Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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