A050791 -id:A050791 - cp's OEIS frontend

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

David Rabahy, Jul 21 2015

nonn

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)

			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.

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

Cf. A001235.

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

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

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 *)

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

A163827 a(n) = 6n^3 + 1, solution z in Diophantine equation x^3 + y^3 = z^3 - 2. It may be considered a Fermat near miss by 2.

Original entry on oeis.org

7, 49, 163, 385, 751, 1297, 2059, 3073, 4375, 6001, 7987, 10369, 13183, 16465, 20251, 24577, 29479, 34993, 41155, 48001, 55567, 63889, 73003, 82945, 93751, 105457, 118099, 131713, 146335, 162001, 178747, 196609, 215623, 235825, 257251, 279937

Offset: 1

Carlos Alves, Aug 05 2009

It is easy to check that with x = 6n^2, y = 6n^3 - 1, and this z = 6n^3 + 1, it satisfies the Diophantine equation x^3 + y^3 = z^3 - 2. Thus these are near-misses for Fermat equation.

For n>2, it seems to be the only solution of x^n + y^n = z^n - 2 (or even that differ by 2 from FLT, see A050787 and A050791 for solutions that differ by 1). As 2 is not a cube, these solutions are not included in the theory for x^3 + y^3 = u^3 + v^3.

			For n=1, a(1)=7 and 7^3 - 2 (=341) = 5^3 + 6^3.
For n=2, a(2)=49 and 49^3 - 2 (=117647) = 24^3 + 47^3.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A050787, A050791.

[6*n^3+1 : n in [1..50]]; // Wesley Ivan Hurt, Jan 09 2017

A163827:=n->6*n^3+1: seq(A163827(n), n=1..50); # Wesley Ivan Hurt, Jan 09 2017

6*Range[40]^3+1 (* or *) LinearRecurrence[{4,-6,4,-1},{7,49,163,385},40] (* Harvey P. Dale, Dec 12 2011 *)

a(n)=6*n^3+1 \\ Charles R Greathouse IV, Nov 29 2014

a(n) = 6n^3+1.
a(1)=7, a(2)=49, a(3)=163, a(4)=385, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). [Harvey P. Dale, Dec 12 2011]
G.f.: (-x^3+9*x^2+21*x+7)/(x-1)^4. [Harvey P. Dale, Dec 12 2011]

A330013 a(n) is the number of solutions with nonnegative (x,y,z) to the cubic Diophantine equation x^3+y^3+z^3 - 3xy*z = n.

Original entry on oeis.org

3, 3, 0, 3, 3, 0, 3, 6, 6, 3, 3, 0, 3, 3, 0, 6, 3, 6, 3, 6, 0, 3, 3, 0, 3, 3, 9, 12, 3, 0, 3, 6, 0, 3, 9, 6, 3, 3, 0, 6, 3, 0, 3, 6, 6, 3, 3, 0, 9, 3, 0, 6, 3, 12, 3, 12, 0, 3, 3, 0, 3, 3, 6, 9, 9, 0, 3, 6, 0, 9, 3, 12, 3, 3, 0, 6, 9, 0, 3, 6, 12, 3, 3, 0, 3

Offset: 1

Bernard Schott, Nov 27 2019

nonn

Some results coming from the Alarcon and Duval reference.
For n = 0, there are infinitely many solutions because every triple (k,k,k) with k >= 0 satisfies the equation.
a(n) = 0 iff 3 divides n and 9 doesn't divide n (equivalent to n is in A016051).
When n belongs to A074232 (complement of A016051), a(n) is always a multiple of 3 because
1) if (a,a,b) [resp. (a,b,b)] with a < b is a primitive solution, then these triples generate 3 solutions with the permutations (a,a,b), (a,b,a), (b,a,a), [resp. (a,b,b), (b,b,a), (b,a,b)] and,
2) if (a,b,c) with a < b < c is a primitive solution, then this triple generates 6 solutions with the permutations (a,b,c), (b,c,a), (c,a,b), (a,c,b), (c,b,a), (b,a,c).
For prime p <> 3, a(p) = a(2*p) = 3.
An inequality: (n/4)^(1/3) <= max(x, y, z) <= (n+2)/3.
This sequence is unbounded.
A261029 gives the number of triples without counting the permutations and, in link, a list of primitive triples up to n = 2000.

