A345812 - OEIS

%I #6 Jul 31 2021 22:27:44

%S 721,754,756,782,792,797,806,808,819,834,847,848,850,860,871,874,876,

%T 877,878,881,884,886,893,902,903,907,909,910,916,917,918,921,929,932,

%U 933,936,937,938,941,942,944,945,955,965,966,968,973,991,994,999,1001

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

%C 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 Likely finite.

%H Sean A. Irvine, <a href="/A345812/b345812.txt">Table of n, a(n) for n = 1..72</a>

%e 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 (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 == 10])

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

%o         print(rets[x])

%Y Cf. A345558, A345802, A345811, A345862.

%K nonn

%O 1,1

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