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 A345800 #6 Jul 31 2021 22:32:58 %S A345800 744,770,805,818,840,842,844,847,866,868,877,880,883,887,894,908,909, %T A345800 910,911,913,915,916,920,940,945,946,948,950,952,954,955,957,961,964, %U A345800 965,972,976,983,987,990,1000,1001,1002,1006,1007,1013,1015,1025,1028,1032 %N A345800 Numbers that are the sum of nine cubes in exactly eight ways. %C A345800 Differs from A345547 at term 9 because 859 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 6^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3. %C A345800 Likely finite. %H A345800 Sean A. Irvine, <a href="/A345800/b345800.txt">Table of n, a(n) for n = 1..93</a> %e A345800 770 is a term because 770 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3. %o A345800 (Python) %o A345800 from itertools import combinations_with_replacement as cwr %o A345800 from collections import defaultdict %o A345800 keep = defaultdict(lambda: 0) %o A345800 power_terms = [x**3 for x in range(1, 1000)] %o A345800 for pos in cwr(power_terms, 9): %o A345800 tot = sum(pos) %o A345800 keep[tot] += 1 %o A345800 rets = sorted([k for k, v in keep.items() if v == 8]) %o A345800 for x in range(len(rets)): %o A345800 print(rets[x]) %Y A345800 Cf. A345547, A345790, A345799, A345801, A345810, A345850. %K A345800 nonn %O A345800 1,1 %A A345800 _David Consiglio, Jr._, Jun 26 2021