OFFSET
0,2
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).
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
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, May 10 2000
STATUS
approved