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A259753 For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.

%I A259753 #82 Oct 16 2015 09:49:28
%S A259753 1,30,71,100,104,485,1169,705,887,1727,421,775,4499,4260,3180,5748,
%T A259753 9719,307,6092,18521,20304,18825,32255,6174,33082,7601,29400,17607,
%U A259753 13457,52487,1727,44794,41772,76328,1801,29707,80999,119789,111226,132105,122730,171071,123117,237275
%N A259753 For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.
%C A259753 It seems to me the sequence can never drop all the way to 1 again.
%C A259753 From _Robert Israel_, Oct 13 2015: (Start)
%C A259753 There are only finitely many n with a(n) = 1.
%C A259753 Such n correspond to solutions of the Diophantine equation x (2 x^2 + 3 x + 3) = z^3.
%C A259753 Since gcd(x, 2 x^2 + 3 x + 3) = 1 or 3, we get two cases:
%C A259753   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 A259753   otherwise x = 9  s^3, z = 3 s t, where 54 s^6 + 9 s^3 + 1 = t^3.
%C A259753 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 A259753 (End)
%H A259753 Robert G. Wilson v, <a href="/A259753/b259753.txt">Table of n, a(n) for n = 1..368</a>
%F A259753 a(n) = A050793(n) - A050792(n). - _Robert G. Wilson v_, Jul 21 2015
%e A259753 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 A259753 Cubes:= {seq(x^3, x=2..10^4)}:
%p A259753 count:= 0:
%p A259753 for z from 1 to 10^4 do
%p A259753   s:= z^3+1;
%p A259753   M:= map(t -> s-t, select(`<`,Cubes,floor(s/2))) intersect Cubes;
%p A259753   for m in M do
%p A259753     count:= count+1;
%p A259753     y:= simplify(m^(1/3));
%p A259753     x:= simplify((s-m)^(1/3));
%p A259753     A[count]:= y-x;
%p A259753   od
%p A259753 od:
%p A259753 seq(A[i],i=1..count); # _Robert Israel_, Oct 13 2015
%t A259753 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 A259753 (Python 3.x)
%o A259753 start = 12
%o A259753 end = 100000
%o A259753 print(start,end)
%o A259753 cubes = []
%o A259753 for i in range(0,end):
%o A259753   cubes.append(i**3)
%o A259753 for z in range(start,end):
%o A259753   target = cubes[z]+1
%o A259753   for x in range(2,z):
%o A259753    t = target-cubes[x]
%o A259753    y = int(t**(1/3)+.5)
%o A259753    if cubes[y] == t:
%o A259753     print(target,y,x,z,1,y-x,'\a')
%o A259753     break
%Y A259753 Cf. A001235.
%Y A259753 Cf. x = A050792, y =  A050793, z = A050791 , x^3+y^3 = A050794.
%K A259753 nonn
%O A259753 1,2
%A A259753 _David Rabahy_, Jul 21 2015