A345836 - OEIS

%I #6 Jul 31 2021 21:33:34

%S 2933,2948,3013,3173,3188,3557,4148,4163,4213,4293,4388,4453,4643,

%T 4772,4837,4883,5012,5123,5188,5203,5268,5333,5363,5378,5398,5428,

%U 5538,5573,5603,5618,5668,5733,5748,5858,5923,6052,6163,6227,6292,6548,6578,6628,6693

%N Numbers that are the sum of eight fourth powers in exactly four ways.

%C Differs from A345579 at term 10 because 4228 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4  = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4  = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4  = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4  = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

%H Sean A. Irvine, <a href="/A345836/b345836.txt">Table of n, a(n) for n = 1..10000</a>

%e 2948 is a term because 2948 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

%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, 8):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v == 4])

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

%o         print(rets[x])

%Y Cf. A345579, A345786, A345826, A345835, A345837, A345846, A346329.

%K nonn

%O 1,1

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