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%I #68 Nov 15 2020 08:24:40
%S 60,72,120,144,180,216,240,288,300,360,420,432,480,504,540,576,600,
%T 648,660,720,780,792,840,864,900,936,960,1008,1020,1080,1140,1152,
%U 1200,1224,1260,1296,1320,1368,1380,1440,1500,1512,1560,1584,1620,1656,1680,1710
%N Numbers whose set of divisors contains a quadruple (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3.
%C The subsequence of numbers of the form 2^i*3^j is 72, 144, 216, 288, 432, 576, 648, 864, 1152, 1296, ...
%C 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).
%C The set of divisors of a(n) contains at least one primitive quadruple.
%C Examples:
%C The set of divisors of a(1) = 60 contains only one primitive quadruple: (3, 4, 5, 6).
%C The set of divisors of a(10) = 360 contains two primitive quadruples: (1, 6, 8, 9) and (3, 4, 5, 6).
%C From _Robert Israel_, Jul 06 2020: (Start)
%C Every multiple of a member of the sequence is in the sequence.
%C The first member of the sequence not divisible by 6 is a(68) = 2380, which has the quadruple (7, 14, 17, 20).
%C The first odd member of the sequence is a(1230) = 43065, which has the quadruple (11, 15, 27, 29). (End)
%D Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.
%H Robert Israel, <a href="/A328149/b328149.txt">Table of n, a(n) for n = 1..10000</a>
%H Fred Richman, <a href="http://math.fau.edu/Richman/cubes.htm">Sums of Three Cubes</a>
%e 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.
%p with(numtheory):
%p for n from 3 to 2000 do :
%p d:=divisors(n):n0:=nops(d):it:=0:
%p for i from 1 to n0-3 do:
%p for j from i+1 to n0-2 do :
%p for k from j+1 to n0-1 do:
%p for m from k+1 to n0 do:
%p if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
%p then
%p it:=it+1:
%p else
%p fi:
%p od:
%p od:
%p od:
%p od:
%p if it>0 then
%p printf(`%d, `,n):
%p else fi:
%p od:
%t 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 *)
%o (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
%Y Cf. A023042, A027750, A096545, A328204, A330893.
%K nonn
%O 1,1
%A _Michel Lagneau_, Jun 07 2020
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