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A345769 Numbers that are the sum of six cubes in exactly seven ways. 7

%I #6 Jul 31 2021 22:49:56

%S 1710,1766,1773,1988,2051,2160,2196,2249,2251,2259,2314,2322,2349,

%T 2375,2417,2424,2480,2492,2513,2520,2531,2539,2548,2564,2565,2574,

%U 2611,2613,2639,2656,2702,2707,2762,2770,2773,2792,2798,2808,2818,2825,2826,2833,2844

%N Numbers that are the sum of six cubes in exactly seven ways.

%C Differs from A345516 at term 4 because 1981 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 12^3 = 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 12^3 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 10^3 = 1^3 + 1^3 + 6^3 + 6^3 + 6^3 + 11^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 10^3 = 3^3 + 3^3 + 7^3 + 7^3 + 8^3 + 9^3 = 3^3 + 4^3 + 6^3 + 6^3 + 9^3 + 9^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 10^3.

%H Sean A. Irvine, <a href="/A345769/b345769.txt">Table of n, a(n) for n = 1..1359</a>

%e 1766 is a term because 1766 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 11^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3.

%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, 6):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v == 7])

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

%o print(rets[x])

%Y Cf. A345181, A345516, A345768, A345770, A345779, A345819.

%K nonn

%O 1,1

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

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