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Numbers that are the sum of five third powers in eight or more ways.
8

%I #6 Aug 05 2021 15:25:47

%S 4392,4915,5139,5256,5321,5624,5643,5678,5741,5769,5797,5832,5860,

%T 5914,6075,6112,6138,6202,6462,6497,6499,6560,6588,6616,6642,6651,

%U 6677,6833,6859,6884,6947,7001,7008,7038,7057,7064,7099,7111,7128,7155,7190,7218,7316

%N Numbers that are the sum of five third powers in eight or more ways.

%H David Consiglio, Jr., <a href="/A345183/b345183.txt">Table of n, a(n) for n = 1..10000</a>

%e 4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3 = 1^3 + 3^3 + 7^3 + 9^3 + 14^3 = 1^3 + 8^3 + 8^3 + 11^3 + 11^3 = 2^3 + 4^3 + 6^3 + 6^3 + 15^3 = 3^3 + 3^3 + 5^3 + 7^3 + 15^3 = 3^3 + 3^3 + 10^3 + 11^3 + 11^3 = 4^3 + 6^3 + 6^3 + 8^3 + 14^3 = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.

%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, 5):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v >= 8])

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

%o print(rets[x])

%Y Cf. A344801, A344944, A345152, A345180, A345184, A345185, A345517.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 10 2021