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A345183
Numbers that are the sum of five third powers in eight or more ways.
8
4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5860, 5914, 6075, 6112, 6138, 6202, 6462, 6497, 6499, 6560, 6588, 6616, 6642, 6651, 6677, 6833, 6859, 6884, 6947, 7001, 7008, 7038, 7057, 7064, 7099, 7111, 7128, 7155, 7190, 7218, 7316
OFFSET
1,1
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..10000
EXAMPLE
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.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 5):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 8])
for x in range(len(rets)):
print(rets[x])
KEYWORD
nonn
AUTHOR
STATUS
approved