A345779 - OEIS

%I #6 Jul 31 2021 22:39:27

%S 1072,1170,1235,1261,1268,1305,1392,1396,1411,1440,1441,1448,1450,

%T 1459,1489,1502,1504,1513,1538,1540,1547,1559,1564,1565,1566,1567,

%U 1576,1592,1593,1594,1600,1602,1606,1620,1621,1625,1626,1628,1629,1639,1658,1664,1667

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

%C Differs from A345525 at term 7 because 1385 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 8^3 + 8^3  = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 10^3  = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3  = 1^3 + 2^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3  = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 + 9^3  = 2^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3  = 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3  = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 8^3 + 8^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345779/b345779.txt">Table of n, a(n) for n = 1..345</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. A345525, A345769, A345778, A345780, A345789, A345829.

%K nonn

%O 1,1

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