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 A309778 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. 4
 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, 827, 886, 1, 1, 1, 1142, 1211, 1, 1355, 1, 1507, 2, 1667, 1, 1, 1, 2011, 1, 1, 1, 1, 2486, 2587, 2690, 2795, 1, 3011, 1, 1, 3350, 1, 3586, 3707, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 COMMENTS 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). There are four cases to examine and three possible values for a(n). a(n) = 1 iff n is a nonhypotenuse number or iff n is in A004144. a(n) >= 2 iff n is a hypotenuse number or iff n is in A009003. 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. a(n) = n^2 - 14 iff n^2 is the sum of two and three positive squares or iff n^2 is in A231632. 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. REFERENCES Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79. LINKS IMO, 1992, Moscow, Second day. Problem 6 EXAMPLE 1 = 1^2, 4 = 2^2 and a(1) = a(2) = 1. 25 = 5^2 = 3^2 + 4^2 and a(5) = 2. 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. CROSSREFS Cf. A018820, A004144, A009003, A231632, A309779. Sequence in context: A284154 A080028 A309228 * A143223 A063993 A115722 Adjacent sequences:  A309775 A309776 A309777 * A309779 A309780 A309781 KEYWORD nonn AUTHOR Bernard Schott, Aug 17 2019 STATUS approved

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Last modified June 15 18:33 EDT 2021. Contains 345049 sequences. (Running on oeis4.)