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 A345185 Numbers that are the sum of five third powers in nine or more ways. 8

%I #6 Aug 05 2021 15:25:51

%S 5860,6112,6138,6462,6497,6588,6651,6859,6947,7001,7038,7057,7064,

%T 7099,7190,7316,7328,7372,7433,7561,7587,7703,7759,7841,7902,8056,

%U 8163,8289,8352,8371,8443,8506,8560,8569,8630,8632,8758,8928,8991,9017,9045,9080,9099

%N Numbers that are the sum of five third powers in nine or more ways.

%H David Consiglio, Jr., <a href="/A345185/b345185.txt">Table of n, a(n) for n = 1..10000</a>

%e 6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3 = 1^3 + 3^3 + 7^3 + 12^3 + 14^3 = 1^3 + 6^3 + 6^3 + 7^3 + 16^3 = 2^3 + 2^3 + 9^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 11^3 + 15^3 = 2^3 + 8^3 + 9^3 + 9^3 + 14^3 = 3^3 + 3^3 + 3^3 + 4^3 + 17^3 = 3^3 + 5^3 + 8^3 + 11^3 + 14^3 = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.

%o (Python)

%o from itertools import combinations_with_replacement as cwr

%o from collections import defaultdict

%o keep = defaultdict(lambda: 0)

%o power_terms = [x**3 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 5):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v >= 9])

%o for x in range(len(rets)):

%o print(rets[x])

%Y Cf. A341891, A344802, A345146, A345183, A345186, A345187, A345518.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 10 2021

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Last modified September 17 17:26 EDT 2024. Contains 375990 sequences. (Running on oeis4.)