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A345151 Numbers that are the sum of four third powers in exactly seven ways. 7
13104, 18928, 19376, 20755, 21203, 22743, 24544, 24570, 24787, 25172, 25928, 27755, 27846, 28917, 29582, 31031, 31248, 31528, 32858, 34056, 34713, 35289, 35317, 35441, 35497, 35712, 36190, 36288, 36610, 36890, 36946, 38080, 39221, 39440, 39464, 39851, 39942 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Differs from A345150 at term 6 because 21896 = 1^3 + 11^3 + 19^3 + 22^3 = 2^3 + 2^3 + 12^3 + 26^3 = 2^3 + 3^3 + 19^3 + 23^3 = 2^3 + 5^3 + 15^3 + 25^3 = 3^3 + 10^3 + 16^3 + 24^3 = 3^3 + 17^3 + 19^3 + 19^3 = 4^3 + 6^3 + 20^3 + 22^3 = 5^3 + 8^3 + 14^3 + 25^3 = 7^3 + 11^3 + 17^3 + 23^3 = 8^3 + 9^3 + 19^3 + 22^3.
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..10000
EXAMPLE
13104 is a term because 13104 = 1^3 + 10^3 + 16^3 + 18^3 = 1^3 + 11^3 + 14^3 + 19^3 = 2^3 + 9^3 + 15^3 + 19^3 = 4^3 + 6^3 + 14^3 + 20^3 = 4^3 + 9^3 + 10^3 + 21^3 = 5^3 + 7^3 + 11^3 + 21^3 = 8^3 + 9^3 + 14^3 + 19^3.
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, 4):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 7])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
Sequence in context: A235335 A184768 A345150 * A252548 A015341 A233689
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)