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A120588
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G.f. is 1 + x*c(x), where c(x) is the g.f. of the Catalan numbers (A000108).
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27
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1, 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152
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OFFSET
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0,4
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COMMENTS
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Previous name was: G.f. satisfies: 3*A(x) = 2 + x + A(x)^2, with A(0) = 1.
This is essentially a duplicate of entry A000108, the Catalan numbers (a(n) = A000108(n-1) for n>0).
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.
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LINKS
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FORMULA
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G.f.: A(x) = 1 + Series_Reversion(1+3*x - (1+x)^2).
Lagrange Inversion yields g.f.: A(x) = Sum_{n>=0} C(2*n,n)/(n+1)*(2+x)^(n+1)/3^(2*n+1).
G.f.: (3 - sqrt(1-4*x))/2. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
a(n) = Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
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
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
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
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EXAMPLE
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A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 +...
A(x)^3 = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 42*x^5 + 126*x^6 + 396*x^7 +..
More generally, given the functional equation:
r*A(x) = r-1 + b*x + A(x)^n
the series solution is:
A(x) = Sum_{i>=0} C(n*i,i)/(n*i-i+1)*(r-1+bx)^(n*i-i+1)/r^(n*i+1)
which can be expressed as:
A(x) = G( (r-1+bx)^(n-1)/r^n ) * (r-1+bx)/r
where G(x) satisfies: G(x) = 1 + x*G(x)^n .
Also we have:
A(x) = 1 + Series_Reversion[ (1 + r*x - (1+x)^n )/b ].
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MATHEMATICA
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a[ n_] := SeriesCoefficient[ 1 + (1 - Sqrt[1 - 4 x]) / 2, {x, 0, n}]; (* Michael Somos, May 18 2015 *)
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PROG
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(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)}
(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 */
(Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (3 - Sqrt(1-4*x))/2 )); // G. C. Greubel, Feb 18 2019
(Sage) ((3-sqrt(1-4*x))/2).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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