OFFSET
1,1
COMMENTS
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)
REFERENCES
Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Fred Richman, Sums of Three Cubes
EXAMPLE
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.
MAPLE
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:
MATHEMATICA
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 *)
PROG
(PARI) 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
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jun 07 2020
STATUS
approved