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.

A010683 Let S(x,y) = number of lattice paths from (0,0) to (x,y) that use the step set { (0,1), (1,0), (2,0), (3,0), ...} and never pass below y = x. Sequence gives S(n-1,n) = number of 'Schröder' trees with n+1 leaves and root of degree 2.

Original entry on oeis.org

1, 2, 7, 28, 121, 550, 2591, 12536, 61921, 310954, 1582791, 8147796, 42344121, 221866446, 1170747519, 6216189936, 33186295681, 178034219986, 959260792775, 5188835909516, 28167068630713, 153395382655222
Offset: 0

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Author

Robert Sulanke (sulanke(AT)diamond.idbsu.edu), N. J. A. Sloane

Keywords

Comments

a(n) is the number of compound propositions "on the negative side" that can be made from n simple propositions.
Convolution of A001003 (the little Schröder numbers) with itself. - Emeric Deutsch, Dec 27 2003
Number of dissections of a convex polygon with n+3 sides that have a triangle over a fixed side (the base) of the polygon. - Emeric Deutsch, Dec 27 2003
a(n-1) = number of royal paths from (0,0) to (n,n), A006318, with exactly one diagonal step on the line y=x. - David Callan, Mar 14 2004
Number of short bushes (i.e., ordered trees with no vertices of outdegree 1) with n+2 leaves and having root of degree 2. Example: a(2)=7 because, in addition to the five binary trees with 6 edges (they do have 4 leaves) we have (i) two edges rb, rc hanging from the root r with three edges hanging from vertex b and (ii) two edges rb, rc hanging from the root r with three edges hanging from vertex c. - Emeric Deutsch, Mar 16 2004
The a(n) equal the Fi2 sums, see A180662, of Schröder triangle A033877. - Johannes W. Meijer, Mar 26 2012
Row sums of A144944 and of A186826. - Reinhard Zumkeller, May 11 2013

Crossrefs

Second right-hand column of triangle A011117.
A177010 has a closely-related g.f..

Programs

  • Haskell
    a010683 = sum . a144944_row  -- Reinhard Zumkeller, May 11 2013
    
  • Magma
    [n le 2 select n else (6*(2*(n-1)^2-1)*Self(n-1) - (n-3)*(2*n-1)*Self(n-2))/((n+1)*(2*n-3)): n in [1..30]]; // G. C. Greubel, Mar 11 2023
  • Maple
    a := proc(n) local k: if n=0 then 1 else (2/n)*add(binomial(n, k)* binomial(n+k+1, k-1), k=1..n) fi: end:
    seq(a(n), n=0..21); # Johannes W. Meijer, Mar 26 2012, revised Mar 31 2015
  • Mathematica
    f[ x_, y_ ]:= f[ x, y ]= Module[ {return}, If[x==0, return =1, If[y==x-1, return =0, return= f[x,y-1] + Sum[f[k, y], {k,0,x-1} ]]]; return];
    (* Do[Print[Table[f[ k, j ], {k, 0, j}]], {j, 10, 0, -1}] *)
    Table[f[x, x+1], {x,0,21}]
    (* Second program: *)
    a[n_] := 2*Hypergeometric2F1[1-n, n+3, 2, -1]; a[0]=1;
    Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 09 2014, after Wolfdieter Lang *)
  • PARI
    x='x+O('x^100); Vec(((1-x)^2-(1+x)*sqrt(1-6*x+x^2))/(8*x^2)) \\ Altug Alkan, Dec 19 2015
    
  • Sage
    a = lambda n: (n+1)*hypergeometric([1-n, -n], [3], 2)
    [simplify(a(n)) for n in range(22)] # Peter Luschny, Nov 19 2014
    

Formula

G.f.: ((1-t)^2-(1+t)*sqrt(1-6*t+t^2))/(8*t^2) = A(t)^2, with o.g.f. A(t) of A001003.
From Wolfdieter Lang, Sep 12 2005: (Start)
a(n) = (2/n)*Sum_{k=1..n} binomial(n, k)*binomial(n+k+1, k-1).
a(n) = 2*hypergeometric2F1([1-n, n+3], [2], -1), n>=1. a(0)=1. (End)
a(n) = ((2*n+1)*LegendreP(n+1,3) - (2*n+3)*LegendreP(n,3)) / (4*n*(n+2)) for n>0. - Mark van Hoeij, Jul 02 2010
From Gary W. Adamson, Jul 08 2011: (Start)
Let M = the production matrix:
1, 2, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 2, 1, 2, 0, 0, ...
1, 2, 1, 2, 1, 0, ...
1, 2, 1, 2, 1, 2, ...
...
a(n) is the upper entry in the vector (M(T))^n * [1,0,0,0,...]; where T is the transpose operation. (End)
D-finite with recurrence: (n+2)*(2*n-1)*a(n) = 6*(2*n^2-1)*a(n-1) - (n-2)*(2*n+1)*a(n-2). - Vaclav Kotesovec, Oct 07 2012
a(n) ~ sqrt(48+34*sqrt(2))*(3+2*sqrt(2))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012
Recurrence (an alternative): (n+2)*a(n) = (4-n)*a(n-4) + 2*(2*n-5)*a(n-3) + 10*(n-1)*a(n-2) + 2*(2*n+1)*a(n-1), n >= 4. - Fung Lam, Feb 18 2014
a(n) = (n+1)*hypergeometric2F1([1-n, -n], [3], 2). - Peter Luschny, Nov 19 2014
a(n) = (A001003(n) + A001003(n+1))/2 = sum(A001003(k) * A001003(n-k), k=0..n). - Johannes W. Meijer, Apr 29 2015

Extensions

Minor edits by Johannes W. Meijer, Mar 26 2012