A345184 Numbers that are the sum of five third powers in exactly eight ways.
4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5914, 6075, 6202, 6499, 6560, 6616, 6642, 6677, 6833, 6884, 7008, 7111, 7128, 7155, 7218, 7344, 7395, 7641, 7696, 7729, 7785, 7813, 7820, 7849, 7883, 8037, 8100, 8243, 8282, 8308, 8315
Offset: 1
Keywords
Examples
4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3 = 1^3 + 3^3 + 7^3 + 9^3 + 14^3 = 1^3 + 8^3 + 8^3 + 11^3 + 11^3 = 2^3 + 4^3 + 6^3 + 6^3 + 15^3 = 3^3 + 3^3 + 5^3 + 7^3 + 15^3 = 3^3 + 3^3 + 10^3 + 11^3 + 11^3 = 4^3 + 6^3 + 6^3 + 8^3 + 14^3 = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.
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 == 8]) for x in range(len(rets)): print(rets[x])
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