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A345146 Numbers that are the sum of four third powers in nine or more ways.

%I A345146 #11 Aug 05 2021 15:27:22
%S A345146 21896,36225,42120,46683,46872,48321,48825,50806,50904,51408,51480,
%T A345146 51506,51688,52208,52416,53200,53865,54971,55575,56385,57113,58338,
%U A345146 58968,59059,60480,60515,60984,62244,62433,65303,66024,66276,66339,66430,67158,67536,67851
%N A345146 Numbers that are the sum of four third powers in nine or more ways.
%H A345146 David Consiglio, Jr., <a href="/A345146/b345146.txt">Table of n, a(n) for n = 1..10000</a>
%e A345146 42120 is a term because 42120 = 1^3 + 19^3 + 22^3 + 27^3  = 2^3 + 3^3 + 13^3 + 33^3  = 2^3 + 6^3 + 17^3 + 32^3  = 3^3 + 3^3 + 20^3 + 31^3  = 3^3 + 17^3 + 20^3 + 29^3  = 3^3 + 13^3 + 14^3 + 32^3  = 6^3 + 15^3 + 16^3 + 31^3  = 7^3 + 17^3 + 23^3 + 27^3  = 11^3 + 13^3 + 21^3 + 29^3.
%o A345146 (Python)
%o A345146 from itertools import combinations_with_replacement as cwr
%o A345146 from collections import defaultdict
%o A345146 keep = defaultdict(lambda: 0)
%o A345146 power_terms = [x**3 for x in range(1, 1000)]
%o A345146 for pos in cwr(power_terms, 4):
%o A345146     tot = sum(pos)
%o A345146     keep[tot] += 1
%o A345146 rets = sorted([k for k, v in keep.items() if v >= 9])
%o A345146 for x in range(len(rets)):
%o A345146     print(rets[x])
%Y A345146 Cf. A025374, A344926, A345119, A345152, A345154, A345155, A345185.
%K A345146 nonn
%O A345146 1,1
%A A345146 _David Consiglio, Jr._, Jun 09 2021