A345558 - OEIS

%I #6 Aug 05 2021 15:17:09

%S 721,754,756,771,782,792,797,806,808,819,832,834,845,847,848,850,860,

%T 862,867,869,871,874,876,877,878,881,884,886,888,893,895,897,902,903,

%U 904,906,907,909,910,911,912,914,916,917,918,919,921,923,925,929,930,932

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

%H Sean A. Irvine, <a href="/A345558/b345558.txt">Table of n, a(n) for n = 1..10000</a>

%e 754 is a term because 754 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3.

%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 >= 10])

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

%o         print(rets[x])

%Y Cf. A345549, A345557, A345603, A345812, A346808.

%K nonn

%O 1,1

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