cp's OEIS Frontend

A345553 Numbers that are the sum of ten cubes in five or more ways.

%I A345553 #6 Aug 05 2021 15:16:50
%S A345553 288,349,382,384,401,403,408,410,414,415,417,421,429,436,440,443,447,
%T A345553 454,455,462,466,473,475,477,480,482,487,492,496,499,501,503,506,508,
%U A345553 510,513,515,518,520,525,527,529,532,534,536,538,539,541,543,544,545,546
%N A345553 Numbers that are the sum of ten cubes in five or more ways.
%H A345553 Sean A. Irvine, <a href="/A345553/b345553.txt">Table of n, a(n) for n = 1..10000</a>
%e A345553 349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.
%o A345553 (Python)
%o A345553 from itertools import combinations_with_replacement as cwr
%o A345553 from collections import defaultdict
%o A345553 keep = defaultdict(lambda: 0)
%o A345553 power_terms = [x**3 for x in range(1, 1000)]
%o A345553 for pos in cwr(power_terms, 10):
%o A345553     tot = sum(pos)
%o A345553     keep[tot] += 1
%o A345553     rets = sorted([k for k, v in keep.items() if v >= 5])
%o A345553     for x in range(len(rets)):
%o A345553         print(rets[x])
%Y A345553 Cf. A345544, A345552, A345554, A345598, A345807, A346804.
%K A345553 nonn
%O A345553 1,1
%A A345553 _David Consiglio, Jr._, Jun 20 2021