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Numbers that are the sum of four third powers in ten or more ways.
7

%I #6 Aug 05 2021 15:27:37

%S 21896,36225,46872,48321,48825,51506,52416,53200,55575,58338,58968,

%T 59059,60480,62244,66024,67536,67851,70434,70525,71155,72819,73808,

%U 76384,76923,77896,78624,78912,81081,81991,85995,87507,88641,90181,90783,91448,91728,92008

%N Numbers that are the sum of four third powers in ten or more ways.

%H David Consiglio, Jr., <a href="/A345155/b345155.txt">Table of n, a(n) for n = 1..10000</a>

%e 21896 is a term 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.

%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 >= 10])

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

%o print(rets[x])

%Y Cf. A025375, A344928, A345121, A345146, A345156, A345187.

%K nonn

%O 1,1

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