cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A345808 Numbers that are the sum of ten cubes in exactly six ways.

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%I A345808 #6 Jul 31 2021 22:27:27
%S A345808 436,447,466,477,480,492,503,508,510,513,515,518,527,529,536,538,539,
%T A345808 541,551,553,560,562,564,569,577,581,590,595,601,602,603,607,608,613,
%U A345808 614,616,618,621,628,634,636,642,643,645,647,649,654,655,659,666,678,679
%N A345808 Numbers that are the sum of ten cubes in exactly six ways.
%C A345808 Differs from A345554 at term 2 because 440 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3  = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 6^3  = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3  = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3  = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3  = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3.
%C A345808 Likely finite.
%H A345808 Sean A. Irvine, <a href="/A345808/b345808.txt">Table of n, a(n) for n = 1..73</a>
%e A345808 440 is a term because 440 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.
%o A345808 (Python)
%o A345808 from itertools import combinations_with_replacement as cwr
%o A345808 from collections import defaultdict
%o A345808 keep = defaultdict(lambda: 0)
%o A345808 power_terms = [x**3 for x in range(1, 1000)]
%o A345808 for pos in cwr(power_terms, 10):
%o A345808     tot = sum(pos)
%o A345808     keep[tot] += 1
%o A345808     rets = sorted([k for k, v in keep.items() if v == 6])
%o A345808     for x in range(len(rets)):
%o A345808         print(rets[x])
%Y A345808 Cf. A345554, A345798, A345807, A345809, A345858.
%K A345808 nonn
%O A345808 1,1
%A A345808 _David Consiglio, Jr._, Jun 26 2021