A060464 -id:A060464 - cp's OEIS frontend

A060465 Value of x of the solution to x^3 + y^3 + z^3 = A060464(n) (numbers not 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 <= |x| <= |y| <= |z|.

0, 0, 0, 1, -1, 0, 0, 0, 1, -2, 7, -1, -511, 1, -1, 0, 1, -11, -2901096694, -1, 0, 0, 0, 1, -283059965, -2736111468807040, -1, 0, 1, 0, 1, 117367, 12602123297335631, 2, -5, 2, -2, 6, -23, 602, 23961292454, -1, -7, 1, -11, 1, -1, 0, 2, 0, 0, 0, 1, 2, 11, -1, 7, 1

Offset: 0

N. J. A. Sloane, Apr 10 2001

Indexed by A060464.
Only primitive solutions where gcd(x,y,z) does not divide n are considered.
From the solution A060464(24) = 30 = -283059965^3 - 2218888517^3 + 2220422932^3 (smallest possible magnitudes according to A. Bogomolny), one has a(24) = -283059965. A solution to A060464(25) = 33 remains to be found. Other values for larger n can be found in the first column of the table on Hisanori Mishima's web page. - M. F. Hasler, Nov 10 2015
In 2019 Brooker found a solution for n = 33 (see A332201 and references there) and later in the same year for n = 42, using the collaborative "Charity Engine". It would be nice to have information on how far it is established that these solutions are the smallest possible. - M. F. Hasler, Feb 24 2020

			For n = 16 the smallest solution is 16 = (-511)^3 + (-1609)^3 + 1626^3, which gives the term -511.
42 = 12602123297335631^3 + 80435758145817515^3 + (-80538738812075974)^3 was found by Andrew Booker and Andrew Sutherland.
74 = 66229832190556^3 + 283450105697727^3 + (-284650292555885)^3 was found by Sander Huisman.

R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Springer, New York, 2004, Section D5, 231-234.

D. J. Bernstein, Three cubes
A. Bogomolny, Finicky Diophantine Equations on cut-the-knot.org, accessed Nov. 10, 2015.
B. Conn, L. Vaserstein, On sums of three integral cubes, Contemp. Math 166 (1994) MR1284068
V. L. Gardiner, R. B. Lazarus, P. R. Stein, Solutions of the diophantine equation x^3+y^3=z^3-d, Math. Comp. 18 (1964) 408-413.
D.R. Heath-Brown, W.M. Lioen and H.J.J. te Riele on Solving the Diophantine Equation x3 + y3 + z3 = k on a Vector Computer
J. C. P. Miller, M. F. C. Woollett, Solutions of the Diophantine Equation x^3+y^3+z^3=k, J. Lond. Math. Soc. 30 (1) (1955) 101-110.
Hisanori Mishima, About n=x^3+y^3+z^3

Cf. A060464, A060466, A060467, A173515.

(* this program is not convenient for hard cases *) nmax = 29; xmin[] = 0; xmax[] = 20; xmin[16] = 500; xmax[16] = 600; xmin[24] = 2901096600; xmax[24] = 2901096700; r[n_, x_] := Reduce[0 <= Abs[x] <= Abs[y] <= Abs[z] && n == x^3 + y^3 + z^3, {y, z}, Integers]; r[n_ /; IntegerQ[n^(1/3)]] := {0, 0, n^(1/3)}; mySort = Sort[#1, Which[Abs[#1[[3]]] <= Abs[#2[[3]]], True, Abs[#1[[3]]] == Abs[#2[[3]]], If[Abs[#1[[2]]] <= Abs[#2[[2]]], True, False], True, False] & ] & ; rep := {x_, y_, z_} /; (x + y == 0 && x > 0) :> {-x, -y, z}; r[n_] := Reap[Do[ sp = r[n, x] /. C[1] -> 1; If[sp =!= False, xyz = {x, y, z} /. {ToRules[sp]} /. rep; If[GCD @@ Flatten[{n, xyz}] == 1, Sow[xyz]]]; sn = r[n, -x] /. C[1] -> 1; If[sn =!= False, xyz = {-x, y, z} /. {ToRules[sn]} /. rep; If[GCD @@ Flatten[{n, xyz}] == 1, Sow[xyz]]], {x, xmin[n], xmax[n]}]][[2, 1]] // Flatten[#, 1] & // mySort // First; A060464 = Select[Range[0, nmax], Mod[#, 9] != 4 && Mod[#, 9] != 5 &]; A060465 = Table[xyz = r[n]; Print[ " n = ", n, " {x,y,z} = ", xyz]; xyz[[1]], {n, A060464}] (* Jean-François Alcover, Jul 10 2012 *)

