

A077427


Primitive period length of (regular) continued fraction of (sqrt(D(n))+1)/2 for D(n)=A077425(n).


5



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

1,3


COMMENTS

The Pell equation x^2  D(n)*y^2 = 4 has (infinitely many integer) solutions if and only if a(n) is odd.


REFERENCES

O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109).


LINKS

Table of n, a(n) for n=1..74.


EXAMPLE

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((k1)/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.


CROSSREFS

Cf. A077426.
Sequence in context: A190698 A283183 A327467 * A107641 A299352 A127671
Adjacent sequences: A077424 A077425 A077426 * A077428 A077429 A077430


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Nov 29 2002


STATUS

approved



