Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003

a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003

Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004

a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006

a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006

Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007

Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011

These numbers could also be called "bronze Fibonacci numbers". Indeed, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio; analogously, for Pell numbers (A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013

Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013

The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014

a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014

For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017

This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).

gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)

Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018

Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020

Also called the 3-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 3 kinds of squares available. (End)

a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023

H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.

L.-N. Machaut, Query 3436, L'Intermédiaire des Mathématiciens, 16 (1909), 62-63. - N. J. A. Sloane, Mar 08 2022

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178. (Annotated scanned copy)

W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.

Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 2.

G.f.: x/(1 - 3*x - x^2).

a(2*n) = 3*A004190(n-1); a(3*n) = 10*A041613(n-1) for n >= 1. (End)

a(n-1) + a(n+1) = A006497(n).

A006497(n)^2 - 13*a(n)^2 = 4(-1)^n. (End)

a(n) = U(n-1, (3/2)i)(-i)^(n-1), i^2 = -1. - Paul Barry, Nov 19 2003

a(n) = Sum_{k=0..n-1} binomial(n-k-1,k)*3^(n-2*k-1). - Paul Barry, Oct 02 2004

a(n) = F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.

Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005 [Or a(n) = [M^(n+1)]_{1,1}. - L. Edson Jeffery, Aug 27 2013]

a(n+1) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(k-j).

a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-j-k).

a(n+1) = Sum_{k=0..floor(n/2)} C(n-k,k)*3^(n-2*k).

a(n) = Sum_{k=0..n} C(k,n-k)*3^(2*k-n). (End)

E.g.f.: exp(3*x/2)*sinh(sqrt(13)*x/2)/(sqrt(13)/2). - Paul Barry, Jun 03 2006

a(n) = (ap^n - am^n)/(ap - am), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2.

Let C = (3 + sqrt(13))/2 = exp arcsinh(3/2) = 3.3027756377... Then C^n, n > 0 = a(n)*(1/C) + a(n+1). Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007

1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... . - Gary W. Adamson, Mar 16 2008

a(n) = ((3 + sqrt(13))^n - (3 - sqrt(13))^n)/(2^n*sqrt(13)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009

a(p) == 13^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009

Limit_{k->oo} a(n+k)/a(k) = (A006497(n) + a(n)*sqrt(13))/2.

Limit_{n->oo} A006497(n)/a(n) = sqrt(13). (End)

Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (sqrt(13)-3)/2. - Vladimir Shevelev, Feb 23 2013

(1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a(n)^2 + 4*(-1)^n))/2;

(2) a^2(n+1) - a(n)*a(n+2) = (-1)^n;

(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);

(4) a(n)/a(n+1) = (sqrt(13)-3)/2 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)

Sum_{n >= 1} 1/( a(2*n) + 1/a(2*n) ) = 1/3; Sum_{n >= 1} 1/( a(2*n + 1) - 1/a(2*n + 1) ) = 1/9. - Peter Bala, Mar 26 2015

Some properties:

(1) a(n)*a(n+1) = 3*Sum_{k=1..n} a(k)^2;

(2) a(n)^2 + a(n+1)^2 = a(2*n+1);

(3) a(n)^2 - a(n-2)^2 = 3*a(n-1)*(a(n) + a(n-2));

(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);

(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;

(6) a(2*n) = a(n)*(3*a(n) + 2*a(n-1));

(7) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;

(8) a(n) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = ap^(2*k-1) + am^(2*k-1), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2;

(9) 131|Sum_{k=n..n+9} a(k), for all positive n. (End)

From Kai Wang, Feb 10 2020: (Start)

a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2 - Catalan's identity.

arctan(1/a(2n)) - arctan(1/a(2n+2)) = arctan(a(2)/a(2n+1)).

arctan(1/a(2n)) = Sum_{m>=n} arctan(a(2)/a(2m+1)).

The same formula holds for Fibonacci numbers and Pell numbers. (End)

a(n+2) = 3^(n+1) + Sum_{k=0..n} a(k)*3^(n-k). - Greg Dresden and Gavron Campbell, Feb 22 2022

G.f. = 1/(1-Sum_{k>=1} P(k)*x^k), P(k)=A000129(k) (with a(0)=1). - Enrique Navarrete, Dec 17 2023

G.f.: x/(1 - 3*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 3 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Sum_{n>=0} a(n)/phi^(3*n) = 1, where phi = A001622 is the golden ratio. - Diego Rattaggi, Apr 07 2025

From the comment on compositions with Pell number of parts, A000129(k), there are A000129(1)=1 type of 1, A000129(2)=2 types of 2, A000129(3)=5 types of 3, A000129(4)=12 types of 4, A000129(5)=29 types of 5 and A000129(6)=70 types of 6.

The following table gives the number of compositions of n=6:

Composition, number of such compositions, number of compositions of this type:

6, 1, 70;

5+1, 2, 58;

4+2, 2, 48;

3+3, 1, 25;

4+1+1, 3, 36;

3+2+1, 6, 60;

2+2+2, 1, 8;

3+1+1+1, 4, 20;

2+2+1+1, 6, 24;

2+1+1+1+1, 5, 10;

1+1+1+1+1+1, 1, 1;

for a total of a(6)=360 compositions of n=6. (End).

a[0]:=0: a[1]:=1: for n from 2 to 35 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n], n=0..30); # Emeric Deutsch, Sep 03 2007

A006190:=-1/(-1+3*z+z**2); # Simon Plouffe in his 1992 dissertation, without the leading 0

seq(combinat[fibonacci](n, 3), n=0..30); # R. J. Mathar, Dec 07 2011

a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (* Robert G. Wilson v, Jan 13 2005 *)

LinearRecurrence[{3, 1}, {0, 1}, 30] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x, 0, 30}], x] (* Harvey P. Dale, Apr 20 2011 *)

Table[If[n==0, a1=1; a0=0, a2=a1; a1=a0; a0=3*a1+a2], {n, 0, 30}] (* Jean-François Alcover, Apr 30 2013 *)

(PARI) a(n)=if(n<1, 0, contfracpnqn(vector(n, i, 2+(i>1)))[2, 1])

(SageMath) [lucas_number1(n, 3, -1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009

(Magma)[ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011

(Haskell)

a006190 n = a006190_list !! n

a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list

(PARI) concat([0], Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013

(GAP) a:=[0, 1];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018

Row sums of Pascal's rhombus (A059317). Also row sums of triangle A054456(n, m).