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A077427
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Primitive period length of (regular) continued fraction of (sqrt(D(n))+1)/2 for D(n)=A077425(n).
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5
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1, 1, 3, 2, 1, 4, 3, 5, 2, 1, 6, 3, 3, 4, 9, 2, 1, 7, 2, 9, 3, 6, 7, 7, 2, 1, 10, 4, 7, 4, 3, 5, 8, 5, 10, 2, 1, 12, 5, 3, 4, 15, 3, 14, 4, 12, 4, 16, 2, 1, 9, 2, 19, 2, 16, 6, 3, 8, 11, 5, 6, 9, 15, 2, 1, 10, 10, 4, 6, 19, 3, 4, 3, 16
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OFFSET
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1,3
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COMMENTS
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The Pell equation x^2 - D(n)*y^2 = -4 has (infinitely many integer) solutions if and only if a(n) is odd.
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REFERENCES
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O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109).
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LINKS
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EXAMPLE
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a(6)=4 because the (periodic) continued fraction for (sqrt(D(6))+1)/2 = (sqrt(33)+1)/2 = 3.372281324... is [3, periodic(2, 1, 2, 5,)] with period length 4. Because these continued fractions are always of the form [b(0),periodic(b(1),b(2),...,b(2),b(1),2*b(0)-1,)] with the symmetric piece b(1),b(2),..., b(2),b(1), Perron op. cit. writes for this b(0),b(1),b(2),...,(b(k/2)) if the period length k is even and b(0),b(1),b(2),...,b((k-1)/2) if the period length is odd. In this example: k=4 and Perron writes 3,2,(1). Another example: D(8)= A077425(8)=41 leads to Perron's 3,1,2 standing for [3,periodic(1,2,2,1,5,)], the continued fraction for (sqrt(41)+1)/2 which has odd period length a(8)=5.
a(4)=2 is even and D(4)=A077425(4)=21, hence x^2 - 21*y^2 = -4 has no nontrivial integer solution.
a(8)=5 is odd and D(8)=A077425(8)=41, hence x^2 - 41*y^2 = -4 is solvable (with nontrivial integers) as well as x^2 - 41*y^2 = +4.
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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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