A344943 - OEIS

%I #7 Jul 31 2021 22:03:06

%S 197779,211059,217154,236675,431155,444019,480739,503539,530659,

%T 548994,564979,568450,571539,602450,602770,621859,625635,625939,

%U 626194,650659,651954,653059,654130,666739,687314,692754,692899,698019,708499,716739,728914,730914

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

%C Differs from A344942 at term 10 because 534130 = 1^4 + 3^4 + 16^4 + 22^4 + 22^4  = 2^4 + 2^4 + 4^4 + 7^4 + 27^4  = 2^4 + 3^4 + 6^4 + 6^4 + 27^4  = 2^4 + 6^4 + 9^4 + 21^4 + 24^4  = 4^4 + 16^4 + 17^4 + 18^4 + 23^4  = 6^4 + 8^4 + 11^4 + 22^4 + 23^4  = 7^4 + 8^4 + 16^4 + 19^4 + 24^4  = 13^4 + 14^4 + 14^4 + 21^4 + 22^4.

%H David Consiglio, Jr., <a href="/A344943/b344943.txt">Table of n, a(n) for n = 1..10000</a>

%e 197779 is a term because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4  = 1^4 + 7^4 + 11^4 + 12^4 + 20^4  = 1^4 + 10^4 + 12^4 + 17^4 + 17^4  = 2^4 + 4^4 + 5^4 + 7^4 + 21^4  = 3^4 + 5^4 + 6^4 + 6^4 + 21^4  = 4^4 + 7^4 + 9^4 + 13^4 + 20^4  = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

%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, 5):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v == 7])

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

%o     print(rets[x])

%Y Cf. A344923, A344941, A344942, A344945, A345181, A345819.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 03 2021