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%I #27 Nov 13 2022 10:36:00
%S 1,2,6,13,29,60,122,241,468,894,1686,3144,5807,10636,19338,34931,
%T 62731,112068,199264,352787,622152,1093260,1914780,3343440,5821645,
%U 10110278,17515566,30276073,52221929,89896332,154461110,264930661,453654108,775598634,1324053522
%N First differences of A001628 (Fibonacci convolution).
%C 2*a(n) = C_{n+3} of Turban reference eq.(2.17), C_{1}= 0 = C_{2}.
%C 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
%H Alois P. Heinz, <a href="/A055243/b055243.txt">Table of n, a(n) for n = 0..1000</a>
%H L. Turban, <a href="http://arxiv.org/abs/cond-mat/0011038">Lattice animals on a staircase and Fibonacci numbers</a>, arXiv:cond-mat/0011038 [cond-mat.stat-mech], 2000; J.Phys. A 33 (2000) 2587-2595.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-5,0,3,1)
%F G.f.: (1-x)/(1-x-x^2)^3. (from Turban reference eq.(2.15)).
%F 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)).
%F From _Peter Bala_, Oct 25 2007 (Start):
%F 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).
%F 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;
%F a(n+2) = a(n+1) + a(n) + A010049(n+3), with a(0) = 1 and a(1) = 2.
%F 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.
%F 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)
%p 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
%t Differences[LinearRecurrence[{3,0,-5,0,3,1},{0,1,3,9,22,51,111},40]] (* _Harvey P. Dale_, Jun 12 2019 *)
%Y Cf. A001628, A000045.
%Y Cf. A010049, A094440, A129707.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, May 10 2000
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