A345525 - OEIS

%I #7 Aug 05 2021 15:22:32

%S 1072,1170,1235,1261,1268,1305,1385,1392,1396,1411,1440,1441,1448,

%T 1450,1459,1489,1496,1502,1504,1513,1515,1538,1540,1547,1552,1557,

%U 1559,1564,1565,1566,1567,1576,1585,1587,1592,1593,1594,1600,1602,1603,1606,1613,1620

%N Numbers that are the sum of seven cubes in seven or more ways.

%H Sean A. Irvine, <a href="/A345525/b345525.txt">Table of n, a(n) for n = 1..10000</a>

%e 1170 is a term because 1170 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3 = 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 7^3.

%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 >= 7])

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

%o         print(rets[x])

%Y Cf. A345484, A345516, A345524, A345526, A345537, A345573, A345779.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021