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.

Showing 1-8 of 8 results.

A106640 Row sums of A059346.

Original entry on oeis.org

1, 1, 4, 11, 36, 117, 393, 1339, 4630, 16193, 57201, 203799, 731602, 2643903, 9611748, 35130195, 129018798, 475907913, 1762457595, 6550726731, 24428808690, 91377474411, 342763939656, 1289070060903, 4859587760076, 18360668311027, 69514565858653, 263693929034909
Offset: 0

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Author

Philippe Deléham, May 26 2005

Keywords

Comments

a(n) = p(n + 1) where p(x) is the unique degree-n polynomial such that p(k) = Catalan(k) for k = 0, 1, ..., n. - Michael Somos, Jan 05 2012
Number of Dyck (n+1)-paths whose minimum ascent length is 1. - David Scambler, Aug 22 2012
From Alois P. Heinz, Jun 29 2014: (Start)
a(n) is the number of ordered rooted trees with n+2 nodes such that the minimal outdegree equals 1. a(2) = 4:
o o o o
| | / \ / \
o o o o o o
| / \ | |
o o o o o
|
o
(End)
Number of non-crossing partitions of {1,2,..,n+1} that contain cyclical adjacencies. a(2) = 4, [12|3, 13|2, 1|23, 123]. - Yuchun Ji, Nov 13 2020

Examples

			1 + x + 4*x^2 + 11*x^3 + 36*x^4 + 117*x^5 + 393*x^6 + 1339*x^7 + 4630*x^8 + ...
a(2) = 4 since p(x) = (x^2 - x + 2) / 2 interpolates p(0) = 1, p(1) = 1, p(2) = 2, and p(3) = 4. - _Michael Somos_, Jan 05 2012
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<3, [1, 1, 4][n+1],
          ((30*n^3-44*n^2-22*n+24)*a(n-1)-(25*n^3-105*n^2+140*n-48)*a(n-2)
           -6*(n-1)*(5*n-4)*(2*n-3)*a(n-3))/(n*(n+2)*(5*n-9)))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Jun 29 2014
  • Mathematica
    max = 30; t = Table[Differences[Table[CatalanNumber[k], {k, 0, max}], n], {n, 0, max}]; a[n_] := Sum[t[[n-k+1, k]], {k, 1, n}]; Array[a, max] (* Jean-François Alcover, Jan 21 2017 *)
  • PARI
    {a(n) = if( n<0, 0, n++; subst( polinterpolate( vector(n, k, binomial( 2*k - 2, k - 1) / k)), x, n + 1))} /* Michael Somos, Jan 05 2012 */
    
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( 2 / (sqrt( 1 - 2*x - 3*x^2 + A) + (1 + x) * sqrt( 1 - 4*x + A)) ,n))} /* Michael Somos, Jan 05 2012 */

Formula

G.f.: (sqrt( 1 - 2*x - 3*x^2 ) / (1 + x) - sqrt( 1 - 4*x )) / (2*x^2) = 2 / (sqrt( 1 - 2*x - 3*x^2 ) + (1 + x) * sqrt( 1 - 4*x )). - Michael Somos, Jan 05 2012
a(n) = A000108(n+1) - A005043(n+1).
a(n) ~ 2^(2*n+2) / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jan 21 2017
a(n) = A000296(n+2) - A247494(n+1); i.e., remove the crossing partitions from the partitions with cyclical adjacencies. - Yuchun Ji, Nov 17 2020

Extensions

Typo in a(20) corrected and more terms from Alois P. Heinz, Jun 29 2014

A228338 Third diagonal of Catalan difference table (A059346).

