A345778 - OEIS

%I #6 Jul 31 2021 22:39:24

%S 955,969,1046,1053,1079,1107,1117,1121,1158,1161,1177,1184,1196,1198,

%T 1216,1222,1242,1254,1272,1280,1287,1291,1294,1297,1298,1310,1324,

%U 1350,1351,1355,1366,1369,1376,1378,1388,1403,1404,1415,1417,1418,1422,1433,1437

%N Numbers that are the sum of seven cubes in exactly six ways.

%C Differs from A345524 at term 5 because 1072 = 1^3 + 1^3 + 1^3 + 5^3 + 6^3 + 6^3 + 8^3  = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 10^3  = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 + 9^3  = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 9^3  = 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 7^3 + 7^3  = 3^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3  = 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345778/b345778.txt">Table of n, a(n) for n = 1..344</a>

%e 969 is a term because 969 = 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.

%o (Python)

%o from itertools import combinations_with_replacement as cwr

%o from collections import defaultdict

%o keep = defaultdict(lambda: 0)

%o power_terms = [x**3 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 7):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v == 6])

%o     for x in range(len(rets)):

%o         print(rets[x])

%Y Cf. A345524, A345768, A345777, A345779, A345788, A345828.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021