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A345153
Numbers that are the sum of four third powers in exactly eight ways.
7
27720, 30429, 31339, 31402, 33579, 34624, 34776, 36162, 40105, 42695, 44037, 44163, 44226, 44947, 45162, 45675, 46277, 46900, 47600, 49042, 50112, 50689, 51058, 51597, 51805, 52227, 52264, 52507, 53144, 54271, 54873, 55692, 55790, 56240, 58032, 58221, 58312
OFFSET
1,1
COMMENTS
Differs from A345152 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
30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3 = 1^3 + 16^3 + 17^3 + 26^3 = 2^3 + 12^3 + 21^3 + 25^3 = 3^3 + 3^3 + 14^3 + 29^3 = 4^3 + 17^3 + 21^3 + 23^3 = 5^3 + 11^3 + 15^3 + 28^3 = 6^3 + 6^3 + 22^3 + 25^3 = 7^3 + 14^3 + 18^3 + 26^3.
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 == 8])
for x in range(len(rets)):
print(rets[x])
KEYWORD
nonn
AUTHOR
STATUS
approved