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%I #28 Jul 18 2023 12:26:13
%S 1,1,3,14,75,433,2636,16668,108399,720431,4871555,33409042,231817448,
%T 1624503716,11480658056,81731416480,585579734959,4219179476875,
%U 30552067317233,222225174139730,1622894404239115,11894991079960721,87472260252499560,645183802300787356,4771926560361458884
%N G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x^2*A(x)^6).
%C More generally, for fixed parameters p and q, if F(x) satisfies:
%C F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
%C then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)).
%H Vaclav Kotesovec, <a href="/A200718/b200718.txt">Table of n, a(n) for n = 0..1000</a>
%H Vaclav Kotesovec, <a href="/A200718/a200718.txt">Recurrence (of order 6)</a>
%F G.f. A(x) satisfies:
%F (1) A(x) = sqrt( (1/x)*Series_Reversion( 2*x^5*(1+x)^2/(1 - 2*x^2*(1+x)^2 - sqrt(1 - 4*x^2*(1+x)^2)) ) ).
%F (2) A(x) = G(x*A(x)^2) where G(x) = A(x/G(x)^2) is the g.f. of A104545 (Motzkin paths of length n having no consecutive (1,0) steps).
%F (3) A(x) = exp( Sum_{n>=1} x^n * A(x)^n/n * [Sum_{k=0..n} C(n,k)^2 * x^k * A(x)^(4*k)] ).
%F (4) A(x) = exp( Sum_{n>=1} x^n * A(x)^n/n * [(1-x/A(x)^4)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2*x^k * A(x)^(4*k)] ).
%F a(n) = Sum_{k=0..floor(n/2)}((binomial(2*n+2*k+1,k)*binomial(2*n+2*k+1,n-2*k))/(2*n+2*k+1)). - _Vladimir Kruchinin_, Mar 11 2016
%e G.f.: A(x) = 1 + x + 3*x^2 + 14*x^3 + 75*x^4 + 433*x^5 + 2636*x^6 +...
%e Related expansions:
%e A(x)^2 = 1 + 2*x + 7*x^2 + 34*x^3 + 187*x^4 + 1100*x^5 + 6784*x^6 +...
%e A(x)^6 = 1 + 6*x + 33*x^2 + 194*x^3 + 1200*x^4 + 7674*x^5 + 50317*x^6 +...
%e A(x)^8 = 1 + 8*x + 52*x^2 + 336*x^3 + 2210*x^4 + 14776*x^5 + 100216*x^6 +...
%e where A(x) = 1 + x*A(x)^2 + x^2*A(x)^6 + x^3*A(x)^8.
%e The logarithm of the g.f. A = A(x) equals the series:
%e log(A(x)) = (1 + x*A^4)*x*A + (1 + 2^2*x*A^4 + x^2*A^8)*x^2*A^2/2 +
%e (1 + 3^2*x*A^4 + 3^2*x^2*A^8 + x^3*A^12)*x^3*A^3/3 +
%e (1 + 4^2*x*A^4 + 6^2*x^2*A^8 + 4^2*x^3*A^12 + x^4*A^16)*x^4*A^4/4 +
%e (1 + 5^2*x*A^4 + 10^2*x^2*A^8 + 10^2*x^3*A^12 + 5^2*x^4*A^16 + x^5*A^20)*x^5*A^5/5 + ...
%e The g.f. of A104545, G(x) = A(x/G(x)^2) where A(x) = G(x*A(x)^2), begins:
%e G(x) = 1 + x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 55*x^7 + 129*x^8 +...
%t a[n_] := Sum[Binomial[2*n + 2*k + 1, k]*Binomial[2*n + 2*k + 1, n - 2*k]/ (2*n + 2*k + 1), {k, 0, n/2}];
%t Table[a[n], {n, 0, 24}] (* _Jean-François Alcover_, Jan 09 2018, after _Vladimir Kruchinin_ *)
%o (PARI) {a(n)=polcoeff(sqrt( (1/x)*serreverse( 2*x^5*(1+x)^2/(1 - 2*x^2*(1+x)^2 - sqrt(1 - 4*x^2*(1+x)^2+O(x^(n+6)))) ) ),n)}
%o (PARI) {a(n)=local(p=1,q=4,A=1+x);for(i=1,n,A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n));polcoeff(A,n)}
%o (PARI) {a(n)=local(p=1,q=4,A=1+x);for(i=1,n,A=exp(sum(m=1,n,x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0,m,binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}
%o (PARI) {a(n)=local(p=1,q=4,A=1+x);for(i=1,n,A=exp(sum(m=1,n,x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}
%o (Maxima)
%o a(n):=sum((binomial(2*n+2*k+1,k)*binomial(2*n+2*k+1,n-2*k))/(2*n+2*k+1),k,0,(n)/2); /* _Vladimir Kruchinin_, Mar 11 2016 */
%Y Cf. A104545, A200716, A200717, A199876, A199877, A198951, A198953, A198957, A192415, A198888, A036765.
%Y Cf. A186241, A199874, A200074, A200075, A200719, A215576.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Nov 21 2011
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