A345811 Numbers that are the sum of ten cubes in exactly nine ways.

632, 651, 658, 688, 695, 714, 736, 740, 745, 752, 773, 778, 780, 790, 795, 799, 801, 812, 813, 815, 816, 818, 821, 823, 825, 841, 843, 849, 851, 852, 853, 855, 856, 857, 858, 864, 866, 873, 880, 882, 883, 885, 890, 891, 892, 899, 905, 908, 913, 922, 924, 926

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345557 at term 7 because 721 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3.

Likely finite.

Examples

			651 is a term because 651 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..79

Crossrefs

Cf. A345557, A345801, A345810, A345812, A345861.

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 9])
    for x in range(len(rets)):
        print(rets[x])