A345855 - OEIS

%I #6 Jul 31 2021 20:00:10

%S 520,535,550,600,615,680,775,790,855,1030,1144,1159,1224,1365,1380,

%T 1399,1445,1540,1555,1605,1635,1685,1700,1768,1795,1815,1830,1860,

%U 1875,1895,1989,2070,2164,2229,2244,2439,2485,2580,2595,2645,2675,2680,2695,2710,2755

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

%C Differs from A345596 at term 21 because 1620 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4  = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4.

%H Sean A. Irvine, <a href="/A345855/b345855.txt">Table of n, a(n) for n = 1..10000</a>

%e 535 is a term because 535 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 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 == 3])

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

%o         print(rets[x])

%Y Cf. A345596, A345805, A345845, A345854, A345856, A346348.

%K nonn

%O 1,1

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