A099463 Bisection of tribonacci numbers.

0, 1, 2, 7, 24, 81, 274, 927, 3136, 10609, 35890, 121415, 410744, 1389537, 4700770, 15902591, 53798080, 181997601, 615693474, 2082876103, 7046319384, 23837527729, 80641778674, 272809183135, 922906855808, 3122171529233

Offset

0,3

Comments

Binomial transform of A099462.

From Paul Barry, Feb 07 2006: (Start)

a(n+1) gives row sums of number triangle A114123 or A184883.

Partial sums are A113300. (End)

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,1,1).

Formula

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

a(n) = Sum_{k=0..n} binomial(n, k)*Sum_{j=0..floor((k-1)/2)} binomial(j, k-2*j-1)*4^j.

From Paul Barry, Feb 07 2006: (Start)

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

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

a(n)/a(n-1) tends to 3.38297576..., the square of the tribonacci constant A058265. - Gary W. Adamson, Feb 28 2006

If p[1]=2, p[2]=3, p[i]=4, (i>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>=1, a(n+1) = det A. - Milan Janjic, May 02 2010

Mathematica

LinearRecurrence[{3, 1, 1}, {0, 1, 2}, 30] (* or *) Join[{0}, Mean/@ Partition[ LinearRecurrence[ {1, 1, 1}, {1, 1, 1}, 60], 2]] (* Harvey P. Dale, Apr 02 2012 *)

Prog

(Magma) [n le 3 select (n-1) else 3*Self(n-1) +Self(n-2) +Self(n-3): n in [1..31]]; // G. C. Greubel, Nov 20 2021
(Sage)
def A184883(n, k): return simplify( hypergeometric([-k, 2*(k-n)], [1], 2) )
def A099463(n): return sum( A184883(n, k) for k in (0..n) )
[0]+[A099463(n-1) for n in (1..40)] # G. C. Greubel, Nov 20 2021

Crossrefs

Cf. A000073, A058265, A099462, A113300, A114123, A184883.

Sequence in context: A038765 A027126 A027128 * A021000 A003480 A020727

Adjacent sequences: A099460 A099461 A099462 * A099464 A099465 A099466

Keyword

easy,nonn

Author

Paul Barry, Oct 16 2004

Status

approved