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Numbers that are the sum of nine fifth powers in exactly two ways.
7

%I #6 Jul 31 2021 19:00:47

%S 4101,4132,4163,4194,4225,4343,4374,4405,4436,4585,4616,4647,4827,

%T 4858,5069,5124,5155,5186,5217,5366,5397,5428,5608,5639,5850,6147,

%U 6178,6209,6389,6420,6631,7170,7201,7225,7256,7287,7318,7412,7467,7498,7529,7709,7740

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

%C Differs from A345619 at term 306 because 52418 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.

%H Sean A. Irvine, <a href="/A346337/b346337.txt">Table of n, a(n) for n = 1..10000</a>

%e 4101 is a term because 4101 = 1^5 + 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.

%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 == 2])

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

%o print(rets[x])

%Y Cf. A345619, A345844, A346327, A346336, A346338, A346347.

%K nonn

%O 1,1

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