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 A345187 Numbers that are the sum of five third powers in ten or more ways. 7
 5860, 6588, 6651, 6859, 6947, 8056, 8289, 8371, 8506, 8569, 8758, 9045, 9080, 9099, 9108, 9227, 9414, 9612, 9801, 9829, 9864, 10009, 10018, 10044, 10277, 10466, 10485, 10522, 10529, 10800, 10963, 10970, 10979, 11008, 11017, 11061, 11089, 11152, 11241, 11385 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS David Consiglio, Jr., Table of n, a(n) for n = 1..10000 EXAMPLE 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. 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, 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]) CROSSREFS Cf. A341897, A344803, A345155, A345185, A345188, A345519. Sequence in context: A219008 A020416 A345185 * A345188 A076808 A272324 Adjacent sequences: A345184 A345185 A345186 * A345188 A345189 A345190 KEYWORD nonn AUTHOR David Consiglio, Jr., Jun 10 2021 STATUS approved

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Last modified August 14 06:25 EDT 2024. Contains 375146 sequences. (Running on oeis4.)