A023042 -id:A023042 - cp's OEIS frontend

A176413 a(n) = 19*3^n.

19, 57, 171, 513, 1539, 4617, 13851, 41553, 124659, 373977, 1121931, 3365793, 10097379, 30292137, 90876411, 272629233, 817887699, 2453663097, 7360989291, 22082967873, 66248903619, 198746710857, 596240132571, 1788720397713, 5366161193139, 16098483579417

Offset: 0

Vincenzo Librandi, Apr 17 2010

Since 19^3 = 3^3+10^3+18^3, the cube of any multiple of 19 can be written as the sum of three positive cubes: (19*k)^3 = (3*k)^3 + (10*k)^3 + (18*k)^3.

Index entries for linear recurrences with constant coefficients, signature (3).

Subsequence of A023042.

Cf. A000244.

Magma
```
[19*3^n: n in [0..250]];
```

19*3^Range[0,30] (* or *) NestList[3#&,19,30] (* Harvey P. Dale, Feb 03 2013 *)

G.f.: 19/(1-3*x). - R. J. Mathar, Aug 24 2011
From Elmo R. Oliveira, Aug 16 2024: (Start)
E.g.f.: 19*exp(3*x).
a(n) = 19*A000244(n).
a(n) = 3*a(n-1) for n > 0. (End)

Comment edited by Jon E. Schoenfield, Jun 20 2010

a(24)-a(25) from Elmo R. Oliveira, Aug 16 2024

A272023 Numbers n such that n^3 = (x^3 + y^3 + z^3) / 3 where x > y > z > 0, is soluble.

Original entry on oeis.org

75, 87, 126, 135, 150, 171, 174, 204, 225, 246, 252, 261, 270, 297, 300, 333, 342, 348, 375, 378, 405, 408, 435, 450, 457, 492, 504, 513, 522, 525, 540, 543, 594, 600, 609, 612, 618, 630, 645, 666, 675, 684, 696, 723, 738, 741, 750, 753, 756, 783, 788, 810, 813, 815, 816, 825, 855, 870, 882, 891, 900, 914

Offset: 1

Altug Alkan, Apr 18 2016

nonn

Numbers whose cube is the average of three distinct positive cubes.
If n is in this sequence, then k*n is also a member of this sequence for all k > 0.
Terms that are not divisible by 3 are 457, 788, 815, 914, ...
Terms that are not members of A023042 are 457, 914, 1078, ...

			75 is a term because 75^3 = (10^3 + 17^3 + 108^3) / 3.
150 = 2*75 is a term.
87 is a term because 87^3 = (13^3 + 54^3 + 122^3) / 3.
457 is a term because 457^3 = (226^3 + 379^3 + 604^3) / 3.

Chai Wah Wu, Table of n, a(n) for n = 1..1000

Cf. A023042, A024975.

A346137 Numbers k such that k^3 = x^3 + y^3 + z^3, x > y > z >= 0, has at least 2 distinct solutions.

Original entry on oeis.org

18, 36, 41, 46, 54, 58, 60, 72, 75, 76, 81, 82, 84, 87, 88, 90, 92, 96, 100, 108, 114, 116, 120, 123, 126, 132, 134, 138, 140, 142, 144, 145, 150, 152, 156, 159, 160, 162, 164, 168, 170, 171, 174, 176, 178, 180, 184, 185, 186, 189, 190, 192, 198, 200, 201, 202, 203

Offset: 1

Sebastian Magee, Jul 30 2021

nonn

This sequence is based on a generalization of Fermat's last theorem with n=3, in which three terms are added. Fermat's Theorem states that there are no solution with only two terms, this sequence shows there are many integers for which there are multiple solutions if three terms are allowed. The sequence is also related to the Taxicab numbers.

			41 is in the sequence because 41^3 = 33^3 + 32^3 + 6^3 = 40^3 + 17^3 + 2^3.

Subsequence of A023042.

Cf. A001235, A346071.

q[k_] := Count[IntegerPartitions[k^3, {3}, Range[0, k-1]^3], ?(UnsameQ @@ # &)] > 1; Select[Range[200], q] (* _Amiram Eldar, Sep 03 2021 *)

from itertools import combinations
from collections import Counter
from sympy import integer_nthroot
def icuberoot(n): return integer_nthroot(n, 3)[0]
def aupto(kmax):
    cubes = [i**3 for i in range(kmax+1)]
    cands, cubesset = (sum(c) for c in combinations(cubes, 3)), set(cubes)
    c = Counter(s for s in cands if s in cubesset)
    return sorted(icuberoot(s) for s in c if c[s] >= 2)
print(aupto(203)) # Michael S. Branicky, Sep 04 2021

A377739 Array of positive integer triples (x,y,z) where the sum of their cubes equals another cubic number.

