|
%I #6 Aug 05 2021 15:20:49
%S 256,347,382,401,408,427,434,438,445,464,471,478,480,490,497,499,502,
%T 504,506,511,516,523,530,532,534,537,560,565,567,569,571,578,586,593,
%U 595,597,600,602,604,605,611,612,616,619,621,623,624,626,628,630,635,642
%N Numbers that are the sum of eight cubes in four or more ways.
%H Sean A. Irvine, <a href="/A345534/b345534.txt">Table of n, a(n) for n = 1..10000</a>
%e 347 is a term because 347 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3.
%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, 8):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v >= 4])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345491, A345522, A345533, A345535, A345543, A345579, A345786.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
|
