cp's OEIS Frontend

A345547 Numbers that are the sum of nine cubes in eight or more ways.

%I A345547 #6 Aug 05 2021 15:19:20
%S A345547 744,770,805,818,840,842,844,847,859,861,866,868,877,880,883,887,894,
%T A345547 896,903,908,909,910,911,913,915,916,920,922,929,935,939,940,945,946,
%U A345547 948,950,952,954,955,957,959,961,964,965,966,971,972,973,976,978,983,985
%N A345547 Numbers that are the sum of nine cubes in eight or more ways.
%H A345547 Sean A. Irvine, <a href="/A345547/b345547.txt">Table of n, a(n) for n = 1..10000</a>
%e A345547 770 is a term because 770 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.
%o A345547 (Python)
%o A345547 from itertools import combinations_with_replacement as cwr
%o A345547 from collections import defaultdict
%o A345547 keep = defaultdict(lambda: 0)
%o A345547 power_terms = [x**3 for x in range(1, 1000)]
%o A345547 for pos in cwr(power_terms, 9):
%o A345547     tot = sum(pos)
%o A345547     keep[tot] += 1
%o A345547     rets = sorted([k for k, v in keep.items() if v >= 8])
%o A345547     for x in range(len(rets)):
%o A345547         print(rets[x])
%Y A345547 Cf. A345505, A345538, A345546, A345548, A345556, A345592, A345800.
%K A345547 nonn
%O A345547 1,1
%A A345547 _David Consiglio, Jr._, Jun 20 2021