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Numbers that are the sum of 5 positive cubes in exactly 1 way.
8

%I #18 May 12 2021 21:03:43

%S 5,12,19,26,31,33,38,40,45,52,57,59,64,68,71,75,78,82,83,89,90,94,96,

%T 97,101,108,109,115,116,120,127,129,131,134,135,136,138,143,145,146,

%U 150,152,153,155,162,164,169,171,172,176,181,183,188,190,192,194,195

%N Numbers that are the sum of 5 positive cubes in exactly 1 way.

%C It appears that this sequence has 11062 terms, the last of which is 1685758. This means that all numbers greater than 1685758 can be written as the sum of five positive cubes in at least two ways. - _T. D. Noe_, Dec 13 2006

%H T. D. Noe, <a href="/A048926/b048926.txt">Table of n, a(n) for n=1..11062</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CubicNumber.html">Cubic Number.</a>

%t Select[ Range[200], Count[ PowersRepresentations[#, 5, 3], r_ /; FreeQ[r, 0]] == 1 &] (* _Jean-François Alcover_, Oct 23 2012 *)

%o (Python)

%o from collections import Counter

%o from itertools import combinations_with_replacement as combs_with_rep

%o def aupto(lim):

%o s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))

%o s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_with_rep(s, 5)))

%o s2counts = Counter(s2)

%o return sorted(k for k in s2counts if s2counts[k] == 1)

%o print(aupto(196)) # _Michael S. Branicky_, May 12 2021

%Y Cf. A003328, A048927.

%Y Cf. A057906 (numbers not the sum of five positive cubes)

%K nonn,fini

%O 1,1

%A _Eric W. Weisstein_