A024975 -id:A024975 - cp's OEIS frontend

A165455 Primes p such that p^2 is a sum of three distinct positive cubes.

53, 59, 71, 163, 167, 251, 523, 577, 613, 643, 773, 787, 811, 827, 863, 881, 883, 919, 937, 1097, 1117, 1301, 1567, 1607, 1709, 1777, 1867, 1873, 1877, 1889, 1931, 2161, 2237, 2309, 2447, 2521, 2591, 2647, 2687, 2719, 2843, 2897, 2969, 3011, 3079, 3163

Offset: 1

Vladimir Joseph Stephan Orlovsky, Sep 20 2009

nonn

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

Cf. A161992, A165454.

lst={};Do[Do[Do[d=Sqrt[a^3+b^3+c^3];If[d<=834&&IntegerQ[d]&&PrimeQ[d], AppendTo[lst,d]],{c,b+1,5!,1}],{b,a+1,5!,1}],{a,5!}];Union@lst

A000040 INTERSECT A165454.

{p: p in A000040 and p^2 in A024975}. [R. J. Mathar, Oct 07 2009]

Extended beyond 827 by R. J. Mathar, Oct 07 2009

Title corrected by Jeppe Stig Nielsen, Jan 26 2015

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.

A385409 a(n) is the smallest positive integer k such that the Diophantine equation x^3 + y^3 + z^3 + w^3 = k^2, where 0 < x < y < z < w has exactly n integer solutions.

Original entry on oeis.org

10, 42, 39, 153, 126, 276, 273, 312, 315, 476, 588, 336, 546, 777, 1053, 756, 1216, 1386, 1560, 1134, 1323, 1488, 1365, 1368, 1344, 1596, 2366, 2496, 2988, 1680, 2548, 1736, 2184, 3003, 3720, 2520, 3185, 3552, 2268, 3564, 4095, 3213, 4578, 4392, 5208, 4004, 4599, 5733

Offset: 1

Zhining Yang, Jun 27 2025

nonn

Conjecture: a(n) exists for all n.

			a(4)=153, because 153^2 = 5^3 + 15^3 + 21^3 + 22^3 = 2^3 + 7^3 + 15^3 + 27^3 = 6^3 + 8^3 + 9^3 + 28^3 = 1^3 + 5^3 + 11^3 + 28^3 and no integer less than 153 has 4 solutions.

Zhining Yang, Table of n, a(n) for n = 1..1341

Cf. A024975, A025419, A377444, A384430, A385354.

s = Table[{k, Length@Select[PowersRepresentations[k^2, 4, 3],
      0 < #[[1]] < #[[2]] < #[[3]] < #[[4]] &]}, {k, 500}];
a = Table[SelectFirst[s, #[[2]] == k &], {k, 10}][[All, 1]]

A272072 Nonnegative integers n such that 10^n is not of the form x^3 + y^3 + z^3 where x > y > z > 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 10

Offset: 1

Altug Alkan, Apr 19 2016

This sequence is finite because 10^n can be written as the sum of three distinct positive cubes for all n > 10.
  Proof: If 10^n = x^3 + y^3 + z^3, then 10^(n+3) = 10^3 * 10^n = 10^3 * (x^3 + y^3 + z^3) = (10*x)^3 + (10*y)^3 + (10*z)^3. It is clear that if x, y and z are distinct, then k*x, k*y and k*z are also distinct for all nonzero values of k. So with 10^n also 10^(n+3) is not in the sequence. Therefore, if we can find an integer n such that 10^n, 10^(n+1) and 10^(n+2) are not in the sequence then all 10^m with m >= n will also not be in the sequence. In the example section is shown that 10^n is not in the sequence for n = 11, 12 and 13. Thus this sequence is finite, and all its terms can be found in limited range 0 <= n <= 10.
Note that 10^n, for positive n == 1 (mod 3), can be written as a sum of three nondistinct positive cubes because 10^(1+3*k) = (10^k)^3 + (10^k)^3 + (2*10^k)^3 for all k >= 0. This applies to 10^n with n = 1, 4, 7, 10.
Also note that 10^0, 10^2, 10^3 and 10^5 cannot be the sum of three positive cubes (see the table for A003072) which applies especially to the nondistinct case. Nondistinct representations for 10^1 and 10^4 have been considered above. 10^n can be written as the sum of three positive cubes (not necessarily distinct) for all n > 5. For n >= 11 the existence of distinct representations has been proved above. For n = 7 and 10 see the nondistinct case given in the 1 (mod 3) comment. For n = 6, 8 and 9 see the example section for distinct cases.

