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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.
1

%I #34 Aug 31 2021 02:44:04

%S 720,864,1440,1728,2160,2592,2880,3456,4320,5184,5760,6480,6912,7776,

%T 8640,10368,11520,12960,13824,15552,17280,19440,20736,23040,23328,

%U 25920,27648,31104,34560,38880,41472,46080,46656,51840,55296,58320,62208,69120,69984,77760

%N 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.

%C 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.

%C Conjecture 1: a(n) == 0 (mod 144).

%C 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.

%C 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.

%C 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

%C For 176 terms we get 87 5's and 89 6's. - _Michel Marcus_, Nov 17 2020

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

%H Chai Wah Wu, <a href="/A337036/b337036.txt">Table of n, a(n) for n = 1..176</a> (n = 1..71 from Michel Marcus)

%H Michel Lagneau, <a href="/A337036/a337036.pdf">Table</a>

%e 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.

%p with(numtheory):

%p for n from 6 by 6 to 200000 do :lst:={}:lst1:={}:it:=0:

%p d:=divisors(n):n0:=nops(d):

%p for i from 1 to n0-3 do:

%p for j from i+1 to n0-2 do:

%p for k from j+1 to n0-1 do:

%p for m from k+1 to n0 do:

%p if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3

%p then

%p it:=it+1:

%p lst:=lst union {d[i]} union {d[j]} union {d[k]} union {d[m]}:

%p else

%p fi:

%p od:

%p od:

%p od:

%p od:

%p n1:=nops(lst):

%p for l from 1 to n1 do:

%p lst1:= lst1 union {d[l]}:

%p od:

%p if lst=lst1 and lst<>{}

%p then

%p x:=tau(n)-n1:printf(`%d %d %d %d %d \n`,n,tau(n),n1,x,it):

%p else fi:

%p od:

%o (PARI) 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

%Y Cf. A335654, A328149, A328204.

%K nonn

%O 1,1

%A _Michel Lagneau_, Aug 12 2020