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A345186
Numbers that are the sum of five third powers in exactly nine ways.
7
6112, 6138, 6462, 6497, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8163, 8352, 8443, 8560, 8630, 8632, 8928, 8991, 9017, 9136, 9143, 9171, 9288, 9316, 9379, 9505, 9566, 9647, 9658, 9675, 9684, 9745, 9773
OFFSET
1,1
COMMENTS
Differs from A345185 at term 1 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3 = 1^3 + 2^3 + 3^3 + 11^3 + 15^3 = 1^3 + 3^3 + 8^3 + 11^3 + 14^3 = 1^3 + 5^3 + 5^3 + 10^3 + 15^3 = 1^3 + 9^3 + 10^3 + 10^3 + 12^3 = 2^3 + 3^3 + 8^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 12^3 + 14^3 = 2^3 + 8^3 + 8^3 + 12^3 + 12^3 = 3^3 + 8^3 + 8^3 + 9^3 + 14^3 = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..10000
EXAMPLE
6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3 = 1^3 + 3^3 + 7^3 + 12^3 + 14^3 = 1^3 + 6^3 + 6^3 + 7^3 + 16^3 = 2^3 + 2^3 + 9^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 11^3 + 15^3 = 2^3 + 8^3 + 9^3 + 9^3 + 14^3 = 3^3 + 3^3 + 3^3 + 4^3 + 17^3 = 3^3 + 5^3 + 8^3 + 11^3 + 14^3 = 8^3 + 8^3 + 8^3 + 11^3 + 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 == 9])
for x in range(len(rets)):
print(rets[x])
KEYWORD
nonn
AUTHOR
STATUS
approved