A345786 - OEIS

%I #6 Jul 31 2021 22:37:14

%S 256,347,382,401,408,427,434,438,445,464,478,480,490,499,502,506,511,

%T 516,523,530,532,534,537,560,565,567,569,571,578,586,593,595,600,602,

%U 604,605,611,612,616,619,621,624,626,643,645,656,660,663,664,668,675,679

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

%C Differs from A345534 at term 11 because 471 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3  = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3  = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3  = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345786/b345786.txt">Table of n, a(n) for n = 1..207</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. A345534, A345776, A345785, A345787, A345796, A345836.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021