A330893 -id:A330893 - cp's OEIS frontend

A330894 Numbers of Pythagorean quadruples contained in the divisors of A330893(n).

1, 1, 2, 2, 2, 1, 3, 1, 3, 2, 4, 3, 2, 2, 1, 4, 1, 2, 3, 2, 7, 4, 2, 2, 8, 2, 1, 4, 4, 2, 3, 7, 3, 2, 5, 2, 2, 4, 6, 2, 5, 2, 11, 6, 4, 1, 4, 1, 6, 2, 4, 12, 2, 5, 1, 4, 6, 4, 2, 5, 6, 4, 1, 2, 3, 4, 17, 6, 2, 3, 6, 1, 5, 6, 1, 3, 4, 6, 6, 13, 1, 2, 4, 8, 4, 4

Offset: 1

Michel Lagneau, May 01 2020

nonn

			a(7) = 3 because A330893(7)=168, and the set of divisors of 168: {1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168} contains three Pythagorean quadruples {2, 3, 6, 7}, {4, 6, 12, 14} and {8, 12, 24, 28}.

Cf. A027750, A169580, A330893

with(numtheory):
for n from 3 to 1700 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]^2 + d[j]^2 + d[k]^2 = d[m]^2
        then
        it:=it+1:
        else
       fi:
      od:
     od:
    od:
    od:
    if it>0 then
    printf(`%d, `,it):
    else fi:
   od:

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[Array[nq, 1638], # > 0 &] (* Giovanni Resta, May 04 2020 *)

A331365 Least k whose set of divisors contains exactly n Pythagorean quadruples, or 0 if no such k exists.

Original entry on oeis.org

42, 84, 168, 252, 672, 756, 420, 504, 2592, 1872, 840, 1008, 1512, 2940, 1680, 2016, 1260, 4536, 3360, 3024, 9450, 4620, 5880, 6552, 9504, 6930, 3780, 8400, 23184, 25704, 2520, 6300, 31752, 8820, 19800, 11088, 10920, 13104, 15840, 19152, 19656, 16632, 38016

Offset: 1

Michel Lagneau, May 03 2020

nonn

a(n) == 0 (mod 6).

			a(3) = 168 because the set of the divisors {1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168} contains 3 Pythagorean quadruples {2, 3, 6, 7}, {4, 6, 12, 14} and {8, 12, 24, 28}.

Chai Wah Wu, Table of n, a(n) for n = 1..1026
Eric Weisstein's World of Mathematics, Pythagorean Quadruples.

Cf. A027750, A169580, A330893, A330894.

with(numtheory):
for n from 1 to 52 do :
ii:=0:
for q from 3 to 10^8 while(ii=0) do:
   d:=divisors(q):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 l from k+1 to n0 do:
       if d[i]^2 + d[j]^2 + d[k]^2 = d[l]^2
        then
        it:=it+1:
        else
       fi:
      od:
     od:
    od:
   od:
    if it = n
     then
     ii:=1: printf(`%d %d \n`,n,q):
     else
    fi:
od:
od:

upto = 38016; 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]]; w = ParallelTable[ {nq@ n, n}, {n, 6 Range[ upto / 6]}]; t=0 Range@ Max[First /@ w]; Do[{q, x} = e; If[q > 0 && t[[q]] == 0, t[[q]] = x], {e, w}]; AppendTo[t, 0]; TakeWhile[t, # > 0 &] (* Giovanni Resta, May 04 2020 *)

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

Original entry on oeis.org

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

Michel Lagneau, Jun 07 2020

nonn

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)

			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.

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

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

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

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:

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 *)

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

A335654 Numbers m such that the elements of all Pythagorean quadruples belonging to the set of divisors are exactly their first k divisors for some k.

Original entry on oeis.org

504, 1008, 1512, 1872, 2016, 3024, 3528, 3744, 4032, 4536, 5616, 6048, 6552, 7056, 7488, 8064, 9072, 9576, 10584, 11232, 12096, 13104, 13608, 14112, 14976, 16128, 16848, 17784, 18144, 19152, 19656, 21168, 21672, 22464, 23688, 24192, 24336, 24696, 26208, 27216

Offset: 1

Michel Lagneau, Jun 16 2020

nonn

Members m in A330893 for which there exists a number k < tau(m) such that the elements of all Pythagorean quadruples included in the set of the divisors of m are the first k divisors of m.
Conjecture 1: a(n) == 0 (mod 72).
Conjecture 2: if the numbers m such that the elements of all Pythagorean quadruples contained in the set of divisors of m are exactly the first k divisors of m, then k = tau(m) - 4 or tau(m) - 5.
The corresponding k of the sequence are given by the sequence {b(n)} = {20, 26, 28, 25, 32, 36, 32, 31, 38, 36, 35, 44, 44, 41,...} and the sequence {c(n)} = {tau(a(n)) - b(n)} = {4, 4, 4, 5, 4, 4, 4, 5, 4, 4, 5, 4, 4, 4, 5, 4, 4, 4, 4, 5, 4, 4, 4,...}. We observe that c(n) = 4 or 5 (see the table in the link). For n = 1, 2,...,400, the statistic observed is 301 occurrences for the number 4 (75.25 %) and 99 occurrences for the number 5 (24.75 %). It is probable that Pr(4) tends to .75 and Pr(5) tends to .25 when n tends into infinity, where Pr(x) is the probability of the occurrence x.
Assumes the elements in the quadruple are distinct. Otherwise 6, 12, 18, 24, ... are also terms. For instance the divisors of 6 are 1,2,3,6 and 1^2 + 2^2 + 2^2 = 3^2. - Chai Wah Wu, Nov 16 2020