Edited and a(24) added by M. F. Hasler, Nov 10 2015
a(25) from Tim Browning and further terms added by Charlie Neder, Mar 09 2019
More terms from Jinyuan Wang, Feb 13 2020

A060466 Value of y of the solution to x^3 + y^3 + z^3 = A060464(n) (numbers not 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 <= |x| <= |y| <= |z|.

Original entry on oeis.org

0, 0, 1, 1, -1, -1, 0, 1, 1, -2, 10, 2, -1609, 2, -2, -2, -2, -14, -15550555555, -1, -1, 0, 1, 1, -2218888517, -8778405442862239, 2, 2, 2, -3, -3, 134476, 80435758145817515, 2, -7, -3, 3, 7, -26, 659, 60702901317, 3, -11, 3, -21, -2, -4, -4, 3, -1, 0, 1, 1, -4, 20, 2, 9, 2

Offset: 0

N. J. A. Sloane, Apr 10 2001

Indexed by A060464.
Only primitive solutions where gcd(x,y,z) does not divide n are considered.
From the solution A060464(24) = 30 = -283059965^3 - 2218888517^3 + 2220422932^3 (smallest possible magnitudes according to A. Bogomolny), one has a(24) = -2218888517. A solution to A060464(25) = 33 remains to be found. Other values for larger n can be found in the second column of the table on Hisanori Mishima's web page. - M. F. Hasler, Nov 10 2015

			For n = 16 the smallest solution is 16 = (-511)^3 + (-1609)^3 + 1626^3, which gives the term -1609.
42 = 12602123297335631^3 + 80435758145817515^3 + (-80538738812075974)^3 was found by Andrew Booker and Andrew Sutherland.
74 = 66229832190556^3 + 283450105697727^3 + (-284650292555885)^3 was found by Sander Huisman.

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

A. Bogomolny, Finicky Diophantine Equations on cut-the-knot.org, accessed Nov. 10, 2015.
A.-S. Elsenhans, J. Jahnel, New sums of three cubes, Math. Comp. 78 (2009) 1227-1230.
K. Koyama, Y. Tsuruoka, H. Sekigawa, On searching for solutions of the Diophantine equation x^3+y^3+z^3=n, Math. Comp. 66 (1997) 841.
Hisanori Mishima, About n=x^3+y^3+z^3
Eric S. Rowland, Known families of integer solutions of x^3+y^3+z^3=n
A. Tyszka, A hypothetical upper bound for the solutions of a Diophantine equation with a finite number of solutions, arXiv:0901.2093 [math.NT], 2009-2014.

Cf. A060465, A060467, A173515.

nmax = 29; A060464 = Select[Range[0, nmax], Mod[#, 9] != 4 && Mod[#, 9] != 5 &]; A060465 = {0, 0, 0, 1, -1, 0, 0, 0, 1, -2, 7, -1, -511, 1, -1, 0, 1, -11, -2901096694, -1, 0, 0, 0, 1}; r[n_, x_] := Reduce[0 <= Abs[x] <= Abs[y] <= Abs[z] && n == x^3 + y^3 + z^3, {y, z}, Integers]; A060466 = Table[y /. ToRules[ Simplify[ r[A060464[[k]], A060465[[k]]] /. C[1] -> 0]], {k, 1, Length[A060464]}] (* Jean-François Alcover, Jul 11 2012 *)

In order to be consistent with A060465, where only primitive solutions are selected, a(18)=2 was replaced with -15550555555, by Jean-François Alcover, Jul 11 2012
Edited and a(24) added by M. F. Hasler, Nov 10 2015
a(25) from Tim Browning and further terms added by Charlie Neder, Mar 09 2019
More terms from Jinyuan Wang, Feb 14 2020

