|
%I #9 Apr 26 2024 03:20:16
%S 56,58,59,61,62,64,65,67,68,70,71,72,73,74,75,76,77,78,79,80,81,82,83,
%T 84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,
%U 105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120
%N Numbers that are the sum of eight squares in six or more ways.
%H Sean A. Irvine, <a href="/A345493/b345493.txt">Table of n, a(n) for n = 1..1000</a>
%F Conjectures from _Chai Wah Wu_, Apr 25 2024: (Start)
%F a(n) = 2*a(n-1) - a(n-2) for n > 11.
%F G.f.: x*(-x^10 + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 - 54*x + 56)/(x - 1)^2. (End)
%e 58 is a term because 58 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2.
%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 >= 6])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345483, A345492, A345494, A345503, A345536.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
|