Original entry on oeis.org

5, 9, 19, 43, 102, 250, 628, 1608, 4181, 11009, 29295, 78655, 212815, 579675, 1588245, 4374285, 12103407, 33628827, 93786969, 262450881, 736710360, 2073834420, 5853011850, 16558618510, 46949351275, 133390812255, 379708642289, 1082797114049, 3092894319078, 8848275403642
Offset: 0

Views

Author

N. J. A. Sloane, Aug 29 2013

Keywords

Crossrefs

Programs

  • Maple
    a := n -> 5*(-1)^n*hypergeom([7/2, -n], [5], 4):
    seq(simplify(a(n)), n=0..29); # Peter Luschny, May 25 2021
  • Mathematica
    CoefficientList[Series[-(x+1)^(5/2)*Sqrt[1-3*x]/(2*x^4)-1/2*(- 1 - x + 3*x^2 + 7*x^3)/x^4, {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 14 2014 *)
  • PARI
    x='x+O('x^50); Vec(-(x+1)^(5/2)*sqrt(1-3*x)/(2*x^4)-1/2*(-1-x+3*x^2+7*x^3)/x^4) \\ G. C. Greubel, May 31 2017

Formula

From Vaclav Kotesovec, Feb 14 2014: (Start)
Recurrence: (n+4)*a(n) = (2*n+7)*a(n-1) + 3*(n-1)*a(n-2).
G.f.: -(x+1)^(5/2)*sqrt(1-3*x)/(2*x^4)-1/2*(-1-x+3*x^2+7*x^3)/x^4.
a(n) ~ 8 * 3^(n+3/2) / (sqrt(Pi) * n^(3/2)). (End)
a(n) = 5*(-1)^n*hypergeom([7/2, -n], [5], 4). - Peter Luschny, May 25 2021

Extensions

Terms a(21) onward added by G. C. Greubel, May 31 2017

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

Views

Author

Keywords

Comments

Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
    a005043_list = 1 : 0 : zipWith div
       (zipWith (*) [1..] (zipWith (+)
           (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
    Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
    Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
    a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
    a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
    Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
    Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
    a[1]:0$
    a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
    
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
    
  • Python
    from functools import cache
    @cache
    def A005043(n: int) -> int:
        if n <= 1: return 1 - n
        return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
    print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
    [simplify(A005043(n)) for n in (0..29)]
    # Peter Luschny, Sep 23 2014
    
  • Sage
    def ms():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield -b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
    A005043 = ms()
    print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016
    

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020

A007317 Binomial transform of Catalan numbers.

Original entry on oeis.org

1, 2, 5, 15, 51, 188, 731, 2950, 12235, 51822, 223191, 974427, 4302645, 19181100, 86211885, 390248055, 1777495635, 8140539950, 37463689775, 173164232965, 803539474345, 3741930523740, 17481709707825, 81912506777200, 384847173838501, 1812610804416698
Offset: 1

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Comments

Partial sums of A002212 (the restricted hexagonal polyominoes with n cells). Number of Schroeder paths (i.e., consisting of steps U=(1,1),D=(1,-1),H=(2,0) and never going below the x-axis) from (0,0) to (2n-2,0), with no peaks at even level. Example: a(3)=5 because among the six Schroeder paths from (0,0) to (4,0) only UUDD has a peak at an even level. - Emeric Deutsch, Dec 06 2003
Number of binary trees of weight n where leaves have positive integer weights. Non-commutative Non-associative version of partitions of n. - Michael Somos, May 23 2005
Appears also as the number of Euler trees with total weight n (associated with even switching class of matrices of order 2n). - David Garber, Sep 19 2005
Number of symmetric hex trees with 2n-1 edges; also number of symmetric hex trees with 2n-2 edges. A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read reference). A hex tree is symmetric if it is identical with its reflection in a bisector through the root. - Emeric Deutsch, Dec 19 2006
The Hankel transform of [1, 2, 5, 15, 51, 188, ...] is [1, 1, 1, 1, 1, ...], see A000012 ; the Hankel transform of [2, 5, 15, 51, 188, 731, ...] is [2, 5, 13, 34, 89, ...], see A001519. - Philippe Deléham, Dec 19 2006
a(n) = number of 321-avoiding partitions of [n]. A partition is 321-avoiding if the permutation obtained from its canonical form (entries in each block listed in increasing order and blocks listed in increasing order of their first entries) is 321-avoiding. For example, the only partition of [5] that fails to be 321-avoiding is 15/24/3 because the entries 5,4,3 in the permutation 15243 form a 321 pattern. - David Callan, Jul 22 2008
The sequence 1,1,2,5,15,51,188,... has Hankel transform A001519. - Paul Barry, Jan 13 2009
From Gary W. Adamson, May 17 2009: (Start)
Equals INVERT transform of A033321: (1, 1, 2, 6, 21, 79, 311, ...).
Equals INVERTi transform of A002212: (1, 3, 10, 36, 137, ...).
Convolved with A026378, (1, 4, 17, 75, 339, ...) = A026376: (1, 6, 30, 144, ...)
(End)
a(n) is the number of vertices of the composihedron CK(n). The composihedra are a sequence of convex polytopes used to define maps of certain homotopy H-spaces. They are cellular quotients of the multiplihedra and cellular covers of the cubes. - Stefan Forcey (sforcey(AT)gmail.com), Dec 17 2009
a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps at level 0 come in 2 colors and those at a higher level come in 3 colors. Example: a(4)=15 because we have 2^3 = 8 paths of shape UHD, 2 paths of shape HUD, 2 paths of shape UDH, and 3 paths of shape UHD; here U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, May 02 2011
REVERT transform of (1, 2, -3, 5, -8, 13, -21, 34, ... ) where the entries are Fibonacci numbers, A000045. Equivalently, coefficients in the series reversion of x(1-x)/(1+x-x^2). This means that the substitution of the gf (1-x-(1-6x+5x^2)^(1/2))/(2(1-x)) for x in x(1-x)/(1+x-x^2) will simplify to x. - David Callan, Nov 11 2012
The number of plane trees with nodes that have positive integer weights and whose total weight is n. - Brad R. Jones, Jun 12 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)] = (x-x^2) / (1 + x - x^2).
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1].
(End)
Starting with offset 0, a(n) is also the number of Schröder paths of semilength n avoiding UH (an up step directly followed by a long horizontal step). Example: a(2)=5 because among the six possible Schröder paths of semilength 2 only UHD contains UH. - Valerie Roitner, Jul 23 2020

