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A345122 Numbers that are the sum of three third powers in exactly ten ways.

%I A345122 #7 Jul 31 2021 23:41:49
%S A345122 34012224,58995000,71319312,72505152,92853216,94118760,95331816,
%T A345122 139755888,147545280,150506000,157464000,159874560,161023680,
%U A345122 164186352,171904032,182393856,184909824,188224128,189771336,191260224,199108125,201342240,202440384,217054720
%N A345122 Numbers that are the sum of three third powers in exactly ten ways.
%C A345122 Differs from A345121 at term 3 because 69190848 = 23^3 + 107^3 + 407^3  = 23^3 + 191^3 + 395^3  = 33^3 + 271^3 + 365^3  = 35^3 + 299^3 + 347^3  = 50^3 + 137^3 + 404^3  = 89^3 + 308^3 + 338^3  = 95^3 + 178^3 + 396^3  = 107^3 + 179^3 + 395^3  = 121^3 + 149^3 + 399^3  = 152^3 + 254^3 + 365^3  = 206^3 + 215^3 + 368^3.
%H A345122 Sean A. Irvine, <a href="/A345122/b345122.txt">Table of n, a(n) for n = 1..2238</a>
%e A345122 34012224 is a term because 34012224 = 35^3 + 215^3 + 287^3  = 38^3 + 152^3 + 311^3  = 40^3 + 113^3 + 318^3  = 44^3 + 245^3 + 266^3  = 71^3 + 113^3 + 317^3  = 99^3 + 191^3 + 295^3  = 101^3 + 226^3 + 276^3  = 117^3 + 185^3 + 295^3  = 161^3 + 215^3 + 269^3  = 172^3 + 213^3 + 266^3.
%o A345122 (Python)
%o A345122 from itertools import combinations_with_replacement as cwr
%o A345122 from collections import defaultdict
%o A345122 keep = defaultdict(lambda: 0)
%o A345122 power_terms = [x**3 for x in range(1, 1000)]
%o A345122 for pos in cwr(power_terms, 3):
%o A345122     tot = sum(pos)
%o A345122     keep[tot] += 1
%o A345122 rets = sorted([k for k, v in keep.items() if v == 10])
%o A345122 for x in range(len(rets)):
%o A345122     print(rets[x])
%Y A345122 Cf. A025330, A344861, A345120, A345121, A345156.
%K A345122 nonn
%O A345122 1,1
%A A345122 _David Consiglio, Jr._, Jun 08 2021
%E A345122 More terms from _Sean A. Irvine_, Jun 08 2021