A328149 Numbers whose set of divisors contains a quadruple (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3.

60, 72, 120, 144, 180, 216, 240, 288, 300, 360, 420, 432, 480, 504, 540, 576, 600, 648, 660, 720, 780, 792, 840, 864, 900, 936, 960, 1008, 1020, 1080, 1140, 1152, 1200, 1224, 1260, 1296, 1320, 1368, 1380, 1440, 1500, 1512, 1560, 1584, 1620, 1656, 1680, 1710

Offset: 1

Author

Michel Lagneau, Jun 07 2020

Keywords

• nonn

Comments

The subsequence of numbers of the form 2^i*3^j is 72, 144, 216, 288, 432, 576, 648, 864, 1152, 1296, ...

The corresponding number of quadruples of the sequence is 1, 1, 2, 2, 2, 2, 3, 3, 2, 6, 2, 4, 4, 2, 3, 4, 4, 3, 2, 10, ... (see the sequence A328204).

The set of divisors of a(n) contains at least one primitive quadruple.

Examples:

The set of divisors of a(1) = 60 contains only one primitive quadruple: (3, 4, 5, 6).

The set of divisors of a(10) = 360 contains two primitive quadruples: (1, 6, 8, 9) and (3, 4, 5, 6).

From Robert Israel, Jul 06 2020: (Start)

Every multiple of a member of the sequence is in the sequence.

The first member of the sequence not divisible by 6 is a(68) = 2380, which has the quadruple (7, 14, 17, 20).

The first odd member of the sequence is a(1230) = 43065, which has the quadruple (11, 15, 27, 29). (End)

Examples

			120 is in the sequence because the set of divisors {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} contains the quadruples {3, 4, 5, 6} and {6, 8, 10, 12}. The first quadruple is primitive.
		

References

• Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Fred Richman, Sums of Three Cubes

Crossrefs

Cf. A023042, A027750, A096545, A328204, A330893.

Programs

Maple

with(numtheory):
for n from 3 to 2000 do :
   d:=divisors(n):n0:=nops(d):it:=0:
    for i from 1 to n0-3 do:
     for j from i+1 to n0-2 do :
      for k from j+1 to n0-1 do:
      for m from k+1 to n0 do:
       if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
        then
        it:=it+1:
        else
       fi:
      od:
     od:
    od:
    od:
    if it>0 then
    printf(`%d, `,n):
    else fi:
   od:

Mathematica

nq[n_] := If[ Mod[n, 6]>0, 0, Block[{t, u, v, c = 0, d = Divisors[n], m}, m = Length@ d; Do[ t = d[[i]]^3 + d[[j]]^3; Do[u = t + d[[h]]^3; If[u > n^3, Break[]]; If[ Mod[n^3, u] == 0 && IntegerQ[v = u^(1/3)] && Mod[n, v] == 0, c++], {h, j+1, m - 1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[ Range@ 1026, nq[#] > 0 &] (* program from Giovanni Resta adapted for the sequence. See A330893 *)

PARI

isok(n) = {my(d=divisors(n), m); if (#d > 3, for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (ispower(d[i]^3+d[j]^3+d[k]^3, 3, &m) && !(n%m), return (1));););););} \\ Michel Marcus, Nov 15 2020