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A345807 Numbers that are the sum of ten cubes in exactly five ways. 6

%I #6 Jul 31 2021 22:27:24

%S 288,349,382,384,401,403,408,410,414,415,417,421,429,443,454,455,462,

%T 475,482,487,496,501,520,543,544,545,546,548,555,556,558,565,566,570,

%U 573,574,575,576,578,580,582,586,587,591,592,593,594,600,609,610,611,617

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

%C Differs from A345553 at term 14 because 436 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345807/b345807.txt">Table of n, a(n) for n = 1..86</a>

%e 349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

%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, 10):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v == 5])

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

%o print(rets[x])

%Y Cf. A345553, A345797, A345806, A345808, A345857.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021

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Last modified September 6 22:26 EDT 2024. Contains 375728 sequences. (Running on oeis4.)