A225887 a(n) = A212205(2*n + 1).

1, 4, 18, 86, 426, 2162, 11166, 58438, 309042, 1648154, 8851206, 47813790, 259585002, 1415431266, 7747200558, 42545600310, 234346445154, 1294260644906, 7165245015510, 39754745775886, 221009855334426, 1230909476804594, 6867024985408638, 38369226561522086

Offset: 0

Author

Michael Somos, May 19 2013

Keywords

• nonn
• easy

Comments

From Peter Bala, Apr 23 2017: (Start)

a(n) is also the number of Schröder paths of semilength n (paths from (0, 0) to (2*n, 0), using only single steps northeast or southeast (steps (1, 1) or (1, -1)) or double steps east (steps (2, 0)), that never fall below the x-axis) in which the (2,0)-steps that are on the horizontal axis come in 3 colors (see Oste and Van der Jeugt, Section 7).

Example: a(2) = 18 because from the origin to the point (4,0) we have 3^2 = 9 paths of type HH, 3 paths of type HUD, 3 paths of type UDH as well as the paths UDUD, UUDD, and UHD.

It follows that the sequence may be calculated as the leading diagonal of the lower triangular array (T(n,k))n,k>=0 defined by the relations: T(n,0) = 1, T(n,k) = T(n,k-1) + T(n-1,k) + T(n-1,k-1) for 1 <= k <= n-1 and T(n,n) = 3*T(n-1,n-1) + T(n,n-1). The array begins: [1], [1, 4], [1, 6, 18], [1, 8, 32, 86], [1, 10, 50, 168, 426]. (End)

Examples

			1 + 4*x + 18*x^2 + 86*x^3 + 426*x^4 + 2162*x^5 + 11166*x^6 + 58438*x^7 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
• Taras Goy and Mark Shattuck, Hessenberg-Toeplitz Matrix Determinants with Schröder and Fine Number Entries, Carpathian Math. Publ., Vol. 15 (2023), No. 2, 420-436. See Theorems 1 and 2.
• R. Oste and J. Van der Jeugt, Motzkin paths, Motzkin polynomials and recurrence relations, Electronic Journal of Combinatorics 22(2) (2015), #P2.8. Section 7.

Crossrefs

Cf. A001003, A006125, A127850, A131090, A151090, A212205, A111966.

Programs

Mathematica

a[ n_] := SeriesCoefficient[ 2 / (1 - 5 x + Sqrt[1 - 6 x + x^2]), {x, 0, n}]

Maxima

a(n):=sum((k+1)*sum(binomial(j,n-k-j)*3^(-n+k+2*j)*2^(n-k-j)*binomial(n+1,j),j,0,n+1),k,0,n)/(n+1); /* Vladimir Kruchinin, Mar 13 2016 */

PARI

{a(n) = if( n<0, 0, polcoeff( 2 / (1 -  5*x + sqrt(1 - 6*x + x^2 + x * O(x^n))), n))}

Formula

G.f.: (-1 + 5*x + sqrt(1 - 6*x + x^2)) / (2 * (x - 6*x^2)) = 2 / (1 - 5*x + sqrt(1 - 6*x + x^2)).

G.f.: A(x) = 1 / (1 - 5*x + (x - 6*x^2) * A(x)) = 1 + x * A(x) * (5 - A(x) * (1 - 6*x)).

INVERT transform of A001003(n+1). INVERT transform is A134425.

HANKEL transform is A006125. HANKEL transform with 1 prepended is A127850(n+1).

BINOMIAL transform of A151090.

Conjecture: (n+1)*a(n) +3*(-4*n-1)*a(n-1) +(37*n-20)*a(n-2) +6*(-n+2)*a(n-3)=0. - R. J. Mathar, May 23 2014

a(n) = Sum_{k=0..n}((k+1)*Sum_{j=0..n+1}(binomial(j,n-k-j)*3^(-n+k+2*j)*2^(n-k-j)*binomial(n+1,j)))/(n+1). - Vladimir Kruchinin, Mar 13 2016

a(n) ~ (1+sqrt(2))^(2*n+5) / (2^(3/4)*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 13 2016

G.f.: 1/(1-3*x -x/(1-x -x/(1-x -x/(1-x - ... )))) (continued fraction) = 1/(1 - 3*x - x*S(x)), where S(x) is the generating function of the large Schröder numbers A001003. - Peter Bala, Apr 23 2017