A060467 Value of z of the solution to x^3 + y^3 + z^3 = A060464(n) (numbers not 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 <= |x| <= |y| <= |z|.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 2, 2, 3, -11, 2, 1626, 2, 3, 3, 3, 16, 15584139827, 3, 3, 3, 3, 3, 2220422932, 8866128975287528, 3, 3, 3, 4, 4, -159380, -80538738812075974, 3, 8, 4, 3, -8, 31, -796, -61922712865, 3, 12, 3, 22, 4, 5, 5, 3, 4, 4, 4, 4, 5, -21, 4, -10, 4

Offset: 0

N. J. A. Sloane, Apr 10 2001

Indexed by A060464.
Only primitive solutions where gcd(x,y,z) does not divide n are considered.
From the solution A060464(24) = 30 = -283059965^3 - 2218888517^3 + 2220422932^3 (smallest possible magnitudes according to A. Bogomolny), one has a(24) = 2220422932. A solution to A060464(25) = 33 remains to be found. Other values for larger n can be found in the last column of the table on Hisanori Mishima's web page. - M. F. Hasler, Nov 10 2015

			For n = 16 the smallest solution is 16 = (-511)^3 + (-1609)^3 + 1626^3, which gives the term 1626.
42 = 12602123297335631^3 + 80435758145817515^3 + (-80538738812075974)^3 was found by Andrew Booker and Andrew Sutherland.
74 = 66229832190556^3 + 283450105697727^3 + (-284650292555885)^3 was found by Sander Huisman.

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

A. Bogomolny, Finicky Diophantine Equations on cut-the-knot.org, accessed Nov. 10, 2015.
A.-S. Elsenhans, J. Jahnel, New sums of three cubes, Math. Comp. 78 (2009) 1227-1230.
K. Koyama, Y. Tsuruoka, H. Sekigawa, On searching for solutions of the Diophantine equation x^3+y^3+z^3=n, Math. Comp. 66 (1997) 841.
Hisanori Mishima, About n=x^3+y^3+z^3
Eric S. Rowland, Known families of integer solutions of x^3+y^3+z^3=n
A. Tyszka, A hypothetical upper bound for the solutions of a Diophantine equation with a finite number of solutions, arXiv:0901.2093 [math.NT], 2009-2014.

Cf. A060464, A060465, A060466.

nmax = 29; A060464 = Select[Range[0, nmax], Mod[#, 9] != 4 && Mod[#, 9] != 5 &]; A060465 = {0, 0, 0, 1, -1, 0, 0, 0, 1, -2, 7, -1, -511, 1, -1, 0, 1, -11, -2901096694, -1, 0, 0, 0, 1}; r[n_, x_] := Reduce[0 <= Abs[x] <= Abs[y] <= Abs[z] && n == x^3 + y^3 + z^3, {y, z}, Integers]; A060467 = Table[z /. ToRules[ Simplify[ r[A060464[[k]], A060465[[k]]] /. C[1] -> 0]], {k, 1, Length[A060464]}] (* Jean-François Alcover, Jul 11 2012 *)

In order to be consistent with A060465, where only primitive solutions are selected, a(18)=2 was replaced with 15584139827, by Jean-François Alcover, Jul 11 2012
Edited and a(24) added by M. F. Hasler, Nov 10 2015
a(25) from Tim Browning and further terms added by Charlie Neder, Mar 09 2019
More terms from Jinyuan Wang, Feb 14 2020
a(32) corrected by XU Pingya, May 11 2020

A156638 Numbers k such that k^2 + 2 == 0 (mod 9).

Original entry on oeis.org

4, 5, 13, 14, 22, 23, 31, 32, 40, 41, 49, 50, 58, 59, 67, 68, 76, 77, 85, 86, 94, 95, 103, 104, 112, 113, 121, 122, 130, 131, 139, 140, 148, 149, 157, 158, 166, 167, 175, 176, 184, 185, 193, 194, 202, 203, 211, 212, 220, 221, 229, 230, 238, 239, 247, 248, 256

Offset: 1

Vincenzo Librandi, Feb 12 2009

From Artur Jasinski, Apr 30 2010: (Start)
Numbers congruent to 4 or 5 mod 9.
Numbers which are not the sum of 3 cubes.
Complement to A060464. (End)
Numbers k such that A010888(k^2) = 7. - V.J. Pohjola, Aug 18 2012

Henri Cohen, Number Theory Volume I: Tools and Diophantine Equations. Springer Verlag (2007) p. 380. - Artur Jasinski, Apr 30 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A060464, A010888, A332437, A334521, A334522.

