A345837 - OEIS

%I #6 Jul 31 2021 21:33:37

%S 4228,4403,4468,5443,5508,5683,6613,6643,6658,6708,6773,6838,6868,

%T 6883,6948,7013,7093,7138,7203,7267,7268,7332,7397,7478,7507,7572,

%U 7588,7828,7858,7923,7988,8113,8133,8228,8353,8418,8533,8547,8548,8612,8723,8788,8852

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

%C Differs from A345580 at term 11 because 6723 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4  = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4  = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4  = 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4  = 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4  = 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

%H Sean A. Irvine, <a href="/A345837/b345837.txt">Table of n, a(n) for n = 1..10000</a>

%e 4403 is a term because 4403 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 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 == 5])

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

%o         print(rets[x])

%Y Cf. A345580, A345787, A345827, A345836, A345838, A345847, A346330.

%K nonn

%O 1,1

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