A345805 - OEIS

%I #6 Jul 31 2021 22:27:17

%S 197,239,246,253,260,267,277,279,281,293,295,298,300,302,303,305,309,

%T 312,316,317,319,324,326,329,330,335,336,338,340,343,344,345,351,352,

%U 354,358,361,362,364,365,368,370,379,386,387,388,392,394,395,396,402,406

%N Numbers that are the sum of ten cubes in exactly three ways.

%C Differs from A345551 at term 2 because 225 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3  = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3  = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345805/b345805.txt">Table of n, a(n) for n = 1..93</a>

%e 225 is a term because 225 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

%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, 1000)]

%o for pos in cwr(power_terms, 10):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v == 3])

%o     for x in range(len(rets)):

%o         print(rets[x])

%Y Cf. A345551, A345795, A345804, A345806, A345855.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021