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A337098
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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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1
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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
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OFFSET
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1,1
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COMMENTS
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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
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REFERENCES
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Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.
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LINKS
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(Python)
from itertools import combinations
from sympy import divisors
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
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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