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A345175
Numbers that are the sum of five third powers in exactly six ways.
7
2430, 2979, 3214, 3249, 3312, 3492, 3520, 3737, 3753, 3788, 3816, 3842, 3942, 3968, 4121, 4185, 4213, 4267, 4355, 4411, 4418, 4446, 4453, 4456, 4465, 4482, 4509, 4563, 4626, 4663, 4670, 4723, 4753, 4896, 4905, 4924, 4938, 4941, 4950, 4960, 4976, 4987, 4994
OFFSET
1,1
COMMENTS
Differs from A345174 at term 20 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3 = 1^3 + 2^3 + 2^3 + 9^3 + 14^3 = 1^3 + 8^3 + 9^3 + 10^3 + 10^3 = 2^3 + 2^3 + 3^3 + 5^3 + 15^3 = 2^3 + 3^3 + 5^3 + 8^3 + 14^3 = 2^3 + 8^3 + 8^3 + 8^3 + 12^3 = 3^3 + 6^3 + 7^3 + 8^3 + 13^3 = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..10000
EXAMPLE
2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3 = 1^3 + 3^3 + 4^3 + 7^3 + 11^3 = 2^3 + 2^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 9^3 + 10^3 = 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 4^3 + 7^3 + 8^3 + 9^3.
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 == 6])
for x in range(len(rets)):
print(rets[x])
KEYWORD
nonn
AUTHOR
STATUS
approved