[9*n/2 - 9/4 - 7*(-1)^n/4 : n in [1..80]]; // Wesley Ivan Hurt, Aug 16 2015

A156638:=n->9*n/2 - 9/4 - 7*(-1)^n/4: seq(A156638(n), n=1..80); # Wesley Ivan Hurt, Aug 16 2015

LinearRecurrence[{1, 1, -1}, {4, 5, 13}, 50] (* Vincenzo Librandi, Mar 01 2012 *)
Flatten[Table[9n - {5, 4}, {n, 30}]] (* Alonso del Arte, Aug 09 2015 *)
Select[Range[300],PowerMod[#,2,9]==7&] (* Harvey P. Dale, Jan 31 2023 *)

a(n) = (18*n - 9 - 7*(-1)^n)/4 \\ David Lovler, Aug 21 2022

For n > 2, a(n) = a(n-2) + 9.
G.f.: x*(4*x^2 + x + 4)/(x^3 - x^2 - x + 1). - Alexander R. Povolotsky, Feb 15 2009
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3), n>3.
a(n) = 9*n/2 - 9/4 - 7*(-1)^n/4.
G.f.: x*(4 + x + 4*x^2)/((1 + x)*(1 - x)^2). (End)
a(n) = -a(-n+1). - Bruno Berselli, Jan 08 2012
E.g.f.: 4 + ((18*x - 9)*exp(x) - 7*exp(-x))/4. - David Lovler, Aug 21 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(Pi/18)*Pi/9. - Amiram Eldar, Sep 26 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1.
Product_{n>=1} (1 + (-1)^n/a(n)) = 2*cos(Pi/9) - 1 (= A332437 - 1). (End)

A274406 Numbers m such that 9 divides m*(m + 1).

Original entry on oeis.org

0, 8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, 116, 117, 125, 126, 134, 135, 143, 144, 152, 153, 161, 162, 170, 171, 179, 180, 188, 189, 197, 198, 206, 207, 215, 216, 224, 225, 233, 234, 242, 243, 251, 252, 260, 261, 269

Offset: 1

Bruno Berselli, Jun 20 2016

Equivalently, numbers congruent to 0 or 8 mod 9.

Terms of A007494 with indices in A047264. Also, terms of A060464 with indices in A047335.

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A008591 (first bisection), A010689 (first differences), A017257 (second bisection).
Cf. similar sequences in which m*(m+1) is divisible by k: A014601 (k=4), A047208 (k=5), A007494 (k=3 and 6), A047335 (k=7), A047521 (k=8), this sequence (k=9).
Cf. A301451: numbers congruent to {1, 7} mod 9; A193910: numbers congruent to {2, 6} mod 9.

[n: n in [0..300] | IsDivisibleBy(n*(n+1),9)];

Select[Range[0, 300], Divisible[# (# + 1), 9] &]

for(n=0, 300, if(n*(n+1)%9==0, print1(n", ")))

[n for n in range(300) if 9.divides(n*(n+1))]

G.f.: x^2*(8 + x)/((1 + x)*(1 - x)^2).
a(n) = (18*n + 7*(-1)^n - 11)/4. Therefore: a(2*m) = 9*m-1, a(2*m+1) = 9*m. It follows that a(j)+a(k) and a(j)*a(k) belong to the sequence if j and k are not both even.
a(n) = -A090570(-n+2).
a(n) = a(n-1) + a(n-2) - a(n-3).
a(2*r+1) + a(2*r+s+1) = a(4*r+s+1) and a(2*r) + a(2*r+2*s+1) = a(4*r+2*s). A particular case provided by these identities: a(n) = a(n - 2*floor(n/6)) + a(2*floor(n/6) + 1).
E.g.f.: 1 + ((9*x - 2)*cosh(x) + 9*(x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2021

A281899 a(n) = n + 6*floor(n/3).

