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 A086246 Expansion of (1 + x - sqrt(1 - 2*x - 3*x^2)) / 2 in powers of x. 10

%I

%S 0,1,1,1,2,4,9,21,51,127,323,835,2188,5798,15511,41835,113634,310572,

%T 853467,2356779,6536382,18199284,50852019,142547559,400763223,

%U 1129760415,3192727797,9043402501,25669818476,73007772802,208023278209,593742784829,1697385471211,4859761676391,13933569346707

%N Expansion of (1 + x - sqrt(1 - 2*x - 3*x^2)) / 2 in powers of x.

%C A variant of the Motzkin numbers: see A001006 for the main entry.

%C Equals row sums of triangle A144218 starting with "1". - _Gary W. Adamson_, Sep 14 2008

%C Starting (1, 1, 1, ...) = inverse binomial transform of A014137: (1, 2, 4, 9, 23, 65, ...). - _Gary W. Adamson_, Apr 02 2009

%C With a(0) = 1 this is the Riordan transform with the Riordan matrix R(n, m) = (-1)^(n-m)*A097805(n, m) (the inverse of A097805) of the Catalan sequence A000108. See a Feb 17 2017 comment on A097805 for Riordan transforms, and the g.f. given below in terms of the Catalan g.f. - _Wolfdieter Lang_, Feb 17 2017

%H Joerg Arndt, <a href="/A086246/b086246.txt">Table of n, a(n) for n = 0..200</a>

%H Paul Barry, <a href="https://arxiv.org/abs/1807.05794">Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences</a>, arXiv:1807.05794 [math.CO], 2018.

%H T. Feil, K. Hutson, R. M. Kretchmar, <a href="http://personal.denison.edu/~kretchmar/pubs/TreeTraversals.pdf">Tree Traversals and Permutations</a>, Congr. Numer. (2005), omitting the leading 0 and with a typo in the last number (303 should be 323), last sentence of chapter 6.

%F Series reversion of g.f. A(x) is -A(-x).

%F a(n) + a(n-1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0), n > 2.

%F G.f. A(x) satisfies 0 = f(x, A(x)) where f(x, y) = x - y - x*y + x^2 + y^2.

%F G.f. A(x) satisfies 0 = f(x, A(x)) where f(x, y) = (y^2 - y^3) - (x^2 + x^3).

%F G.f.: (1 + x - sqrt(1 - 2*x - 3*x^2)) / 2.

%F G.f. A(x) satisfies A(x) = x + C(x*A(x)) where C(x) is g.f. for Catalan numbers A000108 (offset 1).

%F G.f.: (1+x-sqrt(1-2*x-3*x^2))/2 = (x+x/G(0))/2 where G(k) = 1-2*x/(1+x/(1+x/(1-2*x/(1-x/(2-x/G(k+1)))))); (continued fraction). - _Sergei N. Gladkovskii_, Dec 11 2011

%F G.f.: x + x^2*Q(0), where Q(k)= 1 + x/(1 - x - x/(x + 1/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Apr 25 2013

%F G.f.: x*Q(0), where Q(k)= 1 + (4*k+1)*x/((1+x)*(k+1) - x*(1+x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1+x)/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 14 2013

%F 0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * (-3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) if n>0. - _Michael Somos_, Jan 25 2014

%F a(n) ~ 3^(n-1/2)/(2*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Apr 20 2014

%F a(n) = Sum_{k=1..n} binomial(2*k-2, k-1)*(-1)^(n-k)*binomial(n-2, n-k)/k. - _Vladimir Kruchinin_, May 27 2014

%F G.f. if a(0) = 1: C(x/(1+x)) with C the g.f. of A000108 (Catalan). See an above implicit formula. - _Wolfdieter Lang_, Feb 17 2017

%F (3*n-3)*a(n)+(1+2*n)*a(n+1)+(-n-2)*a(n+2)=0 for n >= 1. - _Robert Israel_, May 01 2018

%e G.f. = x + x^2 + x^3 + 2*x^4 + 4*x^5 + 9*x^6 + 21*x^7 + 51*x^8 + 127*x^9 + ...

%p with(PolynomialTools): CoefficientList(convert(taylor((1 + x - sqrt(1 - 2*x - 3*x^2))/2, x = 0, 33), polynom), x); # _Taras Goy_, Aug 07 2017

%t a[ n_] := SeriesCoefficient[ (1 + x - Sqrt[1 - 2 x - 3 x^2])/2, {x, 0, n}] (* _Michael Somos_, Jan 25 2014 *)

%o (PARI) {a(n) = polcoeff( (1 + x - sqrt(1 - 2*x - 3*x^2 + x * O(x^n))) / 2, n)}

%o (PARI) x='x+O('x^99); concat(0, Vec((1+x-(1-2*x-3*x^2)^(1/2))/2)) \\ _Altug Alkan_, May 01 2018

%o (Maxima) a(n):=sum((binomial(2*k-2,k-1)*(-1)^(n-k)*binomial(n-2,n-k))/k,k,1,n); /* _Vladimir Kruchinin_, May 27 2014 */

%Y a(n+2) = A001006(n).

%Y Cf. A000108, A014137, A144218.

%K nonn

%O 0,5

%A _Michael Somos_, Jul 13 2003

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Last modified January 15 19:05 EST 2019. Contains 319170 sequences. (Running on oeis4.)