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Numbers that are the sum of five fourth powers in ten or more ways.
6

%I #16 May 10 2024 02:27:06

%S 954979,1205539,1574850,1713859,1801459,1863859,1877394,1882579,

%T 2071939,2109730,2138419,2142594,2157874,2225859,2288179,2419954,

%U 2492434,2495939,2605314,2663539,2711394,2784499,2835939,2847394,2849859,2880994,2919154,2924674,3007474

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

%H David Consiglio, Jr., <a href="/A341897/b341897.txt">Table of n, a(n) for n = 1..10000</a>

%e 954979 = 1^4 + 2^4 + 11^4 + 19^4 + 30^4

%e = 1^4 + 7^4 + 18^4 + 25^4 + 26^4

%e = 3^4 + 8^4 + 17^4 + 20^4 + 29^4

%e = 4^4 + 8^4 + 13^4 + 25^4 + 27^4

%e = 4^4 + 9^4 + 10^4 + 11^4 + 31^4

%e = 6^4 + 6^4 + 15^4 + 21^4 + 29^4

%e = 7^4 + 10^4 + 18^4 + 19^4 + 29^4

%e = 11^4 + 11^4 + 20^4 + 22^4 + 27^4

%e = 16^4 + 17^4 + 17^4 + 24^4 + 25^4

%e = 18^4 + 19^4 + 20^4 + 23^4 + 23^4

%e so 954979 is a term.

%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, 5):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v >= 10])

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

%o print(rets[x])

%Y Cf. A341891, A341898, A344928, A345187, A345567.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 04 2021