Original entry on oeis.org

0, 1, 2, 9, 10, 11, 18, 19, 20, 27, 28, 29, 36, 37, 38, 45, 46, 47, 54, 55, 56, 63, 64, 65, 72, 73, 74, 81, 82, 83, 90, 91, 92, 99, 100, 101, 108, 109, 110, 117, 118, 119, 126, 127, 128, 135, 136, 137, 144, 145, 146, 153, 154, 155, 162, 163, 164, 171, 172, 173, 180, 181, 182, 189

Offset: 0

Bruno Berselli, Feb 06 2017

Equivalently, numbers that are congruent to {0, 1, 2} mod 9.
Also numbers m such that floor(m/3) = 3*floor(m/9).
The n-th term is 3*n, 3*n-2 or 3*n-4.
For n > 0, numbers k such that 3 | floor(k/3). - Wesley Ivan Hurt, Dec 01 2020

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A002264.
Subsequence of A060464 and A248375.
The first differences are in A105395.
Cf. similar sequences with formula n+i*floor(n/3): A004773 (i=1), A047217 (i=2), A047240 (i=3), A047354 (i=4), A047469 (i=5), this sequence (i=6).
Cf. numbers that are congruent to {0, 1, 2} mod j: the sequences are listed in the previous row for j = 4..9, respectively.

Magma
```
[n+6*(n div 3): n in [0..70]];
```

A281899:=n->n+6*floor(n/3): seq(A281899(n), n=0..100); # Wesley Ivan Hurt, Feb 09 2017

Table[n + 6 Floor[n/3], {n, 0, 70}]
LinearRecurrence[{1,0,1,-1},{0,1,2,9},90] (* Harvey P. Dale, Feb 25 2018 *)

Maxima
```
makelist(n+6*floor(n/3), n, 0, 70);
```

a(n)=n\3*6 + n \\ Charles R Greathouse IV, Feb 07 2017

Python
```
[n+6*int(n/3) for n in range(70)]
```
Sage
```
[n+6*floor(n/3) for n in range(70)]
```

G.f.: x*(1 + x + 7*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = 3*n - 2*(n mod 3). In general, n + 3*h*floor(n/3) = (h+1)*n - h*(n mod 3).
a(n) + a(n+s) = a(2*n+s-1) + 1, where s is nonnegative and not divisible by 3. Example: for s=14, a(n) + a(n+14) = a(2*n+13) + 1; for n=3, a(3) + a(17) = a(19) + 1 = 9 + 47 = 55 + 1 = 56.
a(6*k+r) = 18*k + a(r), where 0 <= r <= 5.
a(n) = 7*A002264(n) + A002264(n+1) + A002264(n+2).

A265187 Nonnegative m for which 2floor(m^2/11) = floor(2m^2/11).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 24, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 55, 56, 57, 59, 60, 61, 62, 64, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 79, 81, 82, 83, 84

Offset: 1

Bruno Berselli, Dec 04 2015

Also, nonnegative m not congruent to 3 or 8 (mod 11).
Integers x >= 0 satisfying k*floor(x^2/11) = floor(k*x^2/11) with k >= 0:
k = 0, 1:  x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...   (A001477);
k = 2:     x = 0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, ... (this sequence);
k = 3:     x = 0, 1, 5, 6, 10, 11, 12, 16, 17, 21, 22, ... (A265188);
k = 4..10: x = 0, 1, 10, 11, 12, 21, 22, 23, 32, 33, ...   (A112654);
k > 10:    x = 0, 11, 22, 33, 44, 55, 66, 77, 88, 99, ...  (A008593).
Primes in sequence: 2, 5, 7, 11, 13, 17, 23, 29, 31, 37, 43, 53, 59, ...

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,1,-1).

Cf. A008593, A112654, A265188.

Cf. similar sequences provided by 2*floor(m^2/h) = floor(2*m^2/h): A005843 (h=2), A001477 (h=3,4), A008854 (h=5), A047266 (h=6), A047299 (h=7), A042965 (h=8), A060464 (h=9), A237415 (h=10), this sequence (h=11), A047263 (h=12).

