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%I #37 Aug 26 2019 05:00:10
%S 1,1,1,1,2,1,1,1,1,2,1,1,155,1,211,1,275,1,1,2,1,1,1,1,611,662,1,1,
%T 827,886,1,1,1,1142,1211,1,1355,1,1507,2,1667,1,1,1,2011,1,1,1,1,2486,
%U 2587,2690,2795,1,3011,1,1,3350,1,3586,3707,1,1,1
%N a(n) is the greatest integer such that, for every positive integer k <= a(n), n^2 can be written as the sum of k positive square integers.
%C The idea for this sequence comes from the 6th problem of the 2nd day of the 33rd International Mathematical Olympiad in Moscow, 1992 (see link).
%C There are four cases to examine and three possible values for a(n).
%C a(n) = 1 iff n is a nonhypotenuse number or iff n is in A004144.
%C a(n) >= 2 iff n is a hypotenuse number or iff n is in A009003.
%C a(n) = 2 iff n^2 is the sum of two positive squares but not the sum of three positive squares or iff n^2 is in A309779.
%C a(n) = n^2 - 14 iff n^2 is the sum of two and three positive squares or iff n^2 is in A231632.
%C Theorem: a square n^2 is the sum of k positive squares for all 1 <= k <= n^2 - 14 iff n^2 is the sum of 2 and 3 positive squares (proof in Kuczma). Consequently: A231632 = A018820.
%D Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79.
%H IMO, 1992, Moscow, <a href="https://www.imo-register.org.uk/1992-report.html">Second day. Problem 6</a>
%e 1 = 1^2, 4 = 2^2 and a(1) = a(2) = 1.
%e 25 = 5^2 = 3^2 + 4^2 and a(5) = 2.
%e The first representations of 169 are 13^2 = 12^2 + 5^2 = 12^2 + 4^2 + 3^2 = 11^2 + 4^2 + 4^2 + 4^2 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2 = 6^2 + 6^2 + 6^2 + 6^2 + 4^2 + 3^2 = ... and a(13) = 13^2 - 14 = 155.
%Y Cf. A018820, A004144, A009003, A231632, A309779.
%K nonn
%O 1,5
%A _Bernard Schott_, Aug 17 2019
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