Original entry on oeis.org

3, 4, 5, 1, 6, 8, 6, 8, 10, 2, 12, 16, 9, 12, 15, 3, 10, 18, 7, 14, 17, 12, 16, 20, 4, 17, 22, 3, 18, 24, 18, 19, 21, 11, 15, 27, 15, 20, 25, 4, 24, 32, 18, 24, 30, 6, 20, 36, 14, 28, 34, 2, 17, 40, 6, 32, 33, 21, 28, 35, 16, 23, 41, 5, 30, 40, 3, 36, 37, 27, 30, 37, 24, 32, 40, 8, 34, 44, 29, 34, 44, 6, 36, 48, 12, 19, 53, 27, 36, 45, 36, 38, 42

Offset: 1

Luke Voyles, Nov 05 2024

nonn

The Shiraishi theorem demonstrated that there were an infinite number of cubic number triples whose sum equaled a cubic number (A226903). Through an analysis of the cubic number triples found by Russell and Gwyther, another way to prove that there are an infinite number of cubic number triples who sum equals a cubic number appeared. For any triple, one can add a zero to the end of the three numbers. The new three numbers will also equal a cubic number. For example, 3^3+4^3+5^3=6^3 can be transformed into 30^3+40^3+50^3=60^3. The number of zeros that are consistently applied to each of the numbers who cubic numbers will always create new cubic numbers. For example, 30^3+40^3+50^3=60^3 can become 300^3+400^3+500^3=600^3 and 3000^3+4000^3+5000^3=6000^3, and so on. Through experiments, the formula holds true for Pythagorean triples and Pythagorean quadruples as well. To apply the method to Pythagorean triples, 3^2+4^2=5^2 can be transformed into 30^2+40^2=50^2, 300^2+400^2=500^2, and so on. For Pythagorean quadruples, 3^2+4^2+12^2=13^2 can be transformed into 30^2+40^2+120^2=130^2 and then to 300^2+400^2+1200^2=1300^2. The property holds even beyond the second and third powers. For example, 3530^4=300^4+1200^4+2720^4+3150^4 just as 353^4=30^4+120^4+272^4+315^4. Additionally, 1440^5=270^5+840^5+1100^5+1330^5 just as 144^5=27^5+84^5+110^5+133^5. Once one set is found, it appears there can be an infinite number of similar sets for any power through this method.

The list of Russell and Gwyther also reveals that the cube of 38 can be represented as the sum of the cubes of nine unique positive integers. This is because 38^3=3^3+4^3+5^3+7^3+14^3+17^3+18^3+24^3+30^3.

			3^3+4^3+5^3=6^3
1^3+6^3+8^3=9^3
6^3+8^3+10^3=12^3
2^3+12^3+16^3=18^3
9^3+12^3+15^3=18^3

A. Russell and C. E. Gwyther, The Partition of Cubes, The Mathematical Gazette, Vol. 21, No. 242 (Feb., 1937), pp. 33-35 (3 pages).

The sum of each cubic number triple produce the sequence A023042. The comments produce another method to produce an infinite number of cubic number triples whose sum equals a cube that the method shown by Shiraishi according to A226903. The comments discuss qualities of Pythagorean triples A103606 and Pythagorean quadruples A096907. The title's structure drew inspiration from A291694.

If a^3+b^3+c^3=d^3, then any specific number k that has a zero as the last digit will make k(d^3) another cubic number through the formula k(a^3)+k(b^3)+k(c^3)=k(d^3)

A385364 Numbers whose cube is the sum of four distinct nonnegative cubes.

Original entry on oeis.org

13, 14, 18, 20, 23, 24, 26, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95

Offset: 1

Robert G. Wilson v, Jul 29 2025

nonn

Numbers v such that v^3 = w^3+x^3+y^3+z^3, w>x>y>z>=0, is soluble.

Inspired by A023042.

Cf. A000578, A003072, A023042, A274334.

fQ[n_] := Length[ Select[ PowersRepresentations[ n^3, 4, 3], Union@# == # && #[[1]] > 0 &]] > 0; Select[ Range@ 95, fQ]

cp's OEIS Frontend

A176413 a(n) = 19*3^n.

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A272023 Numbers n such that n^3 = (x^3 + y^3 + z^3) / 3 where x > y > z > 0, is soluble.

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A346137 Numbers k such that k^3 = x^3 + y^3 + z^3, x > y > z >= 0, has at least 2 distinct solutions.

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A377739 Array of positive integer triples (x,y,z) where the sum of their cubes equals another cubic number.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A385364 Numbers whose cube is the sum of four distinct nonnegative cubes.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica