A345149 Numbers that are the sum of four third powers in exactly six ways.

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

Offset: 1

Author

David Consiglio, Jr., Jun 09 2021

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A025362, A343986, A344921, A345084, A345148, A345151, A345175.

Programs

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])