			3^3+2^3+2^3-3*2*2*3 = 7 so (3,2,2), (2,2,3) and (2,3,2) are solutions and a(7) = 3.
When n=35, (0,1,3) is a primitive solution that generates 6 solutions and (9,9,10) is another primitive solution that generates 3 solutions, so a(35)=6+3=9 (see comments).

Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 9, Problème: étude d'une équation diophantienne cubique, pages 137-138 and 147-152.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A016051, A074232.
Cf. A261029 (primitive triples without the permutations).
Cf. A050787, A050791, A212420 (other cubic Diophantine equations).

a[n_] := Length@ Solve[x^3 + y^3 + z^3 - 3 x y z == n && x >= 0 && y >= 0 && z >= 0, {x, y, z}, Integers]; Array[a, 85] (* Giovanni Resta, Nov 28 2019 *)

If n = 3*k + 1, then (k, k, k+1) is a solution for k >= 0.
If n = 3*k - 1, then (k, k, k-1) is a solution for k >= 1.
If n = 9*k, then (k-1, k, k+1) is a solution for k >= 1.
If n = k^3, then (k, 0, 0) is a solution for k >= 0.
If n = 2*k^3, then (k, k, 0) is a solution for k >= 0.

More terms from Giovanni Resta, Nov 28 2019

A336449 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2z^3 = 24^6.

Original entry on oeis.org

1, 9, 16, 25, 49, 81, 121, 169, 225, 289, -356, 361, 441, 529, 625, 729, 841, -948, 961, 1045, 1089, 1225, 1369, 1521, 1681, -1715, 1849, 1876, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, -3587, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5769, 5929

Offset: 1

XU Pingya, Aug 08 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).

Let x = a^(2*k) - (a^k)*t - t^2, y = a^(2*k) + (a^k)*t - t^2, z = t^2; then x^3 + y^3 + 2*z^3 = 2*a^(6*k). When a = 4, k = 1, t = 2*n + 1; (x, y, z) are primitive solutions of equation. Thus, terms of A016754 are terms of the sequence.

			(-15)^3 + (-27)^3 + 2*25^3 = 11^3 + (-29)^3 + 2*25^3 = 8192, 25 is a term.
(-65)^3 + (449)^3 + 2*(-356)^3 = 8192, -356 is a term.

R. K. Guy, Unsolved Problems in Number Theory, D5.

Cf. A000290, A000578, A003215, A004825, A004826, A016754, A050791, A130472, A195006, A336450.

Clear[t]
t = {};
Do[y = (8192 - x^3 - 2z^3)^(1/3) /. (-1)^(1/3) -> -1;
If[Abs@x <= Abs@y && IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -5929, 5929}, {x, -Round[(Abs[8192 - 2z^3]/3)^(1/2)], Round[(Abs[8192 - 2z^3]/3)^(1/2)]}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 12000}];
Select[v, MemberQ[u, #] &]

A336450 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2z^3 = 25^6.

Original entry on oeis.org

1, -3, 4, 9, 16, 25, 36, 49, -56, 64, 81, 88, -104, 121, 144, -167, 169, 177, 196, -203, -243, -255, 256, 277, 289, 324, 361, -363, 373, -395, -411, 441, 484, 529, 576, 676, 709, -719, 729, 784, 841, 961, 1017, 1024, -1028, 1080, 1089, -1091, 1156, 1296, 1369

Offset: 1

XU Pingya, Aug 08 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).

Let x = a^(2*m) - (a^m)*t - t^2, y = a^(2*m) + (a^m)*t - t^2, z = t^2; then x^3 + y^3 + 2*z^3 = 2*a^(6*m). When a = 5, m = 1, t = 5*n + k(k = {1, 2, 3, 4}); (x, y, z) are primitive solutions of equation. Thus, A047201(n)^2 are terms of the sequence.

