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Numbers that are the sum of seven fifth powers in seven or more ways.
7

%I #6 Jul 31 2021 16:25:26

%S 28608832,35663099,36090526,36620574,46998599,51095638,52541851,

%T 54233651,54827543,54886349,61263643,61634374,63514593,64810976,

%U 65198607,66708676,67887843,70979107,72970305,74002457,74115801,74132607,74487093,75044651,75378359

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

%H Sean A. Irvine, <a href="/A345629/b345629.txt">Table of n, a(n) for n = 1..10000</a>

%e 35663099 is a term because 35663099 = 1^5 + 9^5 + 16^5 + 17^5 + 24^5 + 24^5 + 28^5 = 2^5 + 3^5 + 17^5 + 23^5 + 24^5 + 24^5 + 26^5 = 2^5 + 10^5 + 15^5 + 17^5 + 23^5 + 23^5 + 29^5 = 4^5 + 8^5 + 13^5 + 19^5 + 21^5 + 27^5 + 27^5 = 4^5 + 11^5 + 13^5 + 19^5 + 20^5 + 22^5 + 30^5 = 5^5 + 6^5 + 19^5 + 19^5 + 20^5 + 20^5 + 30^5 = 7^5 + 9^5 + 12^5 + 18^5 + 18^5 + 27^5 + 28^5.

%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, 7):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v >= 7])

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

%o print(rets[x])

%Y Cf. A345573, A345609, A345615, A345630, A345721, A346284.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 22 2021