A062328 Length of period of continued fraction expansion of square root of 3^n+1.

1, 0, 1, 4, 1, 26, 1, 56, 1, 44, 1, 264, 1, 814, 1, 136, 1, 3730, 1, 20968, 1, 2448, 1, 287980, 1, 397238, 1, 2678, 1, 670896, 1, 8110044, 1, 20696, 1, 1066520, 1, 366601254, 1, 277444, 1, 5903828476, 1, 7701738148, 1, 8208058, 1, 30287795640, 1, 253244432640, 1, 11656644672, 1, 2376211301858, 1, 590009437260, 1

Offset: 0

Author

Labos Elemer, Jul 13 2001

Keywords

• nonn
• more

Comments

a(n) = 1 iff n is even. In this case, 3^n + 1 = A002522(3^(n/2)) and the continued fraction expansion of sqrt(3^n+1) is {3^(n/2); 2*3^(n/2), 2*3^(n/2), 2*3^(n/2), 2*3^(n/2), ...}. - Bernard Schott, Sep 25 2019

Examples

			The period of sqrt(244) contains 26 terms: [1, 1, 1, 1, 1, 2, 1, 5, 1, 1, 9, 1, 6, 1, 9, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 30], so a(5) = 26.
		

Crossrefs

Cf. A003285, A034472, A059866, A059926, A059927, A062345, A062682.

Programs

Maple

with(numtheory): [seq(nops(cfrac(sqrt(3^k+1),'periodic','quotients')[2]),k=2..18)];

Mathematica

Table[Length[Last[ContinuedFraction[Sqrt[3^w+1]]]],{w,1,40}] (* corrected by Harvey P. Dale, Dec 05 2014 *)

Formula

a(n) = A003285(A034472(n)). - Bernard Schott, Sep 25 2019

Extensions

More terms from Harvey P. Dale, Dec 05 2014

a(41)-a(42) from Vaclav Kotesovec, Sep 17 2019

a(0), a(43)-a(48) from Chai Wah Wu, Sep 25 2019

a(49)-a(56) from Chai Wah Wu, Oct 03 2019