A097332 Expansion of (1/(1-x))(1+2x/(1-x+sqrt(1-2x-3x^2))).

1, 2, 3, 5, 9, 18, 39, 90, 217, 540, 1375, 3563, 9361, 24872, 66707, 180341, 490913, 1344380, 3701159, 10237541, 28436825, 79288844, 221836403, 622599626, 1752360041, 4945087838, 13988490339, 39658308815, 112666081617

Offset: 0

Author

Paul Barry, Aug 05 2004

Keywords

• easy
• nonn

Comments

Binomial transform of A097331. Binomial transform is A014318. Partial sums of 1+2x/(1-x+sqrt(1-2x-3x^2)) or (1+x+sqrt(1-2x-3x^2))/(1-x+sqrt(1-2x-3x^2)), which is A001006 with an extra leading 1.

Apparently the Motzkin transform of 1, 2, bar(1, -1, -1, 1), where bar() denotes a periodically continued series, as in A057077. - R. J. Mathar, Dec 11 2008

Starting with offset 1 = iterates of M * [1,1,0,0,0,...] where M = a tridiagonal matrix with [1,1,1,...] in the main and superdiagonals and [0,1,1,1,...] in the subdiagonal. - Gary W. Adamson, Jan 08 2009

Hankel transform is A087960(n) = (-1)^binomial(n+1,2). - Paul Barry, Aug 10 2009

Examples

			G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 9*x^4 + 18*x^5 + 39*x^6 + 90*x^7 + 217*x^8 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Emeric Deutsch and Bruce E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, arXiv:math/0407326 [math.CO], 2004; J. Num. Theory 117 (2006), 191-215. [See S_n on page 7.]

Programs

Mathematica

CoefficientList[Series[1/(1-x)*(1+(2x)/(1-x+Sqrt[1-2x-3x^2])),{x,0,40}],x] (* Harvey P. Dale, May 03 2012 *)
a[ n_] := SeriesCoefficient[ (1 + x - Sqrt[1 - 2 x - 3 x^2]) / (2 x (1 - x)), {x, 0, n}]; (* Michael Somos, May 19 2014 *)

PARI

{a(n) = if( n<0, 0, polcoeff( (1 + x - sqrt(1 - 2*x - 3*x^2 + x^2 * O(x^n))) / (2 * x * (1 - x)), n))}; /* Michael Somos, May 19 2014 */

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(n, k)*Sum_{i=0..k} Catalan(k-i)*2^i.

G.f.: 1/(1-x-x/(1+x/(1-x+x/(1-x/(1-x-x/(1+x/(1-x+x/(1-x/(1-x-x/(1+... (continued fraction). - Paul Barry, Aug 10 2009

Conjecture D-finite with recurrence: (n+1)*a(n) - 3*n*a(n-1) + (-n+5)*a(n-2) + 3*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 26 2012

a(n) ~ 3^(n+3/2) / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 13 2014

0 = a(n)*(9*a(n+1) + 6*a(n+2) - 27*a(n+3) + 12*a(n+4)) + a(n+1)*(-12*a(n+1) + 10*a(n+2) + 12*a(n+3) - 7*a(n+4)) + a(n+2)*(-12*a(n+2) + 14*a(n+3) - 6*a(n+4)) + a(n+3)*(a(n+4)). - Michael Somos, May 19 2014