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.
%I A345809 #6 Jul 31 2021 22:27:30 %S A345809 440,473,499,506,525,532,534,567,571,584,588,597,599,604,606,627,637, %T A345809 639,640,656,660,663,669,670,673,680,682,689,691,693,696,701,702,704, %U A345809 707,717,718,719,726,729,735,738,743,744,750,755,761,762,763,770,783,784 %N A345809 Numbers that are the sum of ten cubes in exactly seven ways. %C A345809 Differs from A345555 at term 16 because 623 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3. %C A345809 Likely finite. %H A345809 Sean A. Irvine, <a href="/A345809/b345809.txt">Table of n, a(n) for n = 1..78</a> %e A345809 473 is a term because 473 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3. %o A345809 (Python) %o A345809 from itertools import combinations_with_replacement as cwr %o A345809 from collections import defaultdict %o A345809 keep = defaultdict(lambda: 0) %o A345809 power_terms = [x**3 for x in range(1, 1000)] %o A345809 for pos in cwr(power_terms, 10): %o A345809 tot = sum(pos) %o A345809 keep[tot] += 1 %o A345809 rets = sorted([k for k, v in keep.items() if v == 7]) %o A345809 for x in range(len(rets)): %o A345809 print(rets[x]) %Y A345809 Cf. A345555, A345799, A345808, A345810, A345859. %K A345809 nonn %O A345809 1,1 %A A345809 _David Consiglio, Jr._, Jun 26 2021