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 A345791 #6 Jul 31 2021 22:37:30 %S A345791 984,1080,1136,1171,1192,1197,1204,1223,1269,1273,1280,1306,1318,1325, %T A345791 1332,1333,1337,1344,1356,1360,1369,1370,1374,1377,1379,1404,1406, %U A345791 1415,1416,1422,1425,1430,1432,1438,1442,1444,1445,1456,1476,1481,1486,1488,1494 %N A345791 Numbers that are the sum of eight cubes in exactly nine ways. %C A345791 Differs from A345539 at term 5 because 1185 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 9^3 = 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 9^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 + 8^3 = 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3 + 7^3. %C A345791 Likely finite. %H A345791 Sean A. Irvine, <a href="/A345791/b345791.txt">Table of n, a(n) for n = 1..146</a> %e A345791 1080 is a term because 1080 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3. %o A345791 (Python) %o A345791 from itertools import combinations_with_replacement as cwr %o A345791 from collections import defaultdict %o A345791 keep = defaultdict(lambda: 0) %o A345791 power_terms = [x**3 for x in range(1, 1000)] %o A345791 for pos in cwr(power_terms, 8): %o A345791 tot = sum(pos) %o A345791 keep[tot] += 1 %o A345791 rets = sorted([k for k, v in keep.items() if v == 9]) %o A345791 for x in range(len(rets)): %o A345791 print(rets[x]) %Y A345791 Cf. A345539, A345781, A345790, A345792, A345801, A345841. %K A345791 nonn %O A345791 1,1 %A A345791 _David Consiglio, Jr._, Jun 26 2021