			(-20)^3 + 34^3 + 2*(-3)^3 = 31250, -3 is a term.
(-11)^3 + 29^3 + 2*16^3 = 15^3 + 27^3 + 2*16^3 = 31250, 16 is a term.

R. K. Guy, Unsolved Problems in Number Theory, D5.

Cf. A000290, A000578, A003215, A004825, A004826, A047201, A050791, A130472, A195006, A336449.

Clear[t]
t = {};
Do[y = (31250 - x^3 - 2z^3)^(1/3) /. (-1)^(1/3) -> -1;
If[IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -1369, 1369}, {x, -Round[(Abs[31250 - 2z^3]/3)^(1/2)], Round[(Abs[31250 - 2z^3]/3)^(1/2)]}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 2739}];
Select[v, MemberQ[u, #] &]

A336166 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 2.

Original entry on oeis.org

0, 1, -3, 4, 9, -12, 16, 25, -27, -35, 36, 37, -48, 49, -59, 64, -75, 81, 100, -108, 121, 144, -147, -159, 169, 172, -192, 196, 225, -227, -243, -255, 256, 261, -287, 289, -300, -311, 324, -335, 361, -363, 373, 400, -432, 441, 484, -507, 529, 568, 576, -588

Offset: 1

XU Pingya, Jul 10 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).

Segre shows that 1-(9/2)*A000578(2n), (-3)*A000290(n), and A016754(n) are terms of the sequence.

			(-5)^3 + (-11)^3 + 2 * 9^3 = 2, 9 is a term.
(25)^3 + (-23)^3 + 2 * (-12)^3 = 2, -12 is a term.

R. K. Guy, Unsolved Problems in Number Theory, D5.

Beniamino Segre, On the rational solutions of homogeneous cubic equations in four variables, Math. Notae, 11 (1951), 1-68.

Cf. A000290, A000578, A004825, A004826, A016754, A028387, A050791, A130472, A195006.

t1 = Union[Plus@@@Tuples[Range[-11643, 11643]^3, 2]];
t2 = Table[2 - 2z^3, {z, -588, 588}];
t = Select[t1, MemberQ[t2, #] &];
u = ((2 - t)/2)^(1/3) /. (-1)^(1/3) -> (-1);
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 1176}];
Select[v, MemberQ[u, #] &]

A336226 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 1458.

Original entry on oeis.org

1, -3, 4, 9, -10, -12, 16, 21, 25, 37, -47, -48, 49, 64, -75, -87, 88, 100, 105, 121, 134, -147, 169, 172, -192, 196, -241, -243, 256, 289, -300, 361, -363, 400, 443, 484, -507, 529, 541, -588, 625, 676, -699, 732, -759, -768, 777, 784, 841, -867, 897, 961

Offset: 1

XU Pingya, Jul 17 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).
(11 + 3*n - 9*n^2)^3 + (11 + 3*(n + 1) - 9*(n + 1)^2)^3 + 2*(3*n + 1)^6 = 1458, the numbers of the form (3*n + 1)^2 are terms of the sequence.
(11 - 3*n - 9*n^2)^3 + (11 - 3*(n + 1) - 9*(n + 1)^2)^3 + 2*(3*n + 2)^6 = 1458, the numbers of the form (3*n + 2)^2 are also terms of the sequence.
Thus, A001651(n)^2 are terms of the sequence. There is an infinity of nontrivial solutions to the equation.

			5^3 + 11^3 + 2 * 1^3 = 1458, 1 is a term.
(-1)^3 + (11)^3 + 2 * (4)^3 = 1458, 4 is a term.

R. K. Guy, Unsolved Problems in Number Theory, D5.

Cf. A000290, A000578, A001651, A003215, A004825, A004826, A050791, A130472, A195006.

Clear[t]
t = {};
Do[y = (1458 - x^3 - 2 z^3)^(1/3) /. (-1)^(1/3) -> -1; If[IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -980, 980}, {x, -25319, 25319}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 2000}];
Select[v, MemberQ[u, #] &]

A336230 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 128.

