0,2

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

Number of binary sequences of length n+2 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

L. Turban, Lattice animals on a staircase and Fibonacci numbers, J.Phys. A 33 (2000) 2587-2595.

Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

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)).

Comments from Peter Bala (pbala(AT)toucansurf.com), 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)=2a(n+3)+a(n+2)-2a(n+1)-a(n)+F(n+3), a(0)=1, a(1)=2, a(2)=6, a(3)=13; a(n+2)=a(n+1)+a(n)+A010049(n+3), a(0)=1, 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)

A001628, A000045.

Cf. A010049, A094440, A129707.

Adjacent sequences: A055240 A055241 A055242 this_sequence A055244 A055245 A055246

Sequence in context: A031872 A124677 A034465 this_sequence A075632 A115217 A094687

nonn,easy

Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), May 10 2000