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A345618 Numbers that are the sum of eight fifth powers in ten or more ways. 5

%I #6 Jul 31 2021 16:17:34

%S 15539667,22932525,24393600,24650406,24952961,24953742,25054306,

%T 25142513,25201550,25423794,26001294,26851552,27396567,27988486,

%U 28609075,29309819,29558650,31052406,31794336,32223105,32527286,32610600,32807777,32890541,32998317,33015125

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

%e 15539667 is a term because 15539667 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5.

%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, 8):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v >= 10])

%o for x in range(len(rets)):

%o print(rets[x])

%Y Cf. A345585, A345617, A345627, A345643, A346335.

%K nonn

%O 1,1

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

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