A328149 -id:A328149 - cp's OEIS frontend

A328204 Numbers of quadruples contained in the divisors of A328149(n).

1, 1, 2, 2, 2, 2, 3, 3, 2, 6, 2, 4, 4, 2, 3, 4, 4, 3, 2, 10, 2, 2, 4, 6, 4, 2, 5, 4, 2, 10, 2, 5, 6, 2, 4, 6, 4, 2, 2, 14, 3, 4, 4, 4, 4, 2, 6, 1, 8, 2, 11, 2, 4, 6, 4, 4, 6, 4, 2, 4, 17, 2, 2, 4, 6, 4, 4, 1, 8, 4, 2, 12, 2, 9, 6, 2, 6, 4, 4, 4, 2, 18, 3, 2, 6

Offset: 1

Michel Lagneau, Jun 07 2020

nonn

A quadruple (x, y, z, w) of A328149 is a set of positive integers that satisfy x^3 + y^3 + z^3 = w^3.

			a(7) = 3 because the set of divisors of A328149(7) = 240: {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240} contains the three quadruples {3, 4, 5, 6}, {6, 8, 10, 12} and {12, 16, 20, 24}. The first quadruple is primitive.

Michel Marcus, Table of n, a(n) for n = 1..10000

Cf. A027750, A096545, A328149, A330894.

with(numtheory):
for n from 3 to 3000 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, `,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]]^3 + d[[j]]^3; Do[u = t + d[[h]]^2; 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[Array[nq, 1638], # > 0 &] (* program from Giovanni Resta adapted for the sequence. See A330894 *)

isok(n) = {my(d=divisors(n), nb=0, 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), nb++););););); nb;}
lista(nn) = my(m); for (n=1, nn, if (m=isok(n), print1(m, ", "))); \\ Michel Marcus, Nov 15 2020

A337036 Numbers m such that the elements of all quadruples (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3 contained in the set of divisors of m are exactly the first k divisors of m for some k.

Original entry on oeis.org

720, 864, 1440, 1728, 2160, 2592, 2880, 3456, 4320, 5184, 5760, 6480, 6912, 7776, 8640, 10368, 11520, 12960, 13824, 15552, 17280, 19440, 20736, 23040, 23328, 25920, 27648, 31104, 34560, 38880, 41472, 46080, 46656, 51840, 55296, 58320, 62208, 69120, 69984, 77760

Offset: 1

Michel Lagneau, Aug 12 2020

nonn

Members m in A328149 for which there exists a number k < tau(m) such that the elements of all quadruples satisfying x^3 + y^3 + z^3 = w^3 included in the set of the divisors of m are the first k divisors of m.
Conjecture 1: a(n) == 0 (mod 144).
Conjecture 2: if the numbers m such that the elements of all quadruples (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3 contained in the set of divisors of m are exactly the first k divisors of m, then k = tau(m) - 5 or tau(m) - 6.
The corresponding k of the sequence are given by the sequence {b(n)} = {24, 19, 30, 23, 34, 25, 36, 27, 42, 30, 42, 44, 31, 31, 50, 35, 48, 54, 35, 37, 58, 54, ...} and the sequence {c(n)} = {tau(a(n)) - b(n)} = {6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 6, 5, 5, 6, 5, 6, 6, 5, 5, 6, 6, ...}. We observe that c(n) = 5 or 6 (see the table in the link). For n = 1 to 70, the statistic observed is 34 occurrences for the number 5 (48.57%) and 36 occurrences for the number 6 (51.42%). It is probable that Pr(5) tends to 0.5 and Pr(6) tends to 0.5 as n tends to infinity, where Pr(x) is the probability of the occurrence x.
It appears that assuming (x,y,z,w) to contain distinct elements or not does not matter to the sequence, unlike A335654. - Chai Wah Wu, Nov 16 2020
For 176 terms we get 87 5's and 89 6's. - Michel Marcus, Nov 17 2020

			2592 is in the sequence because the divisors are {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 72, 81, 96, 108, 144, 162, 216, 288, 324, 432, 648, 864, 1296, 2592} and the elements of the 9 quadruples (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3 and belonging to the set of divisors of 2592: (1, 6, 8, 9), (2, 12, 16, 18), (3, 18, 24, 27), (4, 24, 32, 36), (6, 36, 48, 54), (9, 54, 72, 81), (12, 72, 96, 108), (18, 108, 144, 162) and (36, 216, 288, 324) are the first 25 divisors of 2592 with 25 = tau(2592) - 5 = 30 - 5.

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

Chai Wah Wu, Table of n, a(n) for n = 1..176 (n = 1..71 from Michel Marcus)
Michel Lagneau, Table

Cf. A335654, A328149, A328204.

with(numtheory):
for n from 6 by 6 to 200000 do :lst:={}:lst1:={}:it:=0:
   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]^3 + d[j]^3 + d[k]^3 = d[m]^3
        then
        it:=it+1:
        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
     x:=tau(n)-n1:printf(`%d %d %d %d %d \n`,n,tau(n),n1,x,it):
    else fi:
   od:

isok(n) = {my(d=divisors(n), nb=0, s=[]); 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), s = concat(s, [d[i], d[j], d[k], m]));););); s = Set(s); if (#s, for (k=1, #s, if (s[k] != d[k], return (0));); return(1);););} \\ Michel Marcus, Nov 15 2020

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

Michel Lagneau, Aug 15 2020

Observation: a(n) == 0 (mod 12).

Listing primitive tuples (w, x, y, z) enables to compute for some m how many such tuples are in its divisors using the lcm of such tuples. - David A. Corneth, Sep 26 2020

			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.

David A. Corneth, Table of n, a(n) for n = 1..504
Fred Richman, Sums of Three Cubes

Cf. A027750, A095868, A095867, A096545, A096546, A328204, A328149, A331365.

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

a(13)-a(22) from Chai Wah Wu, Sep 25 2020

More terms from David A. Corneth, Sep 26 2020

cp's OEIS Frontend

A328204 Numbers of quadruples contained in the divisors of A328149(n).

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A337036 Numbers m such that the elements of all quadruples (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3 contained in the set of divisors of m are exactly the first k divisors of m for some k.

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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.

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