A136210 Numerators in continued fraction [0; 1, 3, 1, 3, 1, 3, ...].

1, 3, 4, 15, 19, 72, 91, 345, 436, 1653, 2089, 7920, 10009, 37947, 47956, 181815, 229771, 871128, 1100899, 4173825, 5274724, 19997997, 25272721, 95816160, 121088881, 459082803, 580171684, 2199597855, 2779769539, 10538906472

Offset: 1

Author

Gary W. Adamson, Dec 21 2007

Keywords

• nonn
• frac
• easy

Comments

A136210(n)/A136211(n) tends to 0.7912878474... = (sqrt(21) - 3)/2 = continued fraction [0; 1, 3, 1, 3, 1, 3, ...] = the inradius of a right triangle with hypotenuse 5, legs 2 and sqrt(21).

This 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) = 15 = 3*a(3) + a(2) = 3*4 + 3.
a(5) = 19 = a(4) + a(3) = 15 + 4.
T^3 = [19, 72; 24, 91], where [19, 72] = [a(5), a(6)]. [24, 91] = [A136211(5), A136211(6)].
G.f. = x + 3*x^2 + 4*x^3 + 15*x^4 + 19*x^5 + 72*x^6 + 91*x^7 + 345*x^8 + ...
		

Links

• 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=3, b=1.
• Index entries for linear recurrences with constant coefficients, signature (0,5,0,-1).

Crossrefs

Cf. A004253, A004254, A136211.

Programs

Mathematica

a = {1, 3}; Do[If[EvenQ[n], AppendTo[a, 3*a[[ -1]] + a[[ -2]]], AppendTo[a, a[[ -1]] + a[[ -2]]]], {n, 3, 30}]; a (* Stefan Steinerberger, Dec 31 2007 *)
a[n_] := FromContinuedFraction[ Join[{0}, 3 - 2*Array[Mod[#, 2]&, n]]] // Numerator; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, May 15 2014 *)

PARI

{a(n) = (-1)^((n+1) * (n<0)) * polcoeff( x * (1 + 3*x - x^2) / (1 - 5*x^2 + x^4) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, May 15 2014 */

Formula

a(0) = 0, a(1) = 1, a(2n) = 3*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Given the 2 X 2 matrix [1, 3; 1, 4] = T, [a(2n-1), a(2n)] = top row of T^n.

g.f.: x*(1+3*x-x^2)/(1-5*x^2+x^4). - Colin Barker, Jan 04 2012

a(-n) = -(-1)^n * a(n). a(2*n - 1) = A004253(n). a(2*n) = 3 * A004254(n). - Michael Somos, May 15 2014

a(n+1) - a(n-1) = a(n) * (2 - (-1)^n) for all n in Z. - Michael Somos, May 15 2014

Extensions

More terms from Stefan Steinerberger, Dec 31 2007