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A345188
Numbers that are the sum of five third powers in exactly ten ways.
6
5860, 6588, 6651, 6859, 6947, 8056, 8289, 8569, 8758, 9045, 9099, 9227, 9414, 9612, 9829, 10009, 10277, 10485, 10522, 10529, 10800, 10963, 10970, 11008, 11061, 11089, 11241, 11385, 11458, 11656, 11719, 11782, 11817, 11845, 11934, 11990, 12016, 12060, 12088
OFFSET
1,1
COMMENTS
Differs from A345187 at term 8 because 8371 = 1^3 + 1^3 + 11^3 + 11^3 + 16^3 = 1^3 + 4^3 + 5^3 + 12^3 + 17^3 = 1^3 + 8^3 + 9^3 + 11^3 + 16^3 = 3^3 + 3^3 + 4^3 + 15^3 + 15^3 = 3^3 + 3^3 + 8^3 + 8^3 + 18^3 = 3^3 + 3^3 + 3^3 + 5^3 + 19^3 = 3^3 + 7^3 + 9^3 + 9^3 + 17^3 = 4^3 + 6^3 + 6^3 + 11^3 + 17^3 = 5^3 + 9^3 + 10^3 + 11^3 + 15^3 = 6^3 + 6^3 + 12^3 + 13^3 + 13^3 = 8^3 + 8^3 + 9^3 + 9^3 + 16^3.
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..10000
EXAMPLE
6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3 = 1^3 + 4^3 + 6^3 + 13^3 + 14^3 = 1^3 + 5^3 + 8^3 + 8^3 + 16^3 = 1^3 + 10^3 + 10^3 + 11^3 + 12^3 = 2^3 + 2^3 + 9^3 + 12^3 + 14^3 = 2^3 + 3^3 + 8^3 + 11^3 + 15^3 = 3^3 + 8^3 + 8^3 + 11^3 + 14^3 = 3^3 + 3^3 + 5^3 + 10^3 + 16^3 = 5^3 + 5^3 + 8^3 + 10^3 + 15^3 = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.
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 == 10])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved