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Numbers that are the sum of four third powers in exactly nine ways.
7

%I #6 Jul 31 2021 23:29:15

%S 42120,46683,50806,50904,51408,51480,51688,52208,53865,54971,56385,

%T 57113,60515,60984,62433,65303,66276,66339,66430,67158,69048,69832,

%U 69930,71162,72072,72520,72576,72800,73017,77714,77903,79345,79667,79849,80066,80073,81207

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

%C Differs from A345146 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="/A345154/b345154.txt">Table of n, a(n) for n = 1..10000</a>

%e 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 (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 == 9])

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

%o print(rets[x])

%Y Cf. A025365, A344927, A345120, A345146, A345153, A345156, A345186.

%K nonn

%O 1,1

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