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A345149 Numbers that are the sum of four third powers in exactly six ways. 7
6883, 12411, 13923, 14112, 14581, 14896, 14904, 15561, 15876, 16317, 16640, 17208, 17479, 17992, 18739, 18865, 19035, 19080, 19665, 19712, 19763, 19880, 20007, 20384, 20979, 21231, 21420, 21707, 22409, 22617, 23149, 23940, 24355, 25515, 25984, 26208, 26334 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Differs from A345148 at term 3 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.
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..10000
EXAMPLE
6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3 = 2^3 + 4^3 + 14^3 + 14^3 = 3^3 + 7^3 + 7^3 + 17^3 = 3^3 + 10^3 + 13^3 + 13^3 = 4^3 + 10^3 + 10^3 + 15^3 = 7^3 + 8^3 + 8^3 + 16^3.
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 == 6])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
Sequence in context: A064351 A366205 A345148 * A345603 A345862 A062678
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified April 22 01:34 EDT 2024. Contains 371887 sequences. (Running on oeis4.)