A055243 First differences of A001628 (Fibonacci convolution).

1, 2, 6, 13, 29, 60, 122, 241, 468, 894, 1686, 3144, 5807, 10636, 19338, 34931, 62731, 112068, 199264, 352787, 622152, 1093260, 1914780, 3343440, 5821645, 10110278, 17515566, 30276073, 52221929, 89896332, 154461110, 264930661, 453654108, 775598634, 1324053522

Offset: 0

Author

Wolfdieter Lang, May 10 2000

Keywords

• nonn
• easy

Comments

2*a(n) = C_{n+3} of Turban reference eq.(2.17), C_{1}= 0 = C_{2}.

Number of binary sequences of length n+3 such that the sequence has exactly two pairs (which may overlap) of consecutive 1's. - George J. Schaeffer (gschaeff(AT)andrew.cmu.edu), Sep 07 2004

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• L. Turban, Lattice animals on a staircase and Fibonacci numbers, arXiv:cond-mat/0011038 [cond-mat.stat-mech], 2000; J.Phys. A 33 (2000) 2587-2595.
• Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
• Index entries for linear recurrences with constant coefficients, signature (3,0,-5,0,3,1).

Crossrefs

Cf. A001628, A000045.

Cf. A010049, A076791, A094440, A129707.

Programs

Maple

a:= n -> (Matrix([[1,0$4,-1]]). Matrix(6, (i,j)-> if (i=j-1) then 1 elif j=1 then [3,0,-5,0,3,1][i] else 0 fi)^(n))[1,1]: seq(a(n), n=0..30); # Alois P. Heinz, Aug 05 2008

Mathematica

Differences[LinearRecurrence[{3,0,-5,0,3,1},{0,1,3,9,22,51,111},40]] (* Harvey P. Dale, Jun 12 2019 *)

Formula

G.f.: (1-x)/(1-x-x^2)^3. (from Turban reference eq.(2.15)).

a(n)= ((5*n^2+37*n+50)*F(n+1)+4*(n+1)*F(n))/50 with F(n)=A000045(n) (Fibonacci numbers) (from Turban reference eq. (2.17)).

From Peter Bala, Oct 25 2007 (Start):

Since F(-n) = (-1)^(n+1)*F(n), we can use the previous formula to extend the sequence to negative values of n; we find a(-n) = (-1)^n* A129707(n-3).

Recurrence relations: a(n+4) = 2*a(n+3) + a(n+2) - 2*a(n+1) - a(n) + F(n+3), with a(0) = 1, a(1) = 2, a(2) = 6 and a(3) = 13;

a(n+2) = a(n+1) + a(n) + A010049(n+3), with a(0) = 1 and a(1) = 2.

a(n-3) = Sum_{k = 2..floor((n+1)/2)} C(k,2)*C(n-k,k-1) = (1/2)*G''(n,1), where the polynomial G(n,x) := Sum_{k = 1..floor((n+1)/2)} C(n-k,k-1)*x^k = x^((n+1)/2) * F(n, 1/sqrt(x)) and where F(n,x) denotes the n-th Fibonacci polynomial. Since G(n,1) yields the Fibonacci numbers A000045 and G'(n,1) yields the second-order Fibonacci numbers A010049, a(n) may be considered as the sequence of third-order Fibonacci numbers.

For n >= 4, the polynomials Sum_{k = 0..n} C(n,k) * G''(n-k,1)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the remarks in A094440 and A010049. (End)

a(n) = A076791(n+3, 2). - Michael Somos, Sep 24 2024

E.g.f.: exp(x/2)*(5*(25 + 23*x + 5*x^2)*cosh(sqrt(5)*x/2) + sqrt(5)*(29 + 65*x + 10*x^2)*sinh(sqrt(5)*x/2))/125. - Stefano Spezia, Sep 26 2024