

A060925


a(n) = 2a(n1) + 3a(n2), a(0) = 1, a(1) = 4.


12



1, 4, 11, 34, 101, 304, 911, 2734, 8201, 24604, 73811, 221434, 664301, 1992904, 5978711, 17936134, 53808401, 161425204, 484275611, 1452826834, 4358480501, 13075441504, 39226324511, 117678973534, 353036920601
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OFFSET

0,2


COMMENTS

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=1, A[i,i1]=1, and A[i,j]=0 otherwise. Then, for n>=1, a(n1)=charpoly(A,2).  Milan Janjic, Jan 26 2010


LINKS

Harry J. Smith, Table of n, a(n) for n=0,...,200
Index entries for linear recurrences with constant coefficients, signature (2, 3).


FORMULA

Row sums of Lucas convolution triangle A060922.
Inverse binomial transform of A003947.  Philippe Deléham, Jul 23 2005
a(n) = sum_{m=0..n} A060922(n, m) = sum_{j=1..n} (a(j1)*A000204(nj+1)) + A000204(n+1).
a(n) = (5*3^n  (1)^n)/4.
G.f.: (1+2*x)/(1  2*x  3*x^2).
a(2n) = 3a(2n1)  1; a(2n+1) = 3a(2n) + 1.  Philippe Deléham, Jul 23 2005
Binomial transform is A003947.  Paul Barry, May 19 2003


MATHEMATICA

f[n_]:=3/(n+2); x=2; Table[x=f[x]; Denominator[x], {n, 0, 5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 11 2010 *)
LinearRecurrence[{2, 3}, {1, 4}, 30] (* Harvey P. Dale, Mar 07 2014 *)


PROG

(PARI) { for (n=0, 200, write("b060925.txt", n, " ", (5*3^n  (1)^n)/4); ) } \\ Harry J. Smith, Jul 19 2009


CROSSREFS

Sequence in context: A327548 A144791 A180305 * A027045 A243781 A227329
Adjacent sequences: A060922 A060923 A060924 * A060926 A060927 A060928


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Apr 20 2001


EXTENSIONS

Recurrence, now used as definition, from Philippe Deléham, Jul 23 2005
Entry revised by N. J. A. Sloane, Sep 10 2006


STATUS

approved