Examples

			a(3)=5 since {3, (1+2), (1+(1+1)), (2+1), ((1+1)+1)} are the five weighted binary trees of weight 3.
G.f. = x + 2*x^2 + 5*x^3 + 15*x^4 + 51*x^5 + 188*x^6 + 731*x^7 + 2950*x^8 + 12235*x^9 + ... _Michael Somos_, Jan 17 2018
		

References

  • J. Brunvoll et al., Studies of some chemically relevant polygonal systems: mono-q-polyhexes, ACH Models in Chem., 133 (3) (1996), 277-298, Eq. 15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A181768 for another version. - N. J. A. Sloane, Nov 12 2010
First column of triangle A104259. Row sums of absolute values of A091699.
Number of vertices of multiplihedron A121988.
m-th binomial transform of the Catalan numbers: A126930 (m = -2), A005043 (m = -1), A000108 (m = 0), A064613 (m = 2), A104455 (m = 3), A104498 (m = 4) and A154623 (m = 5).

Programs

  • Maple
    G := (1-sqrt(1-4*z/(1-z)))*1/2: Gser := series(G, z = 0, 30): seq(coeff(Gser, z, n), n = 1 .. 26); # Emeric Deutsch, Aug 12 2007
    seq(round(evalf(JacobiP(n-1,1,-n-1/2,9)/n,99)),n=1..25); # Peter Luschny, Sep 23 2014
  • Mathematica
    Rest@ CoefficientList[ InverseSeries[ Series[(y - y^2)/(1 + y - y^2), {y, 0, 26}], x], x] (* then A(x)=y(x); note that InverseSeries[Series[y-y^2, {y, 0, 24}], x] produces A000108(x) *) (* Len Smiley, Apr 10 2000 *)
    Range[0, 25]! CoefficientList[ Series[ Exp[ 3x] (BesselI[0, 2x] - BesselI[1, 2x]), {x, 0, 25}], x] (* Robert G. Wilson v, Apr 15 2011 *)
    a[n_] := Sum[ Binomial[n, k]*CatalanNumber[k], {k, 0, n}]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 07 2012 *)
    Rest[CoefficientList[Series[3/2 - (1/2) Sqrt[(1 - 5 x)/(1 - x)], {x, 0, 40}], x]] (* Vincenzo Librandi, Nov 03 2014 *)
    Table[Hypergeometric2F1[1/2, -n+1, 2, -4], {n, 1, 30}] (* Vaclav Kotesovec, May 12 2022 *)
  • PARI
    {a(n) = my(A); if( n<2, n>0, A=vector(n); for(j=1,n, A[j] = 1 + sum(k=1,j-1, A[k]*A[j-k])); A[n])}; /* Michael Somos, May 23 2005 */
    
