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A345580 Numbers that are the sum of eight fourth powers in five or more ways. 8

%I #6 Jul 31 2021 17:51:37

%S 4228,4403,4468,5443,5508,5683,6613,6643,6658,6708,6723,6773,6788,

%T 6838,6853,6868,6883,6898,6948,6963,7013,7028,7093,7138,7203,7267,

%U 7268,7332,7397,7478,7507,7572,7588,7828,7858,7923,7938,7988,8003,8068,8113,8133,8178

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

%H Sean A. Irvine, <a href="/A345580/b345580.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. A345535, A345571, A345579, A345581, A345589, A345613, A345837.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021

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Last modified August 12 23:59 EDT 2024. Contains 375113 sequences. (Running on oeis4.)