login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A225887
a(n) = A212205(2*n + 1).
2
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
0,2
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)
LINKS
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
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.
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
EXAMPLE
1 + 4*x + 18*x^2 + 86*x^3 + 426*x^4 + 2162*x^5 + 11166*x^6 + 58438*x^7 + ...
MATHEMATICA
a[ n_] := SeriesCoefficient[ 2 / (1 - 5 x + Sqrt[1 - 6 x + x^2]), {x, 0, n}]
PROG
(PARI) {a(n) = if( n<0, 0, polcoeff( 2 / (1 - 5*x + sqrt(1 - 6*x + x^2 + x * O(x^n))), 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 */
KEYWORD
nonn,easy
AUTHOR
Michael Somos, May 19 2013
STATUS
approved