# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a345149 Showing 1-1 of 1 %I A345149 #7 Jul 31 2021 23:29:03 %S A345149 6883,12411,13923,14112,14581,14896,14904,15561,15876,16317,16640, %T A345149 17208,17479,17992,18739,18865,19035,19080,19665,19712,19763,19880, %U A345149 20007,20384,20979,21231,21420,21707,22409,22617,23149,23940,24355,25515,25984,26208,26334 %N A345149 Numbers that are the sum of four third powers in exactly six ways. %C A345149 Differs from A345148 at term 3 because 13104 = 1^3 + 10^3 + 16^3 + 18^3 = 1^3 + 11^3 + 14^3 + 19^3 = 2^3 + 9^3 + 15^3 + 19^3 = 4^3 + 6^3 + 14^3 + 20^3 = 4^3 + 9^3 + 10^3 + 21^3 = 5^3 + 7^3 + 11^3 + 21^3 = 8^3 + 9^3 + 14^3 + 19^3. %H A345149 David Consiglio, Jr., Table of n, a(n) for n = 1..10000 %e A345149 6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3 = 2^3 + 4^3 + 14^3 + 14^3 = 3^3 + 7^3 + 7^3 + 17^3 = 3^3 + 10^3 + 13^3 + 13^3 = 4^3 + 10^3 + 10^3 + 15^3 = 7^3 + 8^3 + 8^3 + 16^3. %o A345149 (Python) %o A345149 from itertools import combinations_with_replacement as cwr %o A345149 from collections import defaultdict %o A345149 keep = defaultdict(lambda: 0) %o A345149 power_terms = [x**3 for x in range(1, 1000)] %o A345149 for pos in cwr(power_terms, 4): %o A345149 tot = sum(pos) %o A345149 keep[tot] += 1 %o A345149 rets = sorted([k for k, v in keep.items() if v == 6]) %o A345149 for x in range(len(rets)): %o A345149 print(rets[x]) %Y A345149 Cf. A025362, A343986, A344921, A345084, A345148, A345151, A345175. %K A345149 nonn %O A345149 1,1 %A A345149 _David Consiglio, Jr._, Jun 09 2021 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE