A345769 Numbers that are the sum of six cubes in exactly seven ways.

1710, 1766, 1773, 1988, 2051, 2160, 2196, 2249, 2251, 2259, 2314, 2322, 2349, 2375, 2417, 2424, 2480, 2492, 2513, 2520, 2531, 2539, 2548, 2564, 2565, 2574, 2611, 2613, 2639, 2656, 2702, 2707, 2762, 2770, 2773, 2792, 2798, 2808, 2818, 2825, 2826, 2833, 2844

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345516 at term 4 because 1981 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 12^3 = 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 12^3 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 10^3 = 1^3 + 1^3 + 6^3 + 6^3 + 6^3 + 11^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 10^3 = 3^3 + 3^3 + 7^3 + 7^3 + 8^3 + 9^3 = 3^3 + 4^3 + 6^3 + 6^3 + 9^3 + 9^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 10^3.

Examples

			1766 is a term because 1766 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 11^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..1359

Crossrefs

Cf. A345181, A345516, A345768, A345770, A345779, A345819.

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, 6):
    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])