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 A346329 Numbers that are the sum of eight fifth powers in exactly four ways. 7

%I #7 Jul 31 2021 19:03:48

%S 391250,392031,455750,519236,604822,622281,672023,672054,672265,

%T 673554,697492,703978,707368,730259,763292,857761,893605,893636,

%U 893816,893847,894027,894058,894452,894628,896729,897151,901380,903839,909124,909597,910411,911403

%N Numbers that are the sum of eight fifth powers in exactly four ways.

%C Differs from A345612 at term 33 because 926372 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5.

%H Sean A. Irvine, <a href="/A346329/b346329.txt">Table of n, a(n) for n = 1..10000</a>

%e 391250 is a term because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.

%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 == 4])

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

%o print(rets[x])

%Y Cf. A345612, A345836, A346281, A346328, A346330, A346339.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jul 13 2021

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