A086246 Expansion of (1 + x - sqrt(1 - 2*x - 3*x^2)) / 2 in powers of x.
0, 1, 1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209, 593742784829, 1697385471211, 4859761676391, 13933569346707
Offset: 0
Keywords
Examples
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 + ...
Links
- Joerg Arndt, Table of n, a(n) for n = 0..200
- Paul Barry, Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences, arXiv:1807.05794 [math.CO], 2018.
- Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
- Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
- Gi-Sang Cheon, Marshall M. Cohen and Nikolaos Pantelidis, Decompositions and eigenvectors of Riordan matrices, Linear Algebra and its Applications, Vol. 642 (2022), 118-138.
- T. Feil, K. Hutson and R. M. Kretchmar, Tree Traversals and Permutations, Congr. Numer. (2005), omitting the leading 0 and with a typo in the last number (303 should be 323), last sentence of chapter 6.
Programs
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Maple
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
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Mathematica
a[ n_] := SeriesCoefficient[ (1 + x - Sqrt[1 - 2 x - 3 x^2])/2, {x, 0, n}] (* Michael Somos, Jan 25 2014 *) a = DifferenceRoot[Function[{y, n}, {(3n-3)*y[n] + (2n+1)*y[n+1] + (-n-2)*y[n+2] == 0, y[0] == 0, y[1] == 1, y[2] == 1}]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Oct 28 2021 *) -
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 */
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PARI
{a(n) = polcoeff( (1 + x - sqrt(1 - 2*x - 3*x^2 + x * O(x^n))) / 2, n)} -
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
Formula
Series reversion of g.f. A(x) is -A(-x).
a(n) + a(n-1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0), n > 2.
G.f. A(x) satisfies 0 = f(x, A(x)) where f(x, y) = x - y - x*y + x^2 + y^2.
G.f. A(x) satisfies 0 = f(x, A(x)) where f(x, y) = (y^2 - y^3) - (x^2 + x^3).
G.f.: (1 + x - sqrt(1 - 2*x - 3*x^2)) / 2.
G.f. A(x) satisfies A(x) = x + C(x*A(x)) where C(x) is g.f. for Catalan numbers A000108 (offset 1).
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
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
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
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
a(n) ~ 3^(n-1/2)/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Apr 20 2014
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
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
D-finite with recurrence: (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
a(n) = A007971(n)/2, n>=2. - R. J. Mathar, Jan 20 2020
Comments