  • PARI
    {a(n) = if( n<1, 0, polcoeff( serreverse( (x - x^2) / (1 + x - x^2) + x * O(x^n)), n))}; /* Michael Somos, May 23 2005 */
    
  • PARI
    /* Offset = 0: */ {a(n)=local(A=1+x);for(i=1,n, A=sum(m=0,n, x^m*sum(k=0,m,A^k)+x*O(x^n))); polcoeff(A,n)} \\ Paul D. Hanna

Formula

(n+2)*a(n+2) = (6n+4)*a(n+1) - 5n*a(n).
G.f.: 3/2-(1/2)*sqrt((1-5*x)/(1-x)) [Gessel-Kim]. - N. J. A. Sloane, Jul 05 2014
G.f. for sequence doubled: (1/(2*x))*(1+x-(1-x)^(-1)*(1-x^2)^(1/2)*(1-5*x^2)^(1/2)).
a(n) = hypergeom([1/2, -n], [2], -4), n=0, 1, 2...; Integral representation as n-th moment of a positive function on a finite interval of the positive half-axis: a(n)=int(x^n*sqrt((5-x)/(x-1))/(2*Pi), x=1..5), n=0, 1, 2... This representation is unique. - Karol A. Penson, Sep 24 2001
a(1)=1, a(n)=1+sum(i=1, n-1, a(i)*a(n-i)). - Benoit Cloitre, Mar 16 2004
a(n) = Sum_{k=0..n} (-1)^k*3^(n-k)*binomial(n, k)*binomial(k, floor(k/2)) [offset 0]. - Paul Barry, Jan 27 2005
G.f. A(x) satisfies 0=f(x, A(x)) where f(x, y)=x-(1-x)(y-y^2). - Michael Somos, May 23 2005
G.f. A(x) satisfies 0=f(x, A(x), A(A(x))) where f(x, y, z)=x(z-z^2)+(x-1)y^2 . - Michael Somos, May 23 2005
G.f. (for offset 0): (-1+x+(1-6*x+5*x^2)^(1/2))/(2*(-x+x^2)).
G.f. =z*c(z/(1-z))/(1-z) = 1/2 - (1/2)sqrt(1-4z/(1-z)), where c(z)=(1-sqrt(1-4z))/(2z) is the Catalan function (follows from Michael Somos' first comment). - Emeric Deutsch, Aug 12 2007
G.f.: 1/(1-2x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-.... (continued fraction). - Paul Barry, Apr 19 2009
a(n) = Sum_{k, 0<=k<=n} A091965(n,k)*(-1)^k. - Philippe Deléham, Nov 28 2009
E.g.f.: exp(3x)*(I_0(2x)-I_1(2x)), where I_k(x) is a modified Bessel function of the first kind. - Emanuele Munarini, Apr 15 2011
If we prefix sequence with an additional term a(0)=1, g.f. is (3-3*x-sqrt(1-6*x+5*x^2))/(2*(1-x)). [See Kim, 2011] - N. J. A. Sloane, May 13 2011
From Gary W. Adamson, Jul 21 2011: (Start)
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows:
2, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 1, 2, 1, 0, 0, ...
1, 1, 1, 2, 1, 0, ...
1, 1, 1, 1, 2, 1, ...
1, 1, 1, 1, 1, 2, ...
... (End)
G.f. satisfies: A(x) = Sum_{n>=0} x^n * (1 - A(x)^(n+1))/(1 - A(x)); offset=0. - Paul D. Hanna, Nov 07 2011
G.f.: 1/x - 1/x/Q(0), where Q(k)= 1 + (4*k+1)*x/((1-x)*(k+1) - x*(1-x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1-x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
G.f.: (1-x - (1-5*x)*G(0))/(2*x*(1-x)), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-x) - 2*x*(1-x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 25 2013
Asymptotics (for offset 0): a(n) ~ 5^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 28 2013
G.f.: G(0)/(1-x), where G(k) = 1 + (4*k+1)*x/((k+1)*(1-x) - 2*x*(1-x)*(k+1)*(4*k+3)/(2*x*(4*k+3) + (2*k+3)*(1-x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 29 2014
a(n) = JacobiP(n-1,1,-n-1/2,9)/n. - Peter Luschny, Sep 23 2014
0 = +a(n)*(+25*a(n+1) -50*a(n+2) +15*a(n+3)) +a(n+1)*(-10*a(n+1) +31*a(n+2) -14*a(n+3)) +a(n+2)*(+2*a(n+2) +a(n+3)) for all n in Z. - Michael Somos, Jan 17 2018
a(n+1) = (2/Pi) * Integral_{x = -1..1} (m + 4*x^2)^n*sqrt(1 - x^2) dx at m = 1. In general, the integral, qua sequence in n, gives the m-th binomial transform of the Catalan numbers. - Peter Bala, Jan 26 2020

