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A345488 Numbers that are the sum of eight squares in one or more ways.

%I A345488 #11 Jun 12 2025 13:37:46
%S A345488 8,11,14,16,17,19,20,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,
%T A345488 38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,
%U A345488 61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79
%N A345488 Numbers that are the sum of eight squares in one or more ways.
%H A345488 Sean A. Irvine, <a href="/A345488/b345488.txt">Table of n, a(n) for n = 1..1000</a>
%H A345488 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F A345488 From _Chai Wah Wu_, Jun 12 2025: (Start)
%F A345488 All integers >= 22 are terms. See A345508 for a similar proof.
%F A345488 a(n) = 2*a(n-1) - a(n-2) for n > 9.
%F A345488 G.f.: x*(-x^8 + x^7 - x^6 + x^5 - x^4 - x^3 - 5*x + 8)/(x - 1)^2. (End)
%e A345488 11 is a term because 11 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2.
%o A345488 (Python)
%o A345488 from itertools import combinations_with_replacement as cwr
%o A345488 from collections import defaultdict
%o A345488 keep = defaultdict(lambda: 0)
%o A345488 power_terms = [x**2 for x in range(1, 1000)]
%o A345488 for pos in cwr(power_terms, 8):
%o A345488     tot = sum(pos)
%o A345488     keep[tot] += 1
%o A345488     rets = sorted([k for k, v in keep.items() if v >= 1])
%o A345488     for x in range(len(rets)):
%o A345488         print(rets[x])
%Y A345488 Cf. A003331, A345478, A345489, A345498, A345508.
%K A345488 nonn
%O A345488 1,1
%A A345488 _David Consiglio, Jr._, Jun 19 2021