A136211 Denominators in continued fraction [0; 1, 3, 1, 3, 1, 3, ...].

1, 4, 5, 19, 24, 91, 115, 436, 551, 2089, 2640, 10009, 12649, 47956, 60605, 229771, 290376, 1100899, 1391275, 5274724, 6665999, 25272721, 31938720, 121088881, 153027601, 580171684, 733199285, 2779769539, 3512968824

Offset: 1

Author

Gary W. Adamson, Dec 21 2007

Keywords

• nonn
• frac
• easy

Comments

A136210(n)/A136211(n) tends to 0.791287847... = [0; 1, 3, 1, 3, 1, 3, ...] = (sqrt(21) - 3)/2 = the inradius of a right triangle with hypotenuse 3, legs 2 and sqrt(21).

The number 0.791287847... = (sqrt(21) - 3)/2 arises in finding a number which is 5 less than its square; the result is: 2.791287847... because (2.791287847...)^2 = 7.791287847... In general the quadratic equation for finding such numbers is x^2 - x = N, so x = (1 + sqrt(1 + 4N))/2. - Alexander R. Povolotsky, Dec 23 2007

Prepending a 1 to the sequence gives [1, 1, 4, 5, 19, 24, ...]. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) with the parameters R = 3 and Q = -1. It is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all natural numbers n and m. - Peter Bala, May 14 2014

Examples

			a(4) = 19 = 3*a(3) + a(2) = 3*5 + 4.
a(5) = 24 = a(4) + a(3) = 19 + 5.
T^3 = [19, 72; 24, 91], where the bottom row [24, 91] = [a(5), a(6)].
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• P. Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6
• J. L. Ramirez, F. Sirvent, A q-Analogue of the Bi-Periodic Fibonacci Sequence, J. Int. Seq. 19 (2016) # 16.4.6, t_n at a=1, b=3.
• Eric Weisstein's World of Mathematics, Lehmer Number
• Index entries for linear recurrences with constant coefficients, signature (0, 5, 0, -1).

Crossrefs

Cf. A136210.

Programs

Mathematica

Denominator[NestList[(3/(3+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
a[n_] := FromContinuedFraction[ Join[{0}, 3 - 2*Array[Mod[#, 2]&, n]]] // Denominator; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, May 15 2014 *)

PARI

x='x + O('x^25); Vec(x*(1+4*x-x^3)/(1-5*x^2+x^4)) \\ G. C. Greubel, Feb 18 2017

Formula

a(1) = 1, a(2) = 4, then for n>2, a(2n) = 3*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Let T = the 2 X 2 matrix [1, 3; 1, 4]. Then T^n = [A136210(2n-1), A136210(2n); a(2n-1), a(2n)].

From R. J. Mathar, May 18 2008: (Start)

O.g.f.: x*(1+4*x-x^3)/(1-5*x^2+x^4).

a(2*n) = A004253(n+1).

a(2*n+1) = A004254(n). (End)

a(n)*a(n+1) = A099025(n). - R. K. Guy, May 18 2008

{-a(n) + 5 a(n + 2) - a(n + 4), a(0) = 1, a(1) = 4, a(2) = 5, a(3) = 19}. - Robert Israel, May 14 2008