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 A345188 Numbers that are the sum of five third powers in exactly ten ways. 6

%I #10 Jul 31 2021 23:16:57

%S 5860,6588,6651,6859,6947,8056,8289,8569,8758,9045,9099,9227,9414,

%T 9612,9829,10009,10277,10485,10522,10529,10800,10963,10970,11008,

%U 11061,11089,11241,11385,11458,11656,11719,11782,11817,11845,11934,11990,12016,12060,12088

%N Numbers that are the sum of five third powers in exactly ten ways.

%C Differs from A345187 at term 8 because 8371 = 1^3 + 1^3 + 11^3 + 11^3 + 16^3 = 1^3 + 4^3 + 5^3 + 12^3 + 17^3 = 1^3 + 8^3 + 9^3 + 11^3 + 16^3 = 3^3 + 3^3 + 4^3 + 15^3 + 15^3 = 3^3 + 3^3 + 8^3 + 8^3 + 18^3 = 3^3 + 3^3 + 3^3 + 5^3 + 19^3 = 3^3 + 7^3 + 9^3 + 9^3 + 17^3 = 4^3 + 6^3 + 6^3 + 11^3 + 17^3 = 5^3 + 9^3 + 10^3 + 11^3 + 15^3 = 6^3 + 6^3 + 12^3 + 13^3 + 13^3 = 8^3 + 8^3 + 9^3 + 9^3 + 16^3.

%H David Consiglio, Jr., <a href="/A345188/b345188.txt">Table of n, a(n) for n = 1..10000</a>

%e 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.

%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 == 10])

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

%o print(rets[x])

%Y Cf. A294744, A341898, A345156, A345186, A345187, A345772.

%K nonn

%O 1,1

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

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Last modified August 13 14:06 EDT 2024. Contains 375142 sequences. (Running on oeis4.)