A005554 Sums of successive Motzkin numbers.

Original entry on oeis.org

1, 2, 3, 6, 13, 30, 72, 178, 450, 1158, 3023, 7986, 21309, 57346, 155469, 424206, 1164039, 3210246, 8893161, 24735666, 69051303, 193399578, 543310782, 1530523638, 4322488212, 12236130298, 34713220977, 98677591278
Offset: 1

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Comments

The Donaghey reference shows that a(n) is the number of n-vertex binary trees such that for each non-root vertex that is incident to exactly two edges, these two edges have opposite slope. It also notes that these trees correspond to Dyck n-paths (A000108) containing no DUDUs and no subpaths of the form UUPDD with P a nonempty Dyck path. For example, a(3)=3 counts UUDUDD, UDUUDD, UUDDUD. - David Callan, Sep 25 2006
Hankel transform of the sequence starting with 2 appears to be 3, 4, 5, 6, 7, ... Gary W. Adamson, May 27 2011

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Enumerates the branch-reduced trees encoded by A080981. Cf. A001006.
First differences are in A102071.
Cf. A014138.
A diagonal of A059346.

Programs

  • Mathematica
    Rest[CoefficientList[Series[(x+x^2)*(1-x-(1-2*x-3*x^2)^(1/2))/(2*x^2), {x, 0, 20}], x]] (* Vaclav Kotesovec, Mar 21 2014 *)
  • Maxima
    a(n):=(2*sum(binomial(n,j)*binomial(n-j+1,n-2*j+2),j,0,(n+2)/2))/n; /* Vladimir Kruchinin, Oct 04 2015 */
    
  • PARI
    a(n) = sum(k=0, (n+2)/2, 2*(binomial(n,k)*binomial(n-k+1,n-2*k+2)/n));
    vector(40, n, if(n==1, 1, a(n-1))) \\ Altug Alkan, Oct 04 2015

Formula

Inverse binomial transform of A014138: (1, 3, 8, 22, 64, 196, ...). - Gary W. Adamson, Nov 23 2007
D-finite with recurrence (n + 1)*a(n) = 2*n*a(n - 1) + (3*n - 9)*a(n - 2).
G.f.: (x+x^2)*M(x) where M(x)=(1 - x - (1 - 2*x - 3*x^2)^(1/2))/(2*x^2) is the g.f. for the Motzkin numbers A001006. - David Callan, Sep 25 2006
a(n) = (-1)^n*2*hypergeometric([2-n,5/2],[4],4), for n>1. - Peter Luschny, Aug 15 2012
a(n) ~ 2*3^(n-1/2)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 21 2014
a(n) = (2*Sum_{j=0..(n+2)/2} (binomial(n,j)*binomial(n-j+1,n-2*j+2)))/n. - Vladimir Kruchinin, Oct 04 2015

Extensions

More terms from James Sellers, Jul 10 2000

A026012 Second differences of Catalan numbers A000108.

