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.

A110440 Triangular array formed by the little Schröder numbers s(n,k).

Original entry on oeis.org

1, 3, 1, 11, 6, 1, 45, 31, 9, 1, 197, 156, 60, 12, 1, 903, 785, 360, 98, 15, 1, 4279, 3978, 2061, 684, 145, 18, 1, 20793, 20335, 11529, 4403, 1155, 201, 21, 1, 103049, 104856, 63728, 27048, 8270, 1800, 266, 24, 1, 518859, 545073, 350136, 161412, 55458, 14202
Offset: 0

Views

Author

Asamoah Nkwanta (nkwanta(AT)jewel.morgan.edu), Aug 08 2005

Keywords

Comments

s(n,k) is the number of unit step restricted paths (i.e., they never go below the x-axis) from the origin (0,0) to (n-1,k-1) using up step U(1,1), three types of level steps L(1,0), L'(1,0), L"(1,0) and two types of down steps D(1,-1), D'(1,-1). s(0,0)=1 and the leftmost column s(n,0) is A001003.
The sequence factors A038255 into a product of Riordan arrays.

Examples

			Triangle starts:
[0]      1;
[1]      3,      1;
[2]     11,      6,      1;
[3]     45,     31,      9,      1;
[4]    197,    156,     60,     12,     1;
[5]    903,    785,    360,     98,    15,     1;
[6]   4279,   3978,   2061,    684,   145,    18,    1;
[7]  20793,  20335,  11529,   4403,  1155,   201,   21,   1;
[8] 103049, 104856,  63728,  27048,  8270,  1800,  266,  24,  1;
[9] 518859, 545073, 350136, 161412, 55458, 14202, 2646, 340, 27, 1;
		

Crossrefs

Cf. A232246 (central terms), A001003 (left column), A065096 (2nd column?), A225887 (row sums?).

Programs

  • Haskell
    a110440 n k = a110440_tabl !! n !! k
    a110440_row n = a110440_tabl !! n
    a110440_tabl = iterate (\xs ->
       zipWith (+) ([0] ++ xs) $
       zipWith (+) (map (* 3) (xs ++ [0]))
                   (map (* 2) (tail xs ++ [0,0]))) [1]
    -- Reinhard Zumkeller, Nov 21 2013
  • Maple
    T := (n, k) -> ((k + 1)/(n + 1))*add(2^(n - m)*binomial(n+1, m+1)*binomial(n+1, m-k), m = 0..n): seq(seq(T(n, k), k = 0..n), n = 0..9); # Peter Luschny, Jan 09 2022
  • Mathematica
    nmax = 9; t[n_, k_] := Sum[(i*(-1)^(k-i+1)*Binomial[k+1, i]*Sum[(-1)^j*2^(n+1-j)*(2n+i-j+1)! / ((n+i-j+1)!*j!*(n-j+1)!), {j, 0, n+1}]), {i, 0, k+1}]; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
  • Maxima
    T(n,k):=sum((i*(-1)^(k-i+1)*binomial(k+1,i)*sum((-1)^j*2^(n+1-j)*(2*n+i-j+1)!/((n+i-j+1)!*j!*(n-j+1)!),j,0,n+1)),i,0,k+1); /* Vladimir Kruchinin, Oct 17 2011 */
    
  • Maxima
    T(n,k):=((k+1)/(n+1)*sum(binomial(j,-n-k+2*j-2)*3^(-n-k+2*j-2)*2^(n+1-j)*binomial(n+1,j),j,ceiling((n+k+2)/2),n+1)); /* Vladimir Kruchinin, Jan 28 2013 */
    
  • PARI
    {T(n, k)= if(n<0|| k>n, 0, polcoeff(polcoeff( 2/(1 -3*x -2*x*y +sqrt( 1 -6*x +x^2 +x*O(x^n)) ), n), k))} \\ Michael Somos, Mar 31 2007
    
  • Sage
    def A110440_triangle(dim):
        T = matrix(ZZ,dim,dim)
        for n in range(dim): T[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                T[n,k] = T[n-1,k-1]+3*T[n-1,k]+2*T[n-1,k+1]
        return T
    A110440_triangle(9) # Peter Luschny, Sep 20 2012
    

Formula

s(n+1,0) = 3s(n,0) + 2s(n,1) and for k > 0: s(n+1,k) = s(n,k-1) + 3s(n,k) + 2s(n,k+1). [Typo fixed by Reinhard Zumkeller, Nov 21 2013]
Riordan array ((1 - 3z - sqrt(1-6z+z^2))/4z*z, (1 - 3z - sqrt(1-6z+z^2))/4z).
Sum_{k>=0} T(m, k)*T(n, k)*2^k = T(m+n, 0) = A001003(m+n+1). - Philippe Deléham, Sep 14 2005
G.f.: 2/( 1 - x*L -2*x*y*U + sqrt( (1 - x*L)^2 - 4*x^2*D*U ) ) where L=3, U = 1, D = 2. - Michael Somos, Mar 31 2007
Sum_{k=0..n} T(n,k)*(2^(k + 1) - 1) = 6^n. - Philippe Deléham, Nov 29 2009
T(n,k) = Sum_{i=0..k + 1} i*(-1)^(k-i+1)*C(k+1,i)*Sum_{j=0..n+1} (-1)^j*2^(n+1-j)*(2*n+i-j+1)!/((n+i-j+1)!*j!*(n-j+1)!). - Vladimir Kruchinin, Oct 17 2011
T(n,k) = ((k+1)/(n+1))*Sum_{j=ceiling((n+k+2)/2)..n + 1} C(j,2*j-n-k-2)*3^(2*j-n-k- 2)*2^(n+1-j)*C(n+1,j). - Vladimir Kruchinin, Jan 28 2013
T(n,k) = ((k+1)/(n+1))*Sum_{m=0..n} 2^(n-m)*C(n+1,m+1)*C(n+1,m-k). - Vladimir Kruchinin, Jan 09 2022