A345630 - OEIS

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

%S 36620574,80552143,81401376,82078424,92347417,93653176,94486699,

%T 94626949,98873875,105674625,110276376,121050874,124732805,125959393,

%U 127808693,129228307,130298618,134581976,144209018,145340799,147245218,147898763,151727082

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

%H Sean A. Irvine, <a href="/A345630/b345630.txt">Table of n, a(n) for n = 1..10000</a>

%e 80552143 is a term because 80552143 = 1^5 + 4^5 + 21^5 + 21^5 + 23^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 23^5 + 32^5 + 32^5 = 1^5 + 8^5 + 16^5 + 19^5 + 27^5 + 28^5 + 34^5 = 3^5 + 12^5 + 13^5 + 14^5 + 28^5 + 31^5 + 32^5 = 3^5 + 14^5 + 17^5 + 18^5 + 18^5 + 27^5 + 36^5 = 4^5 + 11^5 + 13^5 + 22^5 + 23^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 23^5 + 24^5 + 36^5 = 6^5 + 23^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.

%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 >= 8])

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

%o         print(rets[x])

%Y Cf. A345574, A345616, A345629, A345631, A345722, A346285.

%K nonn

%O 1,1

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