A155069 Expansion of (3 - x - sqrt(1 - 6*x + x^2))/2.
1, 1, 2, 6, 22, 90, 394, 1806, 8558, 41586, 206098, 1037718, 5293446, 27297738, 142078746, 745387038, 3937603038, 20927156706, 111818026018, 600318853926, 3236724317174, 17518619320890, 95149655201962, 518431875418926, 2832923350929742, 15521467648875090
Offset: 0
Keywords
Examples
From _Lee A. Newberg_, Mar 30 2010: (Start) For n = 2, the a(2) = 2 branching configurations are ()() and (()()), where each () indicates a hairpin (also termed 1-loop) and each other pair of parentheses indicates a k-loop for k >= 3. For n = 3, the a(3) = 6 branching configurations are ()()(), (()())(), ()(()()), (()()()), ((()())()), and (()(()())). (End) When inserting balanced parentheses into the product x^n: For n = 0, the a(0) = 1 possible term is the empty term. For n = 1, the a(1) = 1 possible term is x. For n = 2, the a(2) = 2 possible terms are xx and (xx). For n = 3, the a(3) = 6 possible terms are xxx, (xx)x, x(xx), (xxx), ((xx)x), and (x(xx)). - _Lee A. Newberg_, Apr 06 2010 G.f. = 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 90*x^5 + 394*x^6 + 1806*x^7 + ...
References
- S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011. see p. 399 Table A.7
- D. E. Knuth, The Art of Computer Programming, Volume 1, Fundamental Algorithms, section 2.2.1: Stacks, Queues, and Deques.
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..1313
- J. Abate and W. Whitt, Integer Sequences from Queueing Theory, J. Int. Seq. 13 (2010), 10.5.5, Theorem 5.
- Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
- Christian Bean, Finding structure in permutation sets, Ph.D. Dissertation, Reykjavík University, School of Computer Science, 2018.
- Arnauld Mesinga Mwafise, Computational and Combinatorial Enumeration of Poset Matrices, 2024. See p. 8.
- D. Sankoff, Simultaneous solution of the RNA folding, alignment and protosequence problems, Siam J. Appl. Math 45(5):810-825 (1985). [From _Lee A. Newberg_, Mar 30 2010]
Programs
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Magma
R
:=PowerSeriesRing(Rationals(), 25); Coefficients(R!( (3-x-Sqrt(1-6*x+x^2))/2 )); // G. C. Greubel, Jun 08 2020 -
Maple
seq(coeff(series((3-x -sqrt(1-6*x+x^2))/2, x, n+1), x, n), n = 0..25); # G. C. Greubel, Jun 08 2020
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Mathematica
CoefficientList[Series[(3 -x -Sqrt[1-6x+x^2])/2, {x, 0, 25}], x] (* Vincenzo Librandi, Nov 13 2014 *)
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Maxima
a(n):=if n<1 then 1 else sum(binomial(n+i-1,i)* binomial(2*n, n-2*i-1),i,0,(n)/2)/(n); /* Vladimir Kruchinin, Nov 13 2014 */
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Sage
a = lambda n: catalan_number(n)*hypergeometric([1/2-n/2,1-n/2,n],[n/2+1,n/2+3/2],1) print([simplify(a(n)) for n in (0..25)]) # Peter Luschny, Nov 14 2014
Formula
G.f.: (3 - x - sqrt(1 -6*x +x^2))/2.
G.f.: 4 / (3 - x + sqrt(1 - 6*x + x^2)). - Michael Somos, Apr 18 2012
a(n) ~ sqrt((sqrt(18)-4)/(4*Pi)) * n^(-3/2) * (3 + sqrt(8))^n, which is, approximately, a(n) ~ 0.1389558648 * n^(-1.5) * 5.828427125^n. - Lee A. Newberg, Apr 06 2010
a(n) ~ (1 + sqrt(2))^(2*n-1) / (2^(3/4)*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 23 2023
a(n) = top left term of M^n, where M = the production matrix:
1, 1, 0, 0, 0, ...
1, 2, 1, 0, 0, ...
1, 2, 2, 1, 0, ...
1, 2, 2, 2, 1, ...
1, 2, 2, 2, 2, 1, ...
...
Top row terms of M^n generates rows of triangle A132372. - Gary W. Adamson, Jul 07 2011
G.f.: A(x)=(3 -x- sqrt(1-6*x+x^2))/2= 2 - G(0); G(k)= 1 + x - 2*x/G(k+1); (continued fraction, 1-step, 1 var.). - Sergei N. Gladkovskii, Jan 04 2012
G.f.: A(x)=(3 -x -sqrt(1-6*x+x^2))/2= G(0); G(k)= := 1 - x/(1 - 2/G(k+1)); (continued fraction, 2-step, 2 var.). - Sergei N. Gladkovskii, Jan 04 2012
D-finite with recurrence: n*a(n) +3*(3-2*n)*a(n-1) +(n-3)*a(n-2)=0. - R. J. Mathar, Jul 24 2012
G.f.: 1 / (1 - x / (1 - x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ... )))))) = 1 + x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ... )))))). - Michael Somos, Jan 03 2013
G.f.: 2 - x - G(0), where G(k)= k+1 - 2*x*(k+1) - 2*x*(k+1)*(k+2)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Jul 14 2013
a(n) = (1/n)*Sum_{i = 0..floor(n/2)} binomial(n+i-1, i)*binomial(2*n, n-2*i-1), n>0, a(0)=1. - Vladimir Kruchinin, Nov 13 2014
a(n) = Catalan(n)*hypergeometric([1/2-n/2, 1-n/2, n], [n/2+1, n/2+3/2], 1). - Peter Luschny, Nov 14 2014
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n-1} a(k) * a(n-k). - Ilya Gutkovskiy, Apr 11 2021
Comments