%I #6 Aug 05 2021 07:18:15
%S 47,48,50,51,53,54,55,56,57,58,59,60,62,63,64,65,66,67,68,69,70,71,72,
%T 73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,
%U 96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111
%N Numbers that are the sum of nine squares in five or more ways.
%H Sean A. Irvine, <a href="/A345502/b345502.txt">Table of n, a(n) for n = 1..1000</a>
%e 48 is a term because 48 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2.
%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**2 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 9):
%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. A345492, A345501, A345503, A345544, A346804.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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