[n: n in [0..100] | 2*Floor(n^2/11) eq Floor(2*n^2/11)];

Select[Range[0, 100], 2 Floor[#^2/11] == Floor[2 #^2/11] &]
Select[Range[0, 100], ! MemberQ[{3, 8}, Mod[#, 11]] &]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 1, 2, 4, 5, 6, 7, 9, 10, 11}, 80]

is(n)=2*(n^2\11) == (2*n^2)\11 \\ Anders Hellström, Dec 05 2015

[n for n in (0..100) if 2*floor(n^2/11) == floor(2*n^2/11)]

G.f.: x^2*(1 + x + 2*x^2 + x^3 + x^4 + x^5 + 2*x^6 + x^7 + x^8)/((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)).

a(n) = a(n-1) + a(n-9) - a(n-10) for n>10.

A332201 Sum of three cubes problem: a(n) = integer x with the least possible absolute value such that n = x^3 + y^3 + z^3 with |x| >= |y| >= |z|, or 0 if no such x exists.

Original entry on oeis.org

0, 1, 1, 1, 0, 0, 2, 2, 2, 2, 2, 3, -11, 0, 0, 2, 2, 2, 3, 3, 3, 16, 0, 0, 2, 3, 3, 3, 3, 3, 2220422932, 0, 0

Offset: 0

M. F. Hasler, Feb 08 2020

It is known that there is no solution for n congruent to +-4 (mod 9), but it is now conjectured that there is a solution (and probably infinitely many such) for all other numbers. The numbers n = 0, 1 and 2 are the only cases for which infinite families of parametric solutions are known, for other n the solutions seem to be sporadic.
Search on this problem was motivated by a statement in Mordell's paper from 1953. Beck et al. found a solution for n = 30 in 1999, and for 52 in 2000. Huisman found a solution for n = 74 in 2016. A solution for 33 was found by Booker in 2019. The number 42 was the last one below 100 for which a solution was found, in late 2019, using a collaborative effort with supercomputers and home computers from volunteers.
For n < 30, we have a(n) = A246869(n+1) for the nonzero values, while A246869(n+1) = 2 for n == 4 or 5 (mod 9) up to there.

			   0 = 0^3 + 0^3 + 0^3,     1 = 1^3 + 0^3 + 0^3,
   2 = 1^3 + 1^3 + 0^3,     3 = 1^3 + 1^3 + 1^3,
   6 = 2^3 - 1^3 - 1^3,     7 = 2^3 - 1^3 + 0^3,
   8 = 2^3 + 0^3 + 0^3,     9 = 2^3 + 1^3 + 0^3,
  10 = 2^3 + 1^3 + 1^3,    11 = 3^3 - 2^3 - 2^3,
  12 = -11^3 + 10^3 + 7^3, 15 = 2^3 + 2^3 - 1^3,
  16 = 2^3 + 2^3 + 0^3,    17 = 2^3 + 2^3 + 1^3,
  18 = 3^3 - 2^3 - 1^3,    19 = 3^3 - 2^3 + 0^3,
  20 = 3^3 - 2^3 + 1^3,    21 = 16^3 - 14^3 - 11^3,
  24 = 2^3 + 2^3 + 2^3,    25 = 3^3 - 1^3 - 1^3,
  26 = 3^3 - 1^3 + 0^3,    27 = 3^3 + 0^3 + 0^3,
  28 = 3^3 + 1^3 + 0^3,    29 = 3^3 + 1^3 + 1^3,
30 = 2220422932^3 - 2218888517^3 - 283059965^3 was discovered by Beck, Pine, Yarbrough and Tarrant in 1999 following an approach suggested by N. Elkies.
33 = 8866128975287528^3 - 8778405442862239^3 - 2736111468807040^3 was found by A. Booker in 2019. It is uncertain whether these are the smallest solutions.

