A345153 - OEIS

%I #7 Jul 31 2021 23:29:10

%S 27720,30429,31339,31402,33579,34624,34776,36162,40105,42695,44037,

%T 44163,44226,44947,45162,45675,46277,46900,47600,49042,50112,50689,

%U 51058,51597,51805,52227,52264,52507,53144,54271,54873,55692,55790,56240,58032,58221,58312

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

%C Differs from A345152 at term 1 because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.

%H David Consiglio, Jr., <a href="/A345153/b345153.txt">Table of n, a(n) for n = 1..10000</a>

%e 30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3  = 1^3 + 16^3 + 17^3 + 26^3  = 2^3 + 12^3 + 21^3 + 25^3  = 3^3 + 3^3 + 14^3 + 29^3  = 4^3 + 17^3 + 21^3 + 23^3  = 5^3 + 11^3 + 15^3 + 28^3  = 6^3 + 6^3 + 22^3 + 25^3  = 7^3 + 14^3 + 18^3 + 26^3.

%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 == 8])

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

%o     print(rets[x])

%Y Cf. A025364, A344925, A345088, A345151, A345152, A345154, A345184.

%K nonn

%O 1,1

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