Rabahy_, Jul 21 2015
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