cp's OEIS Frontend

A330893 Numbers whose set of divisors contains a Pythagorean quadruple.

%I A330893 #38 Nov 16 2020 23:33:30
%S A330893 42,72,84,126,144,156,168,198,210,216,252,288,294,312,330,336,342,360,
%T A330893 378,396,420,432,462,468,504,546,570,576,588,594,624,630,648,660,672,
%U A330893 684,714,720,756,780,792,798,840,864,882,900,924,930,936,966,990,1008,1026
%N A330893 Numbers whose set of divisors contains a Pythagorean quadruple.
%C A330893 A Pythagorean quadruple (x, y, z, m) is a set of positive integers that satisfy x^2 + y^2 + z^2 = m^2.
%C A330893 The corresponding number of quadruples of the sequence is 1, 1, 2, 2, 2, 1, 3, 1, 3, 2, 4, 3, 2, 2, 1, 4, 1, 2, 3, 2, 7, 4, ... (see the sequence A330894).
%C A330893 It is interesting to note that each set of divisors of a(n) contains m primitive Pythagorean quadruples for some n, m = 1, 2,...
%C A330893 Examples:
%C A330893 - The set of divisors of a(1)= 42 contains only one primitive Pythagorean quadruple: (2, 3, 6, 7).
%C A330893 - The set of divisors of a(9) = 210 contains two primitive Pythagorean quadruples: (2, 3, 6, 7) and (2, 5, 14, 15).
%C A330893 - The set of divisors of a(21) = 420 contains three primitive Pythagorean quadruples: (2, 3, 6, 7), (2, 5, 14, 15) and (4, 5, 20, 21).
%C A330893 If k is in the sequence then so is m*k for m > 1.
%C A330893 Assumes the elements (x,y,z,m) in a quadruple are distinct divisors, as otherwise 6 would be in the sequence with 1^2+2^2+2^2=3^2. - _Chai Wah Wu_, Nov 16 2020
%H A330893 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PythagoreanQuadruple.html">Pythagorean Quadruples</a>.
%F A330893 a(n) == 0 (mod 6).
%e A330893 168 is in the sequence because the set of divisors  {1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168} contains the Pythagorean quadruples {2, 3, 6, 7}, {4, 6, 12, 14} and {8, 12, 24, 28}. The first quadruple is primitive.
%p A330893 with(numtheory):
%p A330893 for n from 3 to 1200 do :
%p A330893    d:=divisors(n):n0:=nops(d):it:=0:
%p A330893     for i from 1 to n0-3 do:
%p A330893      for j from i+1 to n0-2 do :
%p A330893       for k from j+1 to n0-1 do:
%p A330893       for m from k+1 to n0 do:
%p A330893        if d[i]^2 + d[j]^2 + d[k]^2 = d[m]^2
%p A330893         then
%p A330893         it:=it+1:
%p A330893         else
%p A330893        fi:
%p A330893       od:
%p A330893      od:
%p A330893     od:
%p A330893     od:
%p A330893     if it>0 then
%p A330893     printf(`%d, `,n):
%p A330893     else fi:
%p A330893    od:
%t A330893 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]]^2 + d[[j]]^2; Do[u = t + d[[h]]^2; If[u > n^2, Break[]]; If[ Mod[n^2, u] == 0 && IntegerQ[v = Sqrt@ u] && Mod[n, v] == 0, c++], {h, j+1, m - 1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[ Range@ 1026, nq[#] > 0 &] (* _Giovanni Resta_, May 04 2020 *)
%o A330893 (PARI) isok(n) = {my(d=divisors(n), x); for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (issquare(d[i]^2 + d[j]^2 + d[k]^2, &x) && !(n % x), return(1)););););} \\ _Michel Marcus_, Nov 16 2020
%Y A330893 Cf. A027750, A169580, A330894, A331365.
%K A330893 nonn
%O A330893 1,1
%A A330893 _Michel Lagneau_, May 01 2020