|
%I #6 Jul 31 2021 21:36:43
%S 2677,2692,2757,2852,2867,2917,2997,3107,3172,3301,3476,3541,3972,
%T 4132,4227,4242,4257,4307,4322,4372,4437,4452,4482,4497,4562,4627,
%U 4737,4756,4851,4866,4867,4931,4996,5077,5106,5107,5122,5187,5252,5282,5317,5347,5362
%N Numbers that are the sum of seven fourth powers in exactly three ways.
%C Differs from A345569 at term 7 because 2932 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.
%H Sean A. Irvine, <a href="/A345825/b345825.txt">Table of n, a(n) for n = 1..10000</a>
%e 2692 is a term because 2692 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.
%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, 7):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v == 3])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345569, A345775, A345815, A345824, A345826, A345835, A346280.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 26 2021
|
