A095263 a(n+3) = 3*a(n+2) - 2*a(n+1) + a(n).

1, 3, 7, 16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732, 62608681, 145547525, 338356945, 786584466, 1828587033, 4250949112, 9882257736, 22973462017, 53406819691

Offset: 1

Author

Gary W. Adamson, May 31 2004

Keywords

• nonn
• easy

Comments

a(n+1) = number of n-tuples over {0,1,2} without consecutive digits. For the general case see A096261.

Diagonal sums of Riordan array (1/(1-x)^3, x/(1-x^3)), A127893. - Paul Barry, Jan 07 2008

The signed variant (-1)^(n+1)*a(n+1) is the bottom right entry of the n-th power of the matrix [[0,1,0],[0,0,1],[-1,-2,-3]]. - Roger L. Bagula, Jul 01 2007

a(n) is the number of generalized compositions of n+1 when there are i^2/2-i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010

Dedrickson (Section 4.1) gives a bijection between colored compositions of n, where each part k has one of binomial(k,2) colors, and 0,1,2 strings of length n-2 without sequential digits (i.e., avoiding 01 and 12). Cf. A052529. - Peter Bala, Sep 17 2013

Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S^2 - S^3; see A291000. - Clark Kimberling, Aug 24 2017

For n>1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 2 blocks (i.e., subintervals) from each interval. For example, for n=6, a(5)=37 since the number of ways to split [6] into intervals and then select 2 blocks from each interval is C(6,2) + C(4,2)*C(2,2) + C(3,2)*C(3,2) + C(2,2)*C(4,2) + C(2,2)*C(2,2)*C(2,2). - Enrique Navarrete, May 20 2022

Examples

			a(9) = 1081 = 3*465 - 2*200 + 86.
M^9 * [1 0 0] = [a(7) a(8) a(9)] = [200 465 1081].
G.f. = x + 3*x^2 + 7*x^3 + 16*x^4 + 37*x^5 + 86*x^6 + 200*x^7 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• C. R. Dedrickson III, Compositions, Bijections, and Enumerations Thesis, Jack N. Averitt College of Graduate Studies, Georgia Southern University, 2012.
• Index entries for linear recurrences with constant coefficients, signature (3,-2,1).

Crossrefs

Cf. A000931, A010912, A034943, A052529, A052530, A097550, A137531.

Cf. A052921 (first differences), A137229 (partial sums).

Column k=3 of A277666.

Programs

Magma

I:=[1,3,7]; [n le 3 select I[n] else 3*Self(n-1) -2*Self(n-2) +Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 12 2021

Maple

A:= gfun:-rectoproc({a(n+3)=3*a(n+2)-2*a(n+1)+a(n),a(1)=1,a(2)=3,a(3)=7},a(n),remember):
seq(A(n),n=1..100); # Robert Israel, Sep 15 2014

Mathematica

a[1]=1; a[2]=3; a[3]=7; a[n_]:= a[n]= 3a[n-1] -2a[n-2] +a[n-3]; Table[a[n], {n, 22}] (* Or *)
a[n_]:= (MatrixPower[{{0,1,2,3}, {1,2,3,0}, {2,3,0,1}, {3,0,1,2}}, n].{{1}, {0}, {0}, {0}})[[2, 1]]; Table[ a[n], {n, 22}] (* Robert G. Wilson v, Jun 16 2004 *)
RecurrenceTable[{a[1]==1,a[2]==3,a[3]==7,a[n+3]==3a[n+2]-2a[n+1]+a[n]},a,{n,30}] (* Harvey P. Dale, Sep 17 2022 *)

Sage

[sum( binomial(n+k+1,3*k+2) for k in (0..(n-1)//2)) for n in (1..30)] # G. C. Greubel, Apr 12 2021

Formula

Let M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 -2 3]; then M^n *[1 0 0] = [a(n-2) a(n-1) a(n)].

a(n)/a(n-1) tends to 2.3247179572..., an eigenvalue of M and a root of the characteristic polynomial. [Is that constant equal to 1 + A060006? - Michel Marcus, Oct 11 2014] [Yes, the limit is the root of the equation -1 + 2*x - 3*x^2 + x^3 = 0, after substitution x = y + 1 we have the equation for y: -1 - y + y^3 = 0, y = A060006. - Vaclav Kotesovec, Jan 27 2015]

Related to the Padovan sequence A000931 as follows : a(n)=A000931(3n+4). Also the binomial transform of A000931(n+4).

From Paul Barry, Jul 06 2004: (Start)

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

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

From Paul Barry, Jan 07 2008: (Start)

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

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k+2,3*k+2).

a(n) = Sum_{k=0..n} binomial(n,k) * Sum_{j=0..floor((k+4)/2)} binomial(j,k-2j+4). (End)

If p[i]=i(i-1)/2 and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=det A. - Milan Janjic, May 02 2010

a(n) = A000931(3*n + 4). - Michael Somos, Sep 18 2012

Extensions

Edited by Paul Barry, Jul 06 2004

Corrected and extended by Robert G. Wilson v, Jun 05 2004