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A345154
Numbers that are the sum of four third powers in exactly nine ways.
7
42120, 46683, 50806, 50904, 51408, 51480, 51688, 52208, 53865, 54971, 56385, 57113, 60515, 60984, 62433, 65303, 66276, 66339, 66430, 67158, 69048, 69832, 69930, 71162, 72072, 72520, 72576, 72800, 73017, 77714, 77903, 79345, 79667, 79849, 80066, 80073, 81207
OFFSET
1,1
COMMENTS
Differs from A345146 at term 1 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
42120 is a term because 42120 = 1^3 + 19^3 + 22^3 + 27^3 = 2^3 + 3^3 + 13^3 + 33^3 = 2^3 + 6^3 + 17^3 + 32^3 = 3^3 + 3^3 + 20^3 + 31^3 = 3^3 + 17^3 + 20^3 + 29^3 = 3^3 + 13^3 + 14^3 + 32^3 = 6^3 + 15^3 + 16^3 + 31^3 = 7^3 + 17^3 + 23^3 + 27^3 = 11^3 + 13^3 + 21^3 + 29^3.
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 == 9])
for x in range(len(rets)):
print(rets[x])
KEYWORD
nonn
AUTHOR
STATUS
approved