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%I #46 Sep 08 2022 08:45:26
%S 1,1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,
%T 9694845,35357670,129644790,477638700,1767263190,6564120420,
%U 24466267020,91482563640,343059613650,1289904147324,4861946401452,18367353072152
%N G.f. is 1 + x*c(x), where c(x) is the g.f. of the Catalan numbers (A000108).
%C Previous name was: G.f. satisfies: 3*A(x) = 2 + x + A(x)^2, with A(0) = 1.
%C This is essentially a duplicate of entry A000108, the Catalan numbers (a(n) = A000108(n-1) for n>0).
%C In order for the g.f. of an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n, where n > 1, it is necessary that the sequence start with [1, d, m*n*(n-1)/2], where d divides m*n*(n-1)/2 (m>0) and that the coefficients are given by r = n + d^2/m, c = r-1 and b = d^3/m. The remaining terms may then be integer and still satisfy: a_n(k) = r*a(k), where a_n(k) is the k-th term of the n-th self-convolution of the sequence.
%H G. C. Greubel, <a href="/A120588/b120588.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: A(x) = 1 + Series_Reversion(1+3*x - (1+x)^2).
%F Lagrange Inversion yields g.f.: A(x) = Sum_{n>=0} C(2*n,n)/(n+1)*(2+x)^(n+1)/3^(2*n+1).
%F G.f.: (3 - sqrt(1-4*x))/2. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
%F a(n) = Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - _Michael Somos_, Jul 23 2011
%F G.f.: 2 - G(0), where G(k)= 2*x*(2*k+1) + k +1 - 2*x*(k+1)*(2*k+3)/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Jul 14 2013
%F G.f.: 2 - G(0), where G(k)= 1 - x/G(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Jul 19 2013
%F a(n) ~ 2^(2*n-2)/(sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Aug 19 2013
%F Given g.f. A(x), A001850(n-1) = coefficient of x^n in A(x)^n if n>0, the derivative of log(A(x)) is the g.f. for A026641. - _Michael Somos_, May 18 2015
%F A(x) = (1 + 2*Sum_{n >= 1} Catalan(n)*x^n)/(1 + Sum_{n >= 1} Catalan(n)*x^n) = (1 + 3/2*Sum_{n >= 1} binomial(2*n,n)*x^n )/(1 + Sum_{n >= 1} binomial(2*n,n)*x^n). - _Peter Bala_, Sep 01 2016
%e A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 +...
%e A(x)^3 = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 42*x^5 + 126*x^6 + 396*x^7 +..
%e More generally, given the functional equation:
%e r*A(x) = r-1 + b*x + A(x)^n
%e the series solution is:
%e A(x) = Sum_{i>=0} C(n*i,i)/(n*i-i+1)*(r-1+bx)^(n*i-i+1)/r^(n*i+1)
%e which can be expressed as:
%e A(x) = G( (r-1+bx)^(n-1)/r^n ) * (r-1+bx)/r
%e where G(x) satisfies: G(x) = 1 + x*G(x)^n .
%e Also we have:
%e A(x) = 1 + Series_Reversion[ (1 + r*x - (1+x)^n )/b ].
%t a[ n_] := SeriesCoefficient[ 1 + (1 - Sqrt[1 - 4 x]) / 2, {x, 0, n}]; (* _Michael Somos_, May 18 2015 *)
%o (PARI) {a(n)=local(A=1+x+x^2+x*O(x^n));for(i=0,n,A=A-3*A+2+x+A^2);polcoeff(A,n)}
%o (PARI) {a(n) = my(A); if( n<1, n==0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* _Michael Somos_, Jul 23 2011 */
%o (Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (3 - Sqrt(1-4*x))/2 )); // _G. C. Greubel_, Feb 18 2019
%o (Sage) ((3-sqrt(1-4*x))/2).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Feb 18 2019
%Y Cf. A120589 (A(x)^2); A120590 - A120607.
%Y Cf. A001850, A026641.
%K nonn,easy
%O 0,4
%A _Paul D. Hanna_, Jun 16 2006, Jan 24 2008
%E New name by _Wolfdieter Lang_, Feb 06 2020
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