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%I #13 Aug 19 2021 16:21:07
%S 1765,1980,2043,2104,2195,2250,2449,2486,2491,2493,2547,2584,2592,
%T 2738,2745,2764,2817,2888,2915,2953,2969,3095,3096,3133,3142,3186,
%U 3188,3240,3275,3277,3310,3366,3403,3422,3459,3464,3466,3483,3529,3583,3608,3627,3653,3664,3671,3690,3697,3707,3725,3744,3746,3781
%N Numbers that are the sum of five positive cubes in exactly five ways.
%C Differs from A343989 at term 7 because 2430 = 1^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 4^3 + 5^3 + 8^3 + 12^3 = 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 4^3 + 5^3 + 8^3 + 9^3 + 10^3.
%H David Consiglio, Jr., <a href="/A343988/b343988.txt">Table of n, a(n) for n = 1..20000</a>
%e 2043 is a term because 2043 = 1^3 + 4^3 + 5^3 + 5^3 + 12^3 = 2^3 + 2^3 + 3^3 + 10^3 + 10^3 = 2^3 + 3^3 + 4^3 + 6^3 + 12^3 = 4^3 + 5^3 + 5^3 + 9^3 + 10^3 = 4^3 + 6^3 + 6^3 + 6^3 + 11^3.
%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(1,50)]
%o for pos in cwr(power_terms,5):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k,v in keep.items() if v == 5])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A294739, A343986, A343989, A344035, A344359, A345175, A345767.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, May 06 2021
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