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

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