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

1,1

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.

Links

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

Example

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.

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, 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])

Crossrefs

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

Sequence in context: A029559 A222553 A345516 * A062916 A241554 A352949

Adjacent sequences: A345766 A345767 A345768 * A345770 A345771 A345772

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved