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A345811
Numbers that are the sum of ten cubes in exactly nine ways.
6
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
1,1
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.
LINKS
EXAMPLE
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.
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, 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])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved