A346332 Numbers that are the sum of eight fifth powers in exactly seven ways.
4104553, 4915506, 6011150, 6027989, 6323394, 6563733, 6622231, 6776363, 6785394, 7982834, 8181481, 8288806, 8658144, 8710484, 8773477, 8932244, 8996669, 9252219, 9253706, 9311478, 9904983, 9976120, 10045233, 10053008, 10193511, 10359767, 10514944, 10541225
Offset: 1
Keywords
Examples
4104553 is a term because 4104553 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 5^5 + 7^5 + 21^5 = 3^5 + 3^5 + 4^5 + 6^5 + 8^5 + 14^5 + 16^5 + 19^5 = 3^5 + 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 18^5 + 18^5 = 3^5 + 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 18^5 + 18^5 = 1^5 + 1^5 + 4^5 + 7^5 + 10^5 + 16^5 + 16^5 + 18^5 = 7^5 + 11^5 + 11^5 + 13^5 + 14^5 + 15^5 + 16^5 + 16^5 = 6^5 + 12^5 + 12^5 + 13^5 + 13^5 + 15^5 + 16^5 + 16^5.
Links
- Sean A. Irvine, 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**5 for x in range(1, 1000)] for pos in cwr(power_terms, 8): 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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