			504 is in the sequence because the divisors are  {1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, 504} and the elements of the 8 Pythagorean quadruples belonging to the set of divisors of 504: (1, 4, 8, 9), (2, 3, 6, 7), (4, 6, 12, 14), (6, 9, 18, 21), (7, 28, 56, 63), (8, 12, 24, 28), (12, 18, 36, 42) and (24, 36, 72, 84) are the first 20 divisors of 504 with 20 = tau(504) - 4 = 24 - 4.

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

Cf. A000005, A330893, A330894, A331365.

with(numtheory):
for n from 6 by 6 to 20000 do :lst:={}:lst1:={}:
   d:=divisors(n):n0:=nops(d):
    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]^2 + d[j]^2 + d[k]^2 = d[m]^2
        then
        lst:=lst union {d[i]} union {d[j]} union {d[k]} union {d[m]}:
        else
       fi:
      od:
     od:
    od:
    od:
    n1:=nops(lst):
     for l from 1 to n1 do:
      lst1:= lst1 union {d[l]}:
     od:
    if lst=lst1 and lst<>{}
     then
     printf(`%d, `,n):
    else fi:
   od:

A360946 Number of Pythagorean quadruples with inradius n.

Original entry on oeis.org

1, 3, 6, 10, 9, 19, 16, 25, 29, 27, 27, 56, 31, 51, 49, 61, 42, 91, 52, 71, 89, 86, 63, 142, 64, 95, 116, 132, 83, 153, 90, 144, 149, 133, 108, 238, 108, 162, 169, 171, 122, 284, 130, 219, 200, 196, 145, 340, 174, 201, 231, 239, 164, 364, 176, 314, 278, 256, 190, 399, 195, 281, 360, 330

Offset: 1

Miguel-Ángel Pérez García-Ortega, Feb 26 2023

nonn

A Pythagorean quadruple is a quadruple (a,b,c,d) of positive integers such that a^2 + b^2 + c^2 = d^2 with a <= b <= c. Its inradius is (a+b+c-d)/2, which is a positive integer.

For every positive integer n, there is at least one Pythagorean quadruple with inradius n.

			For n=1 the a(1)=1 solution is (1,2,2,3).
For n=2 the a(2)=3 solutions are (1,4,8,9), (2,3,6,7) and (2,4,4,6).
For n=3 the a(3)=6 solutions are (1,6,18,19), (2,5,14,15), (2,6,9,11), (3,4,12,13), (3,6,6,9) and (4,4,7,9).

J. M. Blanco Casado, J. M. Sánchez Muñoz, and M. A. Pérez García-Ortega, El Libro de las Ternas Pitagóricas, Preprint 2023.

Miguel-Ángel Pérez García-Ortega, Pythagorean Quadruples (in Spanish).
Wikipedia, Pythagorean quadruple.

Cf. A096907, A096908, A096909, A096910, A097263, A097264, A097265, A097266, A097267

Cf. A169580, A225206, A225207, A230543, A330893, A330894, A331365, A335654, A359741

n=50;
div={};suc={};A={};
Do[A=Join[A,{Range[1,(1+1/Sqrt[3])q]}],{q,1,n}];
Do[suc=Join[suc,{Length[div]}];div={};For [i=1,i<=Length[Extract[A,q]],i++,div=Join[div,Intersection[Divisors[q^2+(Extract[Extract[A,q],i]-q)^2],Range[2(Extract[Extract[A,q],i]-q),Sqrt[q^2+(Extract[Extract[A,q],i]-q)^2]]]]],{q,1,n}];suc=Rest[Join[suc,{Length[div]}]];matriz={{"q"," ","cuaternas"}};For[j=1,j<=n,j++,matriz=Join[matriz,{{j," ",Extract[suc,j]}}]];MatrixForm[Transpose[matriz]]

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A330894 Numbers of Pythagorean quadruples contained in the divisors of A330893(n).

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A331365 Least k whose set of divisors contains exactly n Pythagorean quadruples, or 0 if no such k exists.

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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.

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PARI

A335654 Numbers m such that the elements of all Pythagorean quadruples belonging to the set of divisors are exactly their first k divisors for some k.

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Maple

A360946 Number of Pythagorean quadruples with inradius n.

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Author

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Mathematica