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A345497 Numbers that are the sum of eight squares in ten or more ways.

%I A345497 #12 May 10 2024 08:51:47
%S A345497 70,71,73,74,77,78,79,80,82,83,85,86,87,88,89,90,91,92,93,94,95,96,97,
%T A345497 98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,
%U A345497 115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131
%N A345497 Numbers that are the sum of eight squares in ten or more ways.
%H A345497 Sean A. Irvine, <a href="/A345497/b345497.txt">Table of n, a(n) for n = 1..1000</a>
%H A345497 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F A345497 From _Chai Wah Wu_, May 09 2024: (Start)
%F A345497 All integers >= 85 are terms. Proof: since 594 can be written as the sum of 3 positive squares in 10 ways (see A025427) and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 628 can be written as a sum of 8 positive squares in 10 or more ways. Integers from 85 to 627 are terms by inspection.
%F A345497 a(n) = 2*a(n-1) - a(n-2) for n > 12.
%F A345497 G.f.: x*(-x^11 + x^10 - x^9 + x^8 - 2*x^5 + 2*x^4 - x^3 + x^2 - 69*x + 70)/(x - 1)^2. (End)
%e A345497 71 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 8^2
%e A345497    = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2
%e A345497    = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 + 5^2
%e A345497    = 1^2 + 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 4^2
%e A345497    = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 5^2
%e A345497    = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 7^2
%e A345497    = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 5^2
%e A345497    = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 + 5^2
%e A345497    = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 6^2
%e A345497    = 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
%e A345497    = 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2
%e A345497 so 71 is a term.
%o A345497 (Python)
%o A345497 from itertools import combinations_with_replacement as cwr
%o A345497 from collections import defaultdict
%o A345497 keep = defaultdict(lambda: 0)
%o A345497 power_terms = [x**2 for x in range(1, 1000)]
%o A345497 for pos in cwr(power_terms, 8):
%o A345497     tot = sum(pos)
%o A345497     keep[tot] += 1
%o A345497     rets = sorted([k for k, v in keep.items() if v >= 10])
%o A345497     for x in range(len(rets)):
%o A345497         print(rets[x])
%o A345497 (Python)
%o A345497 def A345397(n): return (70, 71, 73, 74, 77, 78, 79, 80, 82, 83)[n-1] if n<11 else n+74 # _Chai Wah Wu_, May 09 2024
%Y A345497 Cf. A025427, A025429, A345487, A345496, A345540, A346803.
%K A345497 nonn
%O A345497 1,1
%A A345497 _David Consiglio, Jr._, Jun 20 2021