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

1, 30, 71, 100, 104, 485, 1169, 705, 887, 1727, 421, 775, 4499, 4260, 3180, 5748, 9719, 307, 6092, 18521, 20304, 18825, 32255, 6174, 33082, 7601, 29400, 17607, 13457, 52487, 1727, 44794, 41772, 76328, 1801, 29707, 80999, 119789, 111226, 132105, 122730, 171071, 123117, 237275

Offset: 1

Author

David Rabahy, Jul 21 2015

Keywords

• nonn

Comments

It seems to me the sequence can never drop all the way to 1 again.

From Robert Israel, Oct 13 2015: (Start)

There are only finitely many n with a(n) = 1.

Such n correspond to solutions of the Diophantine equation x (2 x^2 + 3 x + 3) = z^3.

Since gcd(x, 2 x^2 + 3 x + 3) = 1 or 3, we get two cases:

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,

otherwise x = 9 s^3, z = 3 s t, where 54 s^6 + 9 s^3 + 1 = t^3.

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.

(End)

Examples

			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.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..368

Crossrefs

Cf. A001235.

Cf. x = A050792, y = A050793, z = A050791 , x^3+y^3 = A050794.

Programs

Maple

Cubes:= {seq(x^3, x=2..10^4)}:
count:= 0:
for z from 1 to 10^4 do
  s:= z^3+1;
  M:= map(t -> s-t, select(`<`,Cubes,floor(s/2))) intersect Cubes;
  for m in M do
    count:= count+1;
    y:= simplify(m^(1/3));
    x:= simplify((s-m)^(1/3));
    A[count]:= y-x;
  od
od:
seq(A[i],i=1..count); # Robert Israel, Oct 13 2015

Mathematica

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 *)

Formula

a(n) = A050793(n) - A050792(n). - Robert G. Wilson v, Jul 21 2015