cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

Views

Author

Michael Somos, May 19 2013

Keywords

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 + ...
		

Crossrefs

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

A134426 Triangle read by rows: T(n,k) is the number of paths of length n in the first quadrant, starting at the origin, ending at height k and consisting of 2 kind of upsteps U=(1,1) (U1 and U2), 3 kind of flatsteps F=(1,0) (F1, F2 and F3) and 1 kind of downsteps D=(1,-1).

Original entry on oeis.org

1, 3, 2, 11, 12, 4, 45, 62, 36, 8, 197, 312, 240, 96, 16, 903, 1570, 1440, 784, 240, 32, 4279, 7956, 8244, 5472, 2320, 576, 64, 20793, 40670, 46116, 35224, 18480, 6432, 1344, 128, 103049, 209712, 254912, 216384, 132320, 57600, 17024, 3072, 256
Offset: 0

Views

Author

Emeric Deutsch, Nov 05 2007

Keywords

Comments

T(n,0) = A001003(n+1) (the little Schroeder numbers).
Row sums yield A134425.

Examples

			T(2,1)=12 because we have 6 paths of shape FU and 6 paths of shape UF.
Triangle starts:
    1;
    3,   2;
   11,  12,   4;
   45,  62,  36,   8;
  197, 312, 240,  96,  16;
		

Crossrefs

Programs

  • Maple
    T:=proc(n,k) options operator,arrow: 2^k*(k+1)*(sum(2^j*binomial(n+1,j)*binomial(n+1, k+1+j),j=0..n-k))/(n+1) end proc: for n from 0 to 8 do seq(T(n, k),k=0..n) end do; # yields sequence in triangular form

Formula

T(n,k) = ((k+1)*2^k/(n+1))*Sum_{j=0..n-k} binomial(n+1, j)*binomial(n+1, k+j+1)*2^j (0 <= k <= n).
G.f.: g/(1-2*t*z*g), where g = 1 + 3*z*g + 2*z^2*g^2 is the g.f. of the little Schroeder numbers 1, 3, 11, 45, 197, ... (A001003).
Showing 1-2 of 2 results.