cp's OEIS Frontend

A345539 Numbers that are the sum of eight cubes in nine or more ways.

%I A345539 #6 Aug 05 2021 15:21:05
%S A345539 984,1080,1136,1171,1185,1192,1197,1204,1223,1243,1262,1269,1273,1280,
%T A345539 1288,1295,1299,1306,1318,1325,1332,1333,1337,1344,1356,1360,1369,
%U A345539 1370,1374,1377,1379,1386,1393,1397,1400,1404,1406,1412,1415,1416,1419,1422,1423
%N A345539 Numbers that are the sum of eight cubes in nine or more ways.
%H A345539 Sean A. Irvine, <a href="/A345539/b345539.txt">Table of n, a(n) for n = 1..10000</a>
%e A345539 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 A345539 (Python)
%o A345539 from itertools import combinations_with_replacement as cwr
%o A345539 from collections import defaultdict
%o A345539 keep = defaultdict(lambda: 0)
%o A345539 power_terms = [x**3 for x in range(1, 1000)]
%o A345539 for pos in cwr(power_terms, 8):
%o A345539     tot = sum(pos)
%o A345539     keep[tot] += 1
%o A345539     rets = sorted([k for k, v in keep.items() if v >= 9])
%o A345539     for x in range(len(rets)):
%o A345539         print(rets[x])
%Y A345539 Cf. A345496, A345527, A345538, A345540, A345548, A345584, A345791.
%K A345539 nonn
%O A345539 1,1
%A A345539 _David Consiglio, Jr._, Jun 20 2021