A330893
Numbers whose set of divisors contains a Pythagorean quadruple.
Original entry on oeis.org
42, 72, 84, 126, 144, 156, 168, 198, 210, 216, 252, 288, 294, 312, 330, 336, 342, 360, 378, 396, 420, 432, 462, 468, 504, 546, 570, 576, 588, 594, 624, 630, 648, 660, 672, 684, 714, 720, 756, 780, 792, 798, 840, 864, 882, 900, 924, 930, 936, 966, 990, 1008, 1026
Offset: 1
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.
-
with(numtheory):
for n from 3 to 1200 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, `,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]]^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 *)
-
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
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
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.
-
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:
A337098
Least k whose set of divisors contains exactly n quadruples (x, y, z, w) such that x^3 + y^3 + z^3 = w^3, or 0 if no such k exists.
Original entry on oeis.org
60, 120, 240, 432, 960, 360, 3840, 1728, 2592, 720, 1800, 2520, 161700, 1440, 6840, 9000, 2160, 2880, 168300, 5040, 41472, 5760, 1520820, 4320, 7200, 11520, 119700, 10080, 682080, 10800, 8640, 14400, 27360, 12960, 373248, 20160, 61560, 17280, 28800, 55440, 171000, 21600
Offset: 1
a(3) = 240 because the set of the divisors {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240} contains 3 quadruples {3, 4, 5, 6}, {6, 8, 10, 12} and {12, 16, 20, 24}. 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.
-
with(numtheory):divisors(240);
for n from 1 to 52 do :
ii:=0:
for q from 6 by 6 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 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 = n
then
ii:=1: printf (`%d %d \n`,n,q):
else
fi:
od:
od:
-
With[{s = Array[Count[Subsets[Divisors[#], {4}]^3, ?(#1 + #2 + #3 == #4 & @@ # &)] &, 10^4]}, Rest@ Values[#][[1 ;; 1 + LengthWhile[Differences@ Keys@ #, # == 1 &] ]] &@ KeySort@ PositionIndex[s][[All, 1]]] (* _Michael De Vlieger, Sep 18 2020 *)
-
from itertools import combinations
from sympy import divisors
def A337098(n):
k = 1
while True:
if n == sum(1 for x in combinations((d**3 for d in divisors(k)),4) if sum(x[:-1]) == x[-1]):
return k
k += 1 # Chai Wah Wu, Sep 25 2020
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
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.
-
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]]
Showing 1-4 of 4 results.
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