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

%I #21 May 11 2024 09:41:57

%S 4097,51446,51477,51688,52469,54570,59221,68252,68905,84213,110494,

%T 131104,151445,212496,300277,325174,325713,355114,422135,422738,

%U 589269,637418,794434,810820,876734,876765,876976,877757,879858,884509,893540,909501,924912,935782,976733,995571,1037784,1083457

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

%C This sequence differs from A342685:

%C 13124675 = 1^5 + 9^5 + 10^5 + 20^5 + 25^5

%C = 2^5 + 5^5 + 12^5 + 23^5 + 23^5

%C = 16^5 + 19^5 + 20^5 + 20^5 + 20^5,

%C so 13124675 is in A342685, but is not in this sequence.

%H David Consiglio, Jr., <a href="/A342686/b342686.txt">Table of n, a(n) for n = 1..20000</a>

%e 51477 = 2^5 + 4^5 + 7^5 + 7^5 + 7^5

%e = 2^5 + 5^5 + 6^5 + 6^5 + 8^5

%e so 51477 is a term of this sequence.

%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, 500)]

%o for pos in cwr(power_terms, 5):

%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. A342685, A342688, A344237, A344643, A344645, A346357.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, May 18 2021