Original entry on oeis.org

1, 4, 9, 25, 49, 81, 121, 169, -224, 225, 289, 361, -383, 441, 504, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, -2219, 2401, 2601, -2687, 2809, 3025, 3249, 3481, -3680, 3721, 3969, 4225, -4283, 4417, 4489, 4761, 5041, 5329, -5459

Offset: 1

XU Pingya, Jul 12 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).

(5 - 4*n^2)^3 + (5 - 4*(n + 1)^2)^3 + 2*(2*n + 1)^6 = 128. A000290(2*n + 1) are terms of the sequence, i.e., there is an infinity of nontrivial solutions to the equation.

			1^3 + 5^3 + 2 * 1^3 = 128, 1 is a term.
(-11)^3 + (-31)^3 + 2 * (25)^3 = 128, 25 is a term.

R. K. Guy, Unsolved Problems in Number Theory, D5.

Cf. A000290, A000578, A003215, A004825, A004826, A050791, A130472, A195006.

Clear[t]
t = {};
Do[y = (128 - x^3 - 2 z^3)^(1/3) /. (-1)^(1/3) -> -1; If[IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -4761, 4761}, {x, -11550, 11550}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 9523}];
Select[v, MemberQ[u, #] &]

A338239 Values z of primitive solutions (x, y, z) to the Diophantine equation 2x^3 + 2y^3 + z^3 = 1.

Original entry on oeis.org

-1, 1, -5, 11, -17, 19, 29, -31, -37, -61, 79, -85, 113, -127, -143, 145, -209, 305, 361, -485, 487, 545, 647, 667, 811, -1091, -1151, 1153, -1235, -1429, -1525, 1597, 1699, -1793, -2249, 2251, -2533, 2627, -2677, 2977, -2981, 3089, -3295, 3739, -3887, 3889

Offset: 1

XU Pingya, Oct 18 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).

When x = (3*c)*t - (9*a)*t^4, y = (9*a)*t^4, z = c - (9*a)*t^3; a*x^3 + a*y^3 + c*z^3 = c^4. Let a = 2, c = 1, then 1 - 18*n^3 and 1 + 18*n^3 are terms of the sequence. Also, -A337928 and A337929 are subsequences.

			2*25^3 + 2*(-64)^3 + 79^3 = 2*164^3 + 2*(-167)^3 + 79^3 = 1, 79 is a term.

Beniamino Segre, On the rational solutions of homogeneous cubic equations in four variables, Math. Notae, 11 (1951), 1-68.

Cf. A000045, A000290, A000578, A003215, A004825, A004826, A047201, A050791, A130472, A195006, A337928, A337929.

Clear[t]
t = {};
Do[y = ((1 - 2x^3 - z^3)/2)^(1/3) /. (-1)^(1/3) -> -1;
 If[IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -4000, 4000}, {x, -Round[(Abs[1 + z^3]/6)^(1/2)], Round[(Abs[1 + z^3]/6)^(1/2)]}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 8001}];
Select[v, MemberQ[u, #] &]

cp's OEIS Frontend

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

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A163827 a(n) = 6n^3 + 1, solution z in Diophantine equation x^3 + y^3 = z^3 - 2. It may be considered a Fermat near miss by 2.

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A330013 a(n) is the number of solutions with nonnegative (x,y,z) to the cubic Diophantine equation x^3+y^3+z^3 - 3*x*y*z = n.

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A336449 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 2*4^6.

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A336450 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 2*5^6.

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A336166 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 2.

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A336226 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 1458.

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A330013 a(n) is the number of solutions with nonnegative (x,y,z) to the cubic Diophantine equation x^3+y^3+z^3 - 3xy*z = n.

A336449 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2z^3 = 24^6.

A336450 Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2z^3 = 25^6.

A338239 Values z of primitive solutions (x, y, z) to the Diophantine equation 2x^3 + 2y^3 + z^3 = 1.