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 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 (list; graph; refs; listen; history; text; internal format)
 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 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((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. 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

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Last modified April 4 05:34 EDT 2020. Contains 333212 sequences. (Running on oeis4.)