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Numbers that are the sum of nine squares in ten or more ways.
4

%I #15 May 10 2024 08:51:59

%S 63,65,68,71,72,74,75,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,

%T 93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,

%U 112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127

%N Numbers that are the sum of nine squares in ten or more ways.

%H Sean A. Irvine, <a href="/A346803/b346803.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).

%F From _Chai Wah Wu_, May 09 2024: (Start)

%F All integers >= 77 are terms. Proof: since 246 can be written as the sum of 4 positive squares in 10 ways (see A025428) and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 280 can be written as a sum of 9 positive squares in 10 or more ways. Integers from 77 to 279 are terms by inspection.

%F a(n) = 2*a(n-1) - a(n-2) for n > 9.

%F G.f.: x*(-x^8 + x^7 - x^6 + x^5 - 2*x^4 + x^2 - 61*x + 63)/(x - 1)^2. (End)

%e 65 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2

%e = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2

%e = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2

%e = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2

%e = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2

%e = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2

%e = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2

%e = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2

%e = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2

%e = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2

%e so 65 is a term.

%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 >= 10])

%o for x in range(len(rets)):

%o print(rets[x])

%o (Python)

%o def A346803(n): return (63, 65, 68, 71, 72, 74, 75)[n-1] if n<8 else n+69 # _Chai Wah Wu_, May 09 2024

%Y Cf. A025428, A025429, A345497, A345549, A346808.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Aug 04 2021