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%I #27 Aug 04 2021 21:05:13
%S 0,1,3,9,28,90,297,1001,3431,11917,41820,147918,526309,1881009,
%T 6744843,24244145,87300092,314765506,1135980801,4102551897,
%U 14823628015,53581222773,193724727804,700551945014,2533702591613,9164618329825,33151607475987,119927166988761
%N a(n) = Sum_{k=0..n} binomial(n,k) * binomial(Fibonacci(k)+1,2).
%C Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 3. - _Herbert Kociemba_, Jun 14 2004
%H Michael De Vlieger, <a href="/A033190/b033190.txt">Table of n, a(n) for n = 0..1792</a>
%H László Németh and László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Nemeth/nemeth8.html">Sequences Involving Square Zig-Zag Shapes</a>, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,-21,20,-5).
%F G.f.: (-x^4+6x^3-5x^2+x)/((1-3x+x^2)*(1-5x+5x^2)).
%F From _Herbert Kociemba_, Jun 14 2004: (Start)
%F a(n) = (1/5)*Sum_{r=1..9} sin(r*Pi/10)*sin(3*r*Pi/10)*(2*cos(r*Pi/10))^(2*n), n >= 1.
%F a(n) = 8*a(n-1) - 21*a(n-2) + 20*a(n-3) - 5*a(n-4), n >= 5. (End)
%F From _Greg Dresden_, Jan 24 2021: (Start)
%F a(2n) = (5*Fibonacci(4*n) + (5^n)*Lucas(2*n))/10 for n > 0.
%F a(2n+1) = (Fibonacci(4*n+2) + (5^n)*Fibonacci(2*n+1))/2 for n >= 0.
%F (End)
%p A033190 := proc(n)
%p add(binomial(n,k)*binomial(combinat[fibonacci](k)+1,2),k=0..n) ;
%p end proc: # _R. J. Mathar_, Feb 18 2016
%t LinearRecurrence[{8,-21,20,-5},{0,1,3,9,28},30] (* _Harvey P. Dale_, Jan 24 2019 *)
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
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