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| COMMENTS
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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) = 3, s(2n) = 5.
The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. [. (1-sqrt(1-4x/(1+(1-t)x))]/)))/2 and inverse x(*(1-x)/[)/(1+( + (t-1))*x(*(1-x)]. )). See A091867 for more info on this family. Here t=- = -4 (mod signs in the results).
Let C(x) = [) = (1 - sqrt(1-4x)]/))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
O.g.f.: G(x) = x*(1-x)/[)/(1 - 5x*(1-x)] = )) = P[(Cinv(x),-5]. ).
Inverse O.g.f.: Ginv(x) = [) = (1 - sqrt(1 - 4*x/(1+5x))]/)))/2 = C[(P(x,5)] ()) (signed A026378). Cf. A030528. (End)
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| FORMULA
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G.f.: x*(1-x)/(1-5*x+5*x^2)= ) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523).
From Paul Barry, Apr 16 2004: (Start)
Binomial transform of Fibonacci(2n+2).
Binomial transform of Fib(2n+2). a(n)=() = (sqrt(5)/2+ + 5/2)^n*(3*sqrt(5)/10+ + 1/2)-() - (5/2- - sqrt(5)/2)^n*(3*sqrt(5)/10- - 1/2). - _Paul Barry_, Apr 16 2004). (End)
a(n) = (1/5)*Sum(_{r, =1, ..9, } sin(3*r*Pi/10))*sin(r*Pi/2)()*(2*cos(r*Pi/10))^(2n)).
LimLimit_{k->infinity} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.
LimLimit_{n->infinity} A020876(n)/A093131(n) = sqrt(5).
LimLimit_{k->infinity} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.
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