A345085 Numbers that are the sum of three third powers in exactly seven ways.

2016496, 4525632, 4783680, 5268024, 6366816, 7451352, 7457120, 8275392, 9063144, 9086104, 9931167, 10036872, 10266138, 10371024, 10973880, 12002472, 12452049, 12983517, 13639816, 13641480, 13818384, 13832729, 14090112, 15081984, 15212016, 15685704, 16131968

Offset: 1

Author

David Consiglio, Jr., Jun 07 2021

Keywords

• nonn

Comments

Differs from A345086 at term 2 because 2562624 = 7^3 + 35^3 + 135^3 = 7^3 + 63^3 + 131^3 = 11^3 + 99^3 + 115^3 = 16^3 + 45^3 + 134^3 = 29^3 + 102^3 + 112^3 = 35^3 + 59^3 + 131^3 = 50^3 + 84^3 + 121^3 = 68^3 + 71^3 + 122^3.

Examples

			2016496 is a term because 2016496 = 5^3 + 71^3 + 117^3 = 9^3 + 65^3 + 119^3 = 18^3 + 20^3 + 125^3 = 46^3 + 96^3 + 99^3 = 53^3 + 59^3 + 117^3 = 65^3 + 89^3 + 99^3 = 82^3 + 84^3 + 93^3.
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..100

Crossrefs

Cf. A025327, A344730, A345084, A345086, A345088, A345151.

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, 3):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 7])
for x in range(len(rets)):
    print(rets[x])