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
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 24.
- Meng-Han Wu, Henryk A. Witek, and Rafał Podeszwa, Clar Covers and Zhang-Zhang Polynomials of Zigzag and Armchair Carbon Nanotubes, MATCH Commun. Math. Comput. Chem. (2025) Vol. 93, 415-462. See p. 437.
- Index entries for linear recurrences with constant coefficients, signature (3,1,1).
Programs
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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
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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 *) -
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
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
Comments