A309778 - OEIS

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.

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

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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

Table of n, a(n) for n=1..64.

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 A353445

Adjacent sequences: A309775 A309776 A309777 * A309779 A309780 A309781

KEYWORD

nonn

AUTHOR

Bernard Schott, Aug 17 2019

STATUS

approved