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editing
approved
editing
approved
allocated for David Consiglio, JrNumbers that are the sum of five third powers in eight or more ways.
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
1,1
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.
(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])
allocated
nonn
David Consiglio, Jr., Jun 10 2021
approved
editing
allocated for David Consiglio, Jr.
allocated
approved