Original entry on oeis.org

1, 2, 6, 19, 62, 207, 704, 2431, 8502, 30056, 107236, 385662, 1396652, 5088865, 18642420, 68624295, 253706790, 941630580, 3507232740, 13105289370, 49114150020, 184560753390, 695267483664, 2625197720454, 9933364416572, 37660791173152, 143048202990504
Offset: 0

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Number of (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n, s(0) = s(2n) = 2.
Number of Dyck paths of semilength n+2 with no initial and no final UD's. Example: a(2)=6 because the only Dyck paths of semilength 4 with no initial and no final UD's are UUDUDUDD, UUDUUDDD, UUUDDUDD, UUUDUDDD, UUDDUUDD, UUUUDDDD. - Emeric Deutsch, Oct 26 2003
Number of branches of length 1 starting from the root in all ordered trees with n+1 edges. Example: a(1)=2 because the tree /\ has two branches of length 1 starting from the root and the path-tree of length 2 has none. a(n) = Sum_{k=0..n+1} (k*A127158(n+1,k)). - Emeric Deutsch, Mar 01 2007
Number of staircase walks from (0,0) to (n,n) that never cross y=x+2. Example: a(3) = 19 because up,up,up,right,right,right is not allowed but the other binomial(6,3)-1 = 19 paths are. - Mark Spindler, Nov 11 2012
Number of standard Young tableaux of skew shape (n+2,n)/(2), for n>=2. - Ran Pan, Apr 07 2015

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 188, 196).

Crossrefs

T(2n, n), where T is the array defined in A026009.

Programs

  • Mathematica
    Differences[Table[CatalanNumber[n], {n, 0, 28}], 2] (* Jean-François Alcover, Sep 28 2012 *)
    Table[Binomial[2n,n]-Binomial[2n,n-3],{n,0,26}] (* Mark Spindler, Nov 11 2012 *)
  • PARI
    a(n) = 3*(3*n^2+3*n+2)*binomial(2*n, n)/((n+1)*(n+2)*(n+3)); /* Joerg Arndt, Aug 19 2012 */

Formula

Expansion of (1+x^1*C^3)*C^1, where C = (1-(1-4*x)^(1/2))/(2*x) is g.f. for Catalan numbers, A000108.
a(n) = 3*(3*n^2+3*n+2)*binomial(2*n, n)/((n+1)*(n+2)*(n+3)). - Emeric Deutsch, Oct 26 2003
a(n) = Sum_{k=0..2} A039599(n,k) = A000108(n) + A000245(n) + A000344(n). - Philippe Deléham, Nov 12 2008
a(n) = binomial(2*n,n)/(n+1)*hypergeom([-2,n+1/2],[n+2],4). - Peter Luschny, Aug 15 2012
a(n) = binomial(2*n,n) - binomial(2n,n-3). - Mark Spindler, Nov 11 2012
D-finite with recurrence (n+3)*a(n) + (-5*n-6)*a(n-1) + 2*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Jun 20 2013
E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(3,2*x)). - Ilya Gutkovskiy, Feb 28 2017
Sum_{n>=0} a(n)/4^n = 6. - Amiram Eldar, Jul 10 2023
a(n) = C(n+2)+C(n)-2*C(n+1), C = A000108. - Alois P. Heinz, Apr 02 2025
Binomial transform of A342912. - Mélika Tebni, Apr 05 2025

A106534 Sum array of Catalan numbers (A000108) read by upward antidiagonals.

Original entry on oeis.org

1, 2, 1, 5, 3, 2, 15, 10, 7, 5, 51, 36, 26, 19, 14, 188, 137, 101, 75, 56, 42, 731, 543, 406, 305, 230, 174, 132, 2950, 2219, 1676, 1270, 965, 735, 561, 429, 12235, 9285, 7066, 5390, 4120, 3155, 2420, 1859, 1430, 51822, 39587, 30302, 23236, 17846, 13726, 10571, 8151, 6292, 4862
Offset: 0

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Philippe Deléham, May 30 2005

Keywords

Comments

The underlying array A is A(n, k) = Sum_{j=0..n} binomial(n, j)*A000108(k+j), n >= 0, k>= 0. See the example section. - Wolfdieter Lang, Oct 04 2019

