A345188 Numbers that are the sum of five third powers in exactly ten ways.
5860, 6588, 6651, 6859, 6947, 8056, 8289, 8569, 8758, 9045, 9099, 9227, 9414, 9612, 9829, 10009, 10277, 10485, 10522, 10529, 10800, 10963, 10970, 11008, 11061, 11089, 11241, 11385, 11458, 11656, 11719, 11782, 11817, 11845, 11934, 11990, 12016, 12060, 12088
Offset: 1
Keywords
Examples
6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3 = 1^3 + 4^3 + 6^3 + 13^3 + 14^3 = 1^3 + 5^3 + 8^3 + 8^3 + 16^3 = 1^3 + 10^3 + 10^3 + 11^3 + 12^3 = 2^3 + 2^3 + 9^3 + 12^3 + 14^3 = 2^3 + 3^3 + 8^3 + 11^3 + 15^3 = 3^3 + 8^3 + 8^3 + 11^3 + 14^3 = 3^3 + 3^3 + 5^3 + 10^3 + 16^3 = 5^3 + 5^3 + 8^3 + 10^3 + 15^3 = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.
Links
- David Consiglio, Jr., Table of n, a(n) for n = 1..10000
Programs
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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, 5): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v == 10]) for x in range(len(rets)): print(rets[x])
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