A345181 Numbers that are the sum of five third powers in exactly seven ways.
4472, 4544, 4600, 4957, 5076, 5113, 5120, 5132, 5165, 5174, 5347, 5354, 5384, 5391, 5410, 5445, 5474, 5481, 5507, 5543, 5617, 5715, 5760, 5834, 5895, 5923, 5984, 5986, 6049, 6128, 6131, 6245, 6280, 6373, 6407, 6434, 6436, 6544, 6553, 6733, 6768, 6831, 6840
Offset: 1
Keywords
Examples
4472 is a term because 4472 = 1^3 + 4^3 + 4^3 + 4^3 + 15^3 = 2^3 + 2^3 + 9^3 + 11^3 + 11^3 = 2^3 + 3^3 + 4^3 + 5^3 + 15^3 = 2^3 + 3^3 + 7^3 + 11^3 + 12^3 = 3^3 + 3^3 + 6^3 + 10^3 + 13^3 = 3^3 + 4^3 + 5^3 + 8^3 + 14^3 = 5^3 + 5^3 + 7^3 + 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 == 7]) for x in range(len(rets)): print(rets[x])
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