A345806 - OEIS

%I #6 Jul 31 2021 22:27:20

%S 225,232,251,258,265,272,284,286,291,307,310,314,321,323,328,342,347,

%T 356,363,366,373,375,377,380,389,391,398,399,405,412,419,422,424,427,

%U 434,438,441,445,450,451,458,459,461,464,469,471,476,478,481,484,488,489

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

%C Differs from A345552 at term 9 because 288 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3  = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3  = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345806/b345806.txt">Table of n, a(n) for n = 1..93</a>

%e 232 is a term because 232 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 2^3 + 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 == 4])

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

%o         print(rets[x])

%Y Cf. A345552, A345796, A345805, A345807, A345856.

%K nonn

%O 1,1

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