			10^6 = 35^3 + 70^3 + 85^3.
10^8 = 196^3 + 312^3 + 396^3.
10^9 = 84^3 + 658^3 + 894^3.
10^11 = 1960^3 + 3120^3 + 3960^3.
10^12 = 9^3 + 2991^3 + 9910^3.
10^13 = 6150^3 + 14575^3 + 18825^3.

Cf. A011557, A024975, A271876.

Select[Range[0, 10], NoneTrue[PowersRepresentations[10^#, 3, 3], #[[3]] > #[[2]] > #[[1]] > 0 &] &] (* Michael De Vlieger, Apr 20 2016, Version 10 *)

A272341 Numbers n such that 6^n is not of the form (x^3 + y^3 + z^3) / 3 where x > y > z > 0.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 8, 9, 12, 15

Offset: 1

Altug Alkan, Apr 26 2016

Numbers n such that 6^n cannot be written as the average of three distinct cubes.
This sequence is complete. So 6^n can be written as the average of three distinct positive cubes for all n > 15.
Proof: If 6^n = (x^3 + y^3 + z^3)/3, then 6^(n+3) = 6^3 * 6^n = 6^3 * (x^3 + y^3 + z^3)/3 = ((6*x)^3 + (6*y)^3 + (6*z)^3)/3. It is clear that if x, y and z are distinct, then k*x, k*y and k*z are also distinct for all nonzero values of k. So with 6^n also 6^(n+3) is not in the sequence. Therefore, if we can find an integer n such that 6^n, 6^(n+1) and 6^(n+2) are not in the sequence then all 6^m with m >= n will also not be in the sequence. In the example section it is shown that 6^n is not in the sequence for n = 4, 11 and 18. So (4+3*u), (11+3*v) and (18+3*w) cannot be member of this sequence for all nonnegative values of u, v and w. If we choose u=4, v=2 and w=0, we obtain the consecutive values that are 16, 17, 18. Thus this sequence is finite, and all its terms can be found in limited range 0 <= n <= 15.

			2 is a term because 3*6^2 = 108 is not a term of A024975.
4 is not a term because 3*6^4 = 3888 = 1^3 + 8^3 + 15^3.
11 is not a term because 3*6^11 = 1088391168 = 48^3 + 124^3 + 1028^3.
18 is not a term because 3*6^18 = 304679870005248 = 38322^3 + 47840^3 + 51790^3.

Cf. A000400, A024975.

Select[Range@ 12, Length[PowersRepresentations[3*6^#, 3, 3] /. {x_, y_, z_} /; Or[x == 0, x == y == z] -> Nothing] == 0 &] (* Michael De Vlieger, Apr 27 2016, Version 10.2 *)

A333821 Numbers k that can be represented in the form k = p^3 - q^3 - r^3, where p, q, r are positive integers satisfying p = q + r.

Original entry on oeis.org

6, 18, 36, 48, 60, 90, 126, 144, 162, 168, 210, 216, 252, 270, 288, 330, 360, 378, 384, 396, 468, 480, 486, 540, 546, 594, 630, 720, 750, 792, 816, 858, 918, 924, 972, 990, 1008, 1026, 1140, 1152, 1170, 1260, 1296, 1344, 1386, 1404, 1518, 1530, 1560, 1620, 1638, 1656, 1680, 1728, 1800

Offset: 1

Antonio Roldán, Apr 06 2020

nonn

An alternative representation of k is k = 3*q*r*(q+r), with q, r positive integers, then k is a multiple of 6.

			60 is in the sequence because 60 = 5^3 - 4^3 - 1^3, with 5 = 4 + 1.

Cf. A000578, A003072, A024975, A078129, A255018, A275154.

ok(n) = {my(i=1, a=0, m=0, j); if(n%6==0, while(a<=n&&m==0, j=1; while(j

a(n) = 6 * A121741(n).

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A165455 Primes p such that p^2 is a sum of three distinct positive cubes.

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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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A385409 a(n) is the smallest positive integer k such that the Diophantine equation x^3 + y^3 + z^3 + w^3 = k^2, where 0 < x < y < z < w has exactly n integer solutions.

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A272072 Nonnegative integers n such that 10^n is not of the form x^3 + y^3 + z^3 where x > y > z > 0.

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A272341 Numbers n such that 6^n is not of the form (x^3 + y^3 + z^3) / 3 where x > y > z > 0.

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A333821 Numbers k that can be represented in the form k = p^3 - q^3 - r^3, where p, q, r are positive integers satisfying p = q + r.

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