A345494 - OEIS

%I #6 Aug 05 2021 07:20:01

%S 56,59,62,64,65,67,68,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,

%T 86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,

%U 106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121

%N Numbers that are the sum of eight squares in seven or more ways.

%H Sean A. Irvine, <a href="/A345494/b345494.txt">Table of n, a(n) for n = 1..1000</a>

%e 59 is a term because 59 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 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, 8):

%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. A345484, A345493, A345495, A345504, A345537.

%K nonn

%O 1,1

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