cp's OEIS Frontend

A345812 Numbers that are the sum of ten cubes in exactly ten ways.

%I A345812 #6 Jul 31 2021 22:27:44
%S A345812 721,754,756,782,792,797,806,808,819,834,847,848,850,860,871,874,876,
%T A345812 877,878,881,884,886,893,902,903,907,909,910,916,917,918,921,929,932,
%U A345812 933,936,937,938,941,942,944,945,955,965,966,968,973,991,994,999,1001
%N A345812 Numbers that are the sum of ten cubes in exactly ten ways.
%C A345812 Differs from A345558 at term 4 because 771 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 9^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 8^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3  = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 8^3  = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 7^3  = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 7^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3  = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3  = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3  = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3.
%C A345812 Likely finite.
%H A345812 Sean A. Irvine, <a href="/A345812/b345812.txt">Table of n, a(n) for n = 1..72</a>
%e A345812 754 is a term because 754 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3.
%o A345812 (Python)
%o A345812 from itertools import combinations_with_replacement as cwr
%o A345812 from collections import defaultdict
%o A345812 keep = defaultdict(lambda: 0)
%o A345812 power_terms = [x**3 for x in range(1, 1000)]
%o A345812 for pos in cwr(power_terms, 10):
%o A345812     tot = sum(pos)
%o A345812     keep[tot] += 1
%o A345812     rets = sorted([k for k, v in keep.items() if v == 10])
%o A345812     for x in range(len(rets)):
%o A345812         print(rets[x])
%Y A345812 Cf. A345558, A345802, A345811, A345862.
%K A345812 nonn
%O A345812 1,1
%A A345812 _David Consiglio, Jr._, Jun 26 2021