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.

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

Author

Michel Lagneau, Aug 15 2020

Keywords

• nonn
• hard

Comments

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

Examples

			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.
		

References

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

Links

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

Crossrefs

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

Programs

Maple

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:

Mathematica

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 *)

Python

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

Extensions

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

More terms from David A. Corneth, Sep 26 2020