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 A346345 Numbers that are the sum of nine fifth powers in exactly ten ways. 6

%I #6 Jul 31 2021 19:01:16

%S 4157156,4492410,4510461,4915538,5005474,5015506,5179747,5219655,

%T 5756794,6323426,6326519,6382443,6423394,6705284,6793170,6861218,

%U 7101038,7147645,7147656,7148679,7266240,7280391,7283268,7314187,7413493,7422352,7531076,7651645,7693425

%N Numbers that are the sum of nine fifth powers in exactly ten ways.

%C Differs from A345627 at term 5 because 4948274 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 8^5 + 9^5 + 15^5 + 21^5 = 1^5 + 3^5 + 5^5 + 5^5 + 8^5 + 8^5 + 8^5 + 15^5 + 21^5 = 1^5 + 2^5 + 2^5 + 5^5 + 5^5 + 11^5 + 11^5 + 17^5 + 20^5 = 8^5 + 9^5 + 9^5 + 10^5 + 10^5 + 10^5 + 12^5 + 16^5 + 20^5 = 4^5 + 8^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 + 16^5 + 19^5 = 4^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 15^5 + 19^5 = 4^5 + 4^5 + 9^5 + 13^5 + 13^5 + 13^5 + 14^5 + 15^5 + 19^5 = 6^5 + 6^5 + 9^5 + 9^5 + 12^5 + 12^5 + 14^5 + 18^5 + 18^5 = 1^5 + 8^5 + 8^5 + 12^5 + 12^5 + 14^5 + 14^5 + 17^5 + 18^5 = 1^5 + 8^5 + 9^5 + 9^5 + 13^5 + 14^5 + 16^5 + 17^5 + 17^5 = 3^5 + 7^5 + 7^5 + 10^5 + 12^5 + 16^5 + 16^5 + 16^5 + 17^5.

%H Sean A. Irvine, <a href="/A346345/b346345.txt">Table of n, a(n) for n = 1..10000</a>

%e 4157156 is a term because 4157156 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 9^5 + 21^5 = 1^5 + 1^5 + 3^5 + 4^5 + 5^5 + 5^5 + 8^5 + 8^5 + 21^5 = 1^5 + 4^5 + 4^5 + 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 19^5 = 1^5 + 4^5 + 4^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 19^5 = 5^5 + 5^5 + 5^5 + 5^5 + 7^5 + 9^5 + 15^5 + 17^5 + 18^5 = 3^5 + 3^5 + 5^5 + 6^5 + 9^5 + 10^5 + 16^5 + 16^5 + 18^5 = 1^5 + 1^5 + 5^5 + 5^5 + 13^5 + 13^5 + 15^5 + 15^5 + 18^5 = 2^5 + 3^5 + 4^5 + 4^5 + 10^5 + 14^5 + 16^5 + 16^5 + 17^5 = 11^5 + 11^5 + 12^5 + 12^5 + 12^5 + 12^5 + 13^5 + 16^5 + 17^5 = 2^5 + 2^5 + 2^5 + 5^5 + 12^5 + 15^5 + 16^5 + 16^5 + 16^5.

%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, 9):

%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. A345627, A345852, A346335, A346344, A346355.

%K nonn

%O 1,1

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

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Last modified December 9 12:31 EST 2023. Contains 367690 sequences. (Running on oeis4.)