A345149 - OEIS

%I #7 Jul 31 2021 23:29:03

%S 6883,12411,13923,14112,14581,14896,14904,15561,15876,16317,16640,

%T 17208,17479,17992,18739,18865,19035,19080,19665,19712,19763,19880,

%U 20007,20384,20979,21231,21420,21707,22409,22617,23149,23940,24355,25515,25984,26208,26334

%N Numbers that are the sum of four third powers in exactly six ways.

%C Differs from A345148 at term 3 because 13104 = 1^3 + 10^3 + 16^3 + 18^3  = 1^3 + 11^3 + 14^3 + 19^3  = 2^3 + 9^3 + 15^3 + 19^3  = 4^3 + 6^3 + 14^3 + 20^3  = 4^3 + 9^3 + 10^3 + 21^3  = 5^3 + 7^3 + 11^3 + 21^3  = 8^3 + 9^3 + 14^3 + 19^3.

%H David Consiglio, Jr., <a href="/A345149/b345149.txt">Table of n, a(n) for n = 1..10000</a>

%e 6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3  = 2^3 + 4^3 + 14^3 + 14^3  = 3^3 + 7^3 + 7^3 + 17^3  = 3^3 + 10^3 + 13^3 + 13^3  = 4^3 + 10^3 + 10^3 + 15^3  = 7^3 + 8^3 + 8^3 + 16^3.

%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, 4):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v == 6])

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

%o     print(rets[x])

%Y Cf. A025362, A343986, A344921, A345084, A345148, A345151, A345175.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 09 2021