OFFSET
1,2
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
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).
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
EXAMPLE
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 + ...
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gary W. Adamson, May 31 2004
EXTENSIONS
Edited by Paul Barry, Jul 06 2004
Corrected and extended by Robert G. Wilson v, Jun 05 2004
STATUS
approved