A345183 - OEIS

A345183

Numbers that are the sum of five third powers in eight or more ways.

4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5860, 5914, 6075, 6112, 6138, 6202, 6462, 6497, 6499, 6560, 6588, 6616, 6642, 6651, 6677, 6833, 6859, 6884, 6947, 7001, 7008, 7038, 7057, 7064, 7099, 7111, 7128, 7155, 7190, 7218, 7316

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

LINKS

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

EXAMPLE

4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3 = 1^3 + 3^3 + 7^3 + 9^3 + 14^3 = 1^3 + 8^3 + 8^3 + 11^3 + 11^3 = 2^3 + 4^3 + 6^3 + 6^3 + 15^3 = 3^3 + 3^3 + 5^3 + 7^3 + 15^3 = 3^3 + 3^3 + 10^3 + 11^3 + 11^3 = 4^3 + 6^3 + 6^3 + 8^3 + 14^3 = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.

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 >= 8])

for x in range(len(rets)):

print(rets[x])