%I #16 Jan 10 2019 22:54:16

%S 1,1,1,3,9,25,77,247,801,2657,8969,30635,105785,368745,1295493,

%T 4582767,16309953,58357313,209798289,757461011,2745281705,9984464761,

%U 36428252541,133293594343,489028250465,1798543861537,6629635284505

%N G.f. satisfies: A(x) = 1 + x*A(x)/(1 - 2*x^2*A(x)^2).

%C Formula may be derived using the Lagrange Inversion theorem (cf. A049124).

%H Vincenzo Librandi, <a href="/A101786/b101786.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1, k)*C(n, 2*k)*2^k/(2*k+1) for n>0, with a(0)=1.

%F G.f.: A(x) = (1/x)*Series_Reversion( x*(1 - 2*x^2)/(1+x - 2*x^2) ).

%F Recurrence: 2*n*(n+1)*(31*n^2 - 127*n + 120)*a(n) = 3*n*(62*n^3 - 285*n^2 + 359*n - 88)*a(n-1) + (62*n^4 - 378*n^3 + 1009*n^2 - 1425*n + 792)*a(n-2) + (n-3)*(682*n^3 - 3135*n^2 + 4133*n - 1272)*a(n-3) - 9*(n-4)*(n-3)*(31*n^2 - 65*n + 24)*a(n-4). - _Vaclav Kotesovec_, Sep 17 2013

%F a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 3/4 + 1/(4*sqrt(3/(35 - (176*2^(2/3))/(9959 + 465*sqrt(465))^(1/3) + 2*(19918 + 930*sqrt(465))^(1/3)))) + 1/2*sqrt(35/6 + (44*2^(2/3))/(3*(9959 + 465*sqrt(465))^(1/3)) - (9959 + 465*sqrt(465))^(1/3)/(3*2^(2/3)) + 127/2*sqrt(3/(35 - (176*2^(2/3))/ (9959 + 465*sqrt(465))^(1/3) + 2*(19918 + 930*sqrt(465))^(1/3)))) = 3.9027270552404829297969 = ... is the root of the equation 9 - 22*d - 2*d^2 - 6*d^3 + 2*d^4 = 0 and c = 0.68546565145612597016100560323891887595749... - _Vaclav Kotesovec_, Sep 17 2013

%e Generated from Fibonacci polynomials (A011973) and coefficients of odd powers of 1/(1-x):

%e a(1) = 1*1/1

%e a(2) = 1*1/1 + 0*1*2/3

%e a(3) = 1*1/1 + 1*3*2/3

%e a(4) = 1*1/1 + 2*6*2/3 + 0*1*2^2/5

%e a(5) = 1*1/1 + 3*10*2/3 + 1*5*2^2/5

%e a(6) = 1*1/1 + 4*15*2/3 + 3*15*2^2/5 + 0*1*2^3/7

%e a(7) = 1*1/1 + 5*21*2/3 + 6*35*2^2/5 + 1*7*2^3/7

%e a(8) = 1*1/1 + 6*28*2/3 + 10*70*2^2/5 + 4*28*2^3/7 + 0*1*2^4/9

%e This process is equivalent to the formula:

%e a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1,k)*C(n,2*k)*2^k/(2*k+1).

%t Flatten[{1,Table[Sum[Binomial[n-k-1,k]*Binomial[n,2*k]*2^k/(2*k+1),{k,0,Floor[(n-1)/2]}],{n,1,20}]}] (* _Vaclav Kotesovec_, Sep 17 2013 *)

%t ShiftedReversion[ser_, n_, sgn_] := CoefficientList[(sgn/x)InverseSeries[Series[x sgn ser, {x, 0, n}]],x];

%t Jacobsthal := (2x^2 - 1)/((x + 1)(2x - 1)); (* with A001045(0) = 1 *)

%t ShiftedReversion[Jacobsthal, 27, -1] (* _Peter Luschny_, Jan 10 2019 *)

%o (PARI) {a(n)=if(n==0,1,sum(k=0,(n-1)\2, binomial(n-k-1,k)*binomial(n,2*k)*2^k/ (2*k+1)))}

%Y Cf. A011973, A001045, A049124, A101785.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Dec 16 2004