Examples

			From _Wolfdieter Lang_, Oct 04 2019: (Start)
The triangle T(n, k) begins:
n\k      0      1      2      3     4     5     6     7     8     9    10 ...
0:       1
1:       2      1
2:       5      3      2
3:      15     10      7      5
4:      51     36     26     19    14
5:     188    137    101     75    56    42
6:     731    543    406    305   230   174   132
7:    2950   2219   1676   1270   965   735   561   429
8:   12235   9285   7066   5390  4120  3155  2420  1859  1430
9:   51822  39587  30302  23236 17846 13726 10571  8151  6292  4862
10: 223191 171369 131782 101480 78244 60398 46672 36101 27950 21658 16796
... reformatted and extended.
-------------------------------------------------------------------------
The array A(n, k) begins:
n\k  0   1    2    3     4     5      6 ...
-------------------------------------------
0:   1   1    2    5    14    42    132 ... A000108
1    2   3    7   19    56   174    561 ... A005807
2:   5  10   26   75   230   735   2420 ...
3:  15  36  101  305   965  3155  10571 ...
4:  51 137  406 1270  4120 13726  46672 ...
5: 188 543 1676 5390 17846 60398 207963 ...
... (End)
		

Crossrefs

Columns: A007317, A002212, see also A045868, A055452-A055455.
Diagonals: A000108, A005807.
Cf. A059346 (Catalan difference array as triangle).

Programs

  • Magma
    function T(n,k)
      if k gt n then return 0;
      elif k eq n then return Catalan(n);
      else return T(n-1, k) + T(n, k+1);
      end if; return T;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 18 2021
  • Maple
    # Uses floating point, precision might have to be adjusted.
    C := n -> binomial(2*n,n)/(n+1);
    H := (n,k) -> hypergeom([k-n,k+1/2],[k+2],-4);
    T := (n,k) -> C(k)*H(n,k);
    seq(print(seq(round(evalf(T(n,k),32)),k=0..n)),n=0..7);
    # Peter Luschny, Aug 16 2012
  • Mathematica
    T[n_, n_] := CatalanNumber[n]; T[n_, k_] /; 0 <= k < n := T[n-1, k] + T[n, k+1]; T[, ] = 0; Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)
  • Sage
    def T(n, k) :
        if k > n : return 0
        if n == k : return binomial(2*n, n)/(n+1)
        return T(n-1, k) + T(n, k+1)
    A106534 = lambda n,k: T(n, k)
    for n in (0..5): [A106534(n,k) for k in (0..n)] # Peter Luschny, Aug 16 2012
    

Formula

T(n, k) = 0 if k > n; T(n, n) = A000108(n); T(n, k) = T(n-1, k) + T(n, k+1) if 0 <= k < n.
T(n, k) = binomial(2*k,k)/(k+1)*hypergeometric([k-n, k+1/2], [k+2], -4). - Peter Luschny, Aug 16 2012
T(n, k) = A(n-k, k) = Sum_{j=0..n-k} binomial(n-k, j)*A000108(k+j), n >= 0, k = 0..n. - Wolfdieter Lang, Oct 03 2019
G.f.: (sqrt(1-4*x*y)-sqrt((5*x-1)/(x-1)))/(2*x*(x*y-y+1)). - Vladimir Kruchinin, Jan 12 2024

A228339 Fourth differences of Catalan numbers (A000108).

Original entry on oeis.org

3, 9, 30, 102, 352, 1230, 4344, 15483, 55626, 201246, 732564, 2681223, 9861342, 36428320, 135100620, 502841295, 1877678370, 7032454470, 26410804020, 99437742720, 375260126904, 1419223506516, 5378236459328, 20419260060462, 77659985060772, 295844249258796, 1128738495397128, 4312685074680465, 16500218817839274, 63209983347693924
Offset: 0

Views

Author

N. J. A. Sloane, Aug 29 2013

Keywords

Crossrefs

Programs

  • Mathematica
    Differences[Table[CatalanNumber[n], {n, 0, 30}], 4] (* Amiram Eldar, Jul 10 2023 *)

Formula

From Amiram Eldar, Jul 10 2023: (Start)
a(n) = 9*(9*n^4 + 54*n^3 + 135*n^2 + 122*n + 40) * n! * binomial(2*n, n)/(n+5)!.
Sum_{n>=0} a(n)/4^n = 38. (End)
Showing 1-8 of 8 results.