A345854 - OEIS

%I #6 Jul 31 2021 20:00:07

%S 265,280,295,310,325,340,345,355,360,375,390,405,420,425,440,455,470,

%T 485,505,565,580,585,595,630,645,660,665,695,710,725,745,760,805,820,

%U 835,840,870,885,889,900,904,919,920,934,935,949,950,964,965,969,984,999

%N Numbers that are the sum of ten fourth powers in exactly two ways.

%C Differs from A345595 at term 20 because 520 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4  = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4  = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

%H Sean A. Irvine, <a href="/A345854/b345854.txt">Table of n, a(n) for n = 1..10000</a>

%e 280 is a term because 280 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

%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**4 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 10):

%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. A345595, A345804, A345844, A345853, A345855, A346347.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021