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 A345782 #6 Jul 31 2021 22:39:38 %S A345782 1704,1711,1800,1837,1863,1926,1938,1963,2008,2019,2045,2053,2059, %T A345782 2078,2113,2143,2161,2171,2176,2217,2223,2250,2260,2266,2276,2286, %U A345782 2295,2304,2313,2315,2331,2350,2354,2357,2374,2404,2412,2413,2442,2444,2446,2447,2511 %N A345782 Numbers that are the sum of seven cubes in exactly ten ways. %C A345782 Differs from A345506 at term 3 because 1774 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 11^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 9^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 11^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 9^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3. %C A345782 Likely finite. %H A345782 Sean A. Irvine, <a href="/A345782/b345782.txt">Table of n, a(n) for n = 1..328</a> %e A345782 1711 is a term because 1711 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3. %o A345782 (Python) %o A345782 from itertools import combinations_with_replacement as cwr %o A345782 from collections import defaultdict %o A345782 keep = defaultdict(lambda: 0) %o A345782 power_terms = [x**3 for x in range(1, 1000)] %o A345782 for pos in cwr(power_terms, 7): %o A345782 tot = sum(pos) %o A345782 keep[tot] += 1 %o A345782 rets = sorted([k for k, v in keep.items() if v == 10]) %o A345782 for x in range(len(rets)): %o A345782 print(rets[x]) %Y A345782 Cf. A345506, A345772, A345781, A345792, A345832. %K A345782 nonn %O A345782 1,1 %A A345782 _David Consiglio, Jr._, Jun 26 2021