OFFSET
1,1
COMMENTS
This sequence differs from A343704 at term 20 because 1252 = 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 10^3 = 3^3 + 3^3 + 7^3 + 7^3 + 8^3 = 3^3 + 4^3 + 6^3 + 6^3 + 9^3. Thus this term is in A343704 but not in this sequence.
Comment from D. S. McNeil, May 13 2021: (Start)
If we weaken positive cubes to nonnegative cubes, Deshouillers, Hennecart, and Landreau (2000) give numerical and heuristic evidence that all numbers past 7373170279850 are representable as the sum of 4 nonnegative cubes.
So if they are right, then eventually we can just take some N and represent each of (N-1^3, N-2^3, N-3^3, N-4^3) as the sum of four cubes and then take 1^3, 2^3, 3^3, or 4^3 as our fifth cube, giving at least four 5-cube representations for N.
So it is very likely that the set of numbers representable by the sum of 5 positive cubes in exactly three ways is finite. (End)
It is conjectured that the number of ways of writing N as a sum of 5 positive cubes grows like C(N)*N^(2/3), where C(N) depends on N but is bounded away from zero by an absolute constant (Vaughan, 1981; Vaughan and Wooley, 2002). So the number will exceed 3 as soon as N is large enough, and so it is very likely that this sequence is finite. However, at present this is an open question. - N. J. A. Sloane, May 15 2021 (based on correspondence with Robert Vaughan and Trevor Wooley).
REFERENCES
R. C. Vaughan, The Hardy-Littlewood Method, Cambridge University Press, 1981.
R. C. Vaughan, Trevor Wooley (2002), Waring's Problem: A Survey. In Michael A. Bennet, Bruce C. Berndt, Nigel Boston, Harold G. Diamond, Adolf J. Hildebrand, Walter Philipp (eds.). Number Theory for the Millennium. III. Natick, MA: A. K. Peters, pp. 301-340.
LINKS
David Consiglio, Jr. and Sean A. Irvine, Table of n, a(n) for n = 1..18984
Jean-Marc Deshouillers, François Hennecart, and Bernard Landreau, 7373170279850, Math. Comp. 69 (2000), pp. 421-439. Appendix by I. Gusti Putu Purnaba.
EXAMPLE
827 is a term of this sequence because 827 = 1^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 5^3 + 7^3 + 7^3 = 2^3 + 3^3 + 4^3 + 6^3 + 8^3.
MATHEMATICA
Select[Range@2000, Length@Select[PowersRepresentations[#, 5, 3], FreeQ[#, 0]&]==3&] (* Giorgos Kalogeropoulos, Apr 26 2021 *)
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 50)]#n
for pos in cwr(power_terms, 5):#m
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 3])#s
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David Consiglio, Jr., Apr 26 2021
STATUS
approved