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A337036
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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.
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1
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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
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OFFSET
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1,1
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COMMENTS
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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
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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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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.
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MAPLE
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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:
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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