cp's OEIS Frontend

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

%I A345769 #6 Jul 31 2021 22:49:56
%S A345769 1710,1766,1773,1988,2051,2160,2196,2249,2251,2259,2314,2322,2349,
%T A345769 2375,2417,2424,2480,2492,2513,2520,2531,2539,2548,2564,2565,2574,
%U A345769 2611,2613,2639,2656,2702,2707,2762,2770,2773,2792,2798,2808,2818,2825,2826,2833,2844
%N A345769 Numbers that are the sum of six cubes in exactly seven ways.
%C A345769 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 A345769 Sean A. Irvine, <a href="/A345769/b345769.txt">Table of n, a(n) for n = 1..1359</a>
%e A345769 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 A345769 (Python)
%o A345769 from itertools import combinations_with_replacement as cwr
%o A345769 from collections import defaultdict
%o A345769 keep = defaultdict(lambda: 0)
%o A345769 power_terms = [x**3 for x in range(1, 1000)]
%o A345769 for pos in cwr(power_terms, 6):
%o A345769     tot = sum(pos)
%o A345769     keep[tot] += 1
%o A345769     rets = sorted([k for k, v in keep.items() if v == 7])
%o A345769     for x in range(len(rets)):
%o A345769         print(rets[x])
%Y A345769 Cf. A345181, A345516, A345768, A345770, A345779, A345819.
%K A345769 nonn
%O A345769 1,1
%A A345769 _David Consiglio, Jr._, Jun 26 2021