Search: a018820 -id:a018820
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A309778
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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.
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+10
4
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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
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1,5
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COMMENTS
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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.
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REFERENCES
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Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79.
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A231632
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Squares that are also sums of 2 and 3 nonzero squares.
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+10
2
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169, 225, 289, 625, 676, 841, 900, 1156, 1225, 1369, 1521, 1681, 2025, 2500, 2601, 2704, 2809, 3025, 3364, 3600, 3721, 4225, 4624, 4900, 5329, 5476, 5625, 6084, 6724, 7225, 7569, 7921, 8100, 8281, 9025, 9409, 10000, 10201, 10404, 10816, 11025, 11236, 11881, 12100, 12321, 12769, 13225, 13456, 13689, 14161
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OFFSET
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1,1
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COMMENTS
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All terms == {0, 1} (mod 4).
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 (see A309778 and proof in Kuczma) . Consequently this is a duplicate of A018820. - Bernard Schott, Aug 17 2019
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REFERENCES
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Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79.
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LINKS
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EXAMPLE
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169 = 13^2 = 5^2 + 12^2 = 3^2 + 4^2 + 12^2;
225 = 15^2 = 9^2 + 12^2 = 2^2 + 5^2 + 14^2.
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CROSSREFS
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KEYWORD
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dead
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AUTHOR
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STATUS
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approved
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A341329
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Numbers k such that k^2 is the sum of m nonzero squares for all 1 <= m <= k^2 - 14.
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+10
2
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13, 15, 17, 25, 26, 29, 30, 34, 35, 37, 39, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120, 122, 123, 125, 130, 135, 136, 137
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OFFSET
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1,1
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COMMENTS
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Numbers k such that k^2 is in A018820. Note that k^2 is never the sum of k^2 - 13 positive squares.
A square k^2 is the sum of m positive squares for all 1 <= m <= k^2 - 14 if k^2 is the sum of 2 and 3 positive squares (see A309778 and proof in Kuczma).
Numbers k not of the form 5*2^e such that k has at least one prime factor congruent to 1 modulo 4.
Has density 1 over all positive integers.
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REFERENCES
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Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79.
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LINKS
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EXAMPLE
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13 is a term: 169 = 13^2 = 5^2 + 12^2 = 3^2 + 4^2 + 12^2 = 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 = ... = 3^2 + 2^2 + 2^2 + 1^2 + 1^2 + ... + 1^2 (sum of 155 positive squares, with 152 (1^2)'s), but 169 cannot be represented as the sum of 156 positive squares.
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PROG
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(PARI) isA341329(n) = setsearch(Set(factor(n)[, 1]%4), 1) && !(n/5 == 2^valuation(n, 2))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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