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%I #41 Feb 20 2023 21:49:15
%S 766,810,827,829,865,883,981,1018,1025,1044,1070,1105,1108,1142,1145,
%T 1161,1168,1226,1233,1259,1289,1350,1368,1424,1431,1439,1441,1457,
%U 1487,1492,1494,1529,1531,1538,1550,1555,1568,1583,1587,1592,1593,1594,1609,1611,1613,1639,1648,1665,1672,1674,1688,1707,1711
%N Numbers that are the sum of five positive cubes in exactly three ways.
%C 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.
%C Comment from _D. S. McNeil_, May 13 2021: (Start)
%C 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.
%C 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.
%C 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)
%C 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).
%D R. C. Vaughan, The Hardy-Littlewood Method, Cambridge University Press, 1981.
%D 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.
%H David Consiglio, Jr. and Sean A. Irvine, <a href="/A343705/b343705.txt">Table of n, a(n) for n = 1..18984</a>
%H Jean-Marc Deshouillers, François Hennecart, and Bernard Landreau, <a href="https://doi.org/10.1090/S0025-5718-99-01116-3">7373170279850</a>, Math. Comp. 69 (2000), pp. 421-439. Appendix by I. Gusti Putu Purnaba.
%e 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.
%t Select[Range@2000,Length@Select[PowersRepresentations[#,5,3],FreeQ[#,0]&]==3&] (* _Giorgos Kalogeropoulos_, Apr 26 2021 *)
%o (Python)
%o from itertools import combinations_with_replacement as cwr
%o from collections import defaultdict
%o keep = defaultdict(lambda: 0)
%o power_terms = [x**3 for x in range(1,50)]#n
%o for pos in cwr(power_terms,5):#m
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k,v in keep.items() if v == 3])#s
%o for x in range(len(rets)):
%o print(rets[x])
%Y Equivalent sequences for 1 way: A048926; 2 ways: A048927; 1 or more ways: A003328; 3 or more ways: A343704.
%Y Cf. A003327.
%K nonn,easy
%O 1,1
%A _David Consiglio, Jr._, Apr 26 2021
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