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Numbers that are the sum of ten cubes in three or more ways.
7

%I #6 Aug 05 2021 15:16:41

%S 197,225,232,239,246,251,253,258,260,265,267,272,277,279,281,284,286,

%T 288,291,293,295,298,300,302,303,305,307,309,310,312,314,316,317,319,

%U 321,323,324,326,328,329,330,335,336,338,340,342,343,344,345,347,349,351

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

%H Sean A. Irvine, <a href="/A345551/b345551.txt">Table of n, a(n) for n = 1..10000</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. A345510, A345542, A345550, A345552, A345596, A345805.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021