Michael Beck, Eric Pine, Wayne Tarrant and Kim Yarbrough Jensen, New integer representations as the sum of three cubes, Math. Comp. 76 (2007) pp. 1683-1690, DOI:10.1090/S0025-5718-07-01947-3.
A. R. Booker, Cracking the problem with 33, Res. Number Theory 5 no. 26 (2019), DOI:10.1007/s40993-019-0162-1.
V. L. Gardiner, R. B. Lazarus and P. R. Stein: Solutions of the Diophantine Equation x^3 + y^3 = z^3 - d, Math.Comp. 18, No. 87 (1964), pp. 408-413. DOI: 10.2307/2003763.
Jon Grantham and P. G. Walsh, Representing integers as a sum of three cubes, arXiv preprint, arXiv:2211.12149 [math.NT], 2022.
B. Haran, Sum of Three Cubes - Numberphile, YouTube playlist (featuring videos from Nov 06 2015, May 31 2016 (about 74), Mar 12 2019 (about 33), Sep 2019 (about 42)).
Sander G. Huisman, Newer sums of three cubes, arXiv:1604.07746 [math.NT] (2016).
J. C. P. Miller and M. F. C. Woollett, Solutions of the Diophantine Equation x^3+y^3+z^3=k, Journal of the London Mathematical Society, s1-30 (1955) pp. 101-110. DOI: 10.1112/jlms/s1-30.1.101.
L. J. Mordell, Integer Solutions of the Equation x^2+y^2+z^2+2xyz = n, Journal London Math. Soc. s1-28 no. 4 (1953) pp. 500-510.
Bjorn Poonen, Undecidability in number theory, Notices Amer. Math. Soc. 55 (2008), no. 3, pp. 344-350.

Cf. A060464, A060465, A060466, A060467, A246869.

apply( A332201(n,L=oo)={!bittest(48,n%9)&& for(c=0,L, my(t1=c^3-n, t2=c^3+n, a); for(b=0,c,((ispower(t1-b^3,3,&a)&&abs(a)<=c)||(ispower(t1+b^3,3,&a)&&abs(a)<=c))&&return(c); ispower(t2-b^3,3,&a) && abs(a)<=c && return(-c)))}, [0..29])

a(n) = 0 for n == 4 or n == 5 (mod 9).
a(n) <= k if |n - k^3| < 3 or |n - 2*k^3| < 2 or n = 3*k^3 for some k.
a(n) = A246869(n+1) for all n < 30 with a(n) > 0.

a(31) = a(32) = 0 added by Jinyuan Wang, Feb 15 2020

A334521 Numbers that are not congruent to +- mod 9 and are neither a cube nor twice a cube.

Original entry on oeis.org

3, 6, 7, 9, 10, 11, 12, 15, 17, 18, 19, 20, 21, 24, 25, 26, 28, 29, 30, 33, 34, 35, 36, 37, 38, 39, 42, 43, 44, 45, 46, 47, 48, 51, 52, 53, 55, 56, 57, 60, 61, 62, 63, 65, 66, 69, 70, 71, 72, 73, 74, 75, 78, 79, 80, 81, 82, 83, 84, 87, 88, 89, 90, 91, 92, 93, 96, 97, 98, 99, 100, 101, 102, 105, 106, 107, 108, 109

Offset: 1

N. J. A. Sloane, May 08 2020

nonn

Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.

Cf. A000578, A060464, A156638.

Complement of A334522.

A334522 Numbers that are either congruent to +-4 mod 9, or else are cubes or twice cubes.

Original entry on oeis.org

1, 2, 4, 5, 8, 13, 14, 16, 22, 23, 27, 31, 32, 40, 41, 49, 50, 54, 58, 59, 64, 67, 68, 76, 77, 85, 86, 94, 95, 103, 104, 112, 113, 121, 122, 125, 128, 130, 131, 139, 140, 148, 149, 157, 158, 166, 167, 175, 176, 184, 185, 193, 194, 202, 203, 211, 212, 216, 220, 221, 229, 230, 238, 239, 247, 248, 250, 256, 257, 265

Offset: 1

N. J. A. Sloane, May 08 2020

nonn

Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.

Cf. A000578, A060464, A156638.

Complement of A334521.

cp's OEIS Frontend

A060465 Value of x of the solution to x^3 + y^3 + z^3 = A060464(n) (numbers not 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 <= |x| <= |y| <= |z|.

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A060466 Value of y of the solution to x^3 + y^3 + z^3 = A060464(n) (numbers not 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 <= |x| <= |y| <= |z|.

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A060467 Value of z of the solution to x^3 + y^3 + z^3 = A060464(n) (numbers not 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 <= |x| <= |y| <= |z|.

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A156638 Numbers k such that k^2 + 2 == 0 (mod 9).

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A274406 Numbers m such that 9 divides m*(m + 1).

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A281899 a(n) = n + 6*floor(n/3).

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A265187 Nonnegative m for which 2floor(m^2/11) = floor(2m^2/11).