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Numbers that are the sum of ten fifth powers in four or more ways.
6

%I #6 Jul 31 2021 15:58:22

%S 55543,55574,55785,56566,58667,63318,72349,73002,85186,86506,87287,

%T 87529,88310,103134,111498,113599,114591,118250,119031,120351,120382,

%U 120593,121374,123475,128126,134475,134878,135201,137157,142008,142219,143000,143211,143506

%N Numbers that are the sum of ten fifth powers in four or more ways.

%H Sean A. Irvine, <a href="/A345636/b345636.txt">Table of n, a(n) for n = 1..10000</a>

%e 55574 is a term because 55574 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5.

%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**5 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 10):

%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. A345597, A345621, A345635, A345637, A346349.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021