A072263 a(n) = 3*a(n-1) + 5*a(n-2), with a(0)=2, a(1)=3.

2, 3, 19, 72, 311, 1293, 5434, 22767, 95471, 400248, 1678099, 7035537, 29497106, 123669003, 518492539, 2173822632, 9113930591, 38210904933, 160202367754, 671661627927, 2815996722551, 11806298307288, 49498878534619, 207528127140297

Offset: 0

Author

Miklos Kristof, Jul 08 2002

Keywords

• nonn
• easy

Comments

Inverse binomial transform of A087130. - Johannes W. Meijer, Aug 01 2010

Pisano period lengths: 1, 3, 4, 6, 4, 12, 3, 12, 12, 12, 120, 12, 12, 3, 4, 24, 288, 12, 72, 12... - R. J. Mathar, Aug 10 2012

This is the Lucas sequence V(3,-5). - Bruno Berselli, Jan 09 2013

Examples

			a(5)=5*b(4)+b(6): 1293=5*57+1008.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Wikipedia, Lucas sequence: Specific names.
• Index entries for linear recurrences with constant coefficients, signature (3,5).

Crossrefs

Cf. A072264, A152187, A197189.

Appears in A179606 and A015523. - Johannes W. Meijer, Aug 01 2010

Programs

GAP

a:=[2,3];; for n in [3..40] do a[n]:=3*a[n-1]+5*a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

Magma

I:=[2,3]; [n le 2 select I[n] else 3*Self(n-1) +5*Self(n-2): n in [1..40]]; // G. C. Greubel, Jan 14 2020

Maple

seq(coeff(series((2-3*x)/(1-3*x-5*x^2), x, n+1), x, n), n = 0..40); # G. C. Greubel, Jan 14 2020

Mathematica

LinearRecurrence[{3,5},{2,3},40] (* Harvey P. Dale, Nov 23 2018 *)

PARI

my(x='x+O('x^40)); Vec((2-3*x)/(1-3*x-5*x^2)) \\ G. C. Greubel, Jan 14 2020

Sage

[lucas_number2(n,3,-5) for n in range(0, 16)] # Zerinvary Lajos, Apr 30 2009

Formula

a(n) = 2*A015523(n+1) - 3*A015523(n).

a(n) = ((3 + sqrt(29))/2)^n + ((3 - sqrt(29))/2)^n.

G.f.: (2-3*x)/(1-3*x-5*x^2). - R. J. Mathar, Feb 06 2010

From Johannes W. Meijer, Aug 01 2010: (Start)

Limit_{k -> Infinity} a(n+k)/a(k) = (A072263(n) + A015523(n)*sqrt(29))/2.

Limit_{n -> Infinity} (A072263(n)/A015523(n)) = sqrt(29). (End)

G.f.: G(0), where G(k)= 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013

a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 29*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015

a(n) = 5^((n-1)/2)*( 2*sqrt(5)*Fibonacci(n+1, 3/sqrt(5)) - 3*Fibonacci(n, 3/sqrt(5)) ). - G. C. Greubel, Jan 14 2020

Extensions

Offset changed and terms added by Johannes W. Meijer, Jul 19 2010