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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A358371 Number of leaves in the n-th standard ordered rooted tree.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 4, 1, 3, 2, 3, 2, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 5, 2, 2, 3, 4, 2, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 5, 2, 4, 3, 4, 3, 4, 4, 5, 4, 4, 4, 5, 4, 5, 5, 6, 2, 3, 2, 3, 3, 4, 4, 5, 3, 3, 3, 4, 3, 4, 4, 5, 2, 4, 3, 4, 3, 4
Offset: 1

Views

Author

Gus Wiseman, Nov 13 2022

Keywords

Comments

We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

Examples

			The standard ordered rooted tree ranking begins:
  1: o        10: (((o))o)   19: (((o))(o))
  2: (o)      11: ((o)(o))   20: (((o))oo)
  3: ((o))    12: ((o)oo)    21: ((o)((o)))
  4: (oo)     13: (o((o)))   22: ((o)(o)o)
  5: (((o)))  14: (o(o)o)    23: ((o)o(o))
  6: ((o)o)   15: (oo(o))    24: ((o)ooo)
  7: (o(o))   16: (oooo)     25: (o(oo))
  8: (ooo)    17: ((((o))))  26: (o((o))o)
  9: ((oo))   18: ((oo)o)    27: (o(o)(o))
For example, the 25th ordered tree is (o,(o,o)) because the 24th composition is (1,4) and the 3rd composition is (1,1). Hence a(25) = 3.
		

Crossrefs

The triangle counting trees by this statistic is A001263, unordered A055277.
The version for unordered trees is A109129, nodes A061775, edges A196050.
The nodes are counted by A358372.
A000081 counts unordered rooted trees, ranked by A358378.
A000108 counts ordered rooted trees.
A358374 ranks ordered identity trees, counted by A032027.
A358375 ranks ordered binary trees, counted by A126120

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    srt[n_]:=If[n==1,{},srt/@stc[n-1]];
    Table[Count[srt[n],{},{0,Infinity}],{n,100}]

A097610 Triangle read by rows: T(n,k) is number of Motzkin paths of length n and having k horizontal steps.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 2, 0, 6, 0, 1, 0, 10, 0, 10, 0, 1, 5, 0, 30, 0, 15, 0, 1, 0, 35, 0, 70, 0, 21, 0, 1, 14, 0, 140, 0, 140, 0, 28, 0, 1, 0, 126, 0, 420, 0, 252, 0, 36, 0, 1, 42, 0, 630, 0, 1050, 0, 420, 0, 45, 0, 1, 0, 462, 0, 2310, 0, 2310, 0, 660, 0, 55, 0, 1, 132, 0, 2772, 0
Offset: 0

Views

Author

Emeric Deutsch, Aug 30 2004

Keywords

Comments

Row sums are the Motzkin numbers (A001006). Column 0 gives the aerated Catalan numbers (A000108).
Let P_n(x) = Sum_{k=0..n} T(n,k)*x^k. P_0(x) = 1, P_1(x) = x, P_n(x) = x*P_(n-1)(x) + Sum_{j=0..n-2} P_j(x)*P_(n-2-j)(x); essentially the same as A124027. - Philippe Deléham, Oct 03 2007
G. J. Chaitin's numbers of s-expressions of size n are given by the coefficients of polynomials p(k, x) satisfying: p(k, x) = Sum_{j=2..k-1} p(j, x)*p(k-j, x). The coefficients of these polynomials also give (essentially) the triangle shown here. - Roger L. Bagula, Oct 31 2006
Exponential Riordan array [Bessel_I(1,2x)/x,x]. - Paul Barry, Mar 24 2010
Diagonal sums are the aerated large Schroeder numbers. - Paul Barry, Apr 21 2010
Non-vanishing antidiagonals are rows of A060693. - Tom Copeland, Feb 03 2016
These polynomials are related to the Gegenbauer polynomials which in turn are specializations of the Jacobi polynomials. The o.g.f. of the Gegenbauer polynomials is 1 / [1-2tx+x^2]^a. For the generating function Gb(x,h1,h2,a) = [x / (1 + h1 x + h2 x^2)]^a, the compositional inverse in x is Gbinv(x,h1,h2,a) = [(1-h1*y) - sqrt[(1-h1*y)^2-4h2*y^2]]/(2*h2*y) with y = x^(1/a). The polynomials of this entry are generated by Gbinv(x,t,1,1). The Legendre polynomials are related to the o.g.f. Gb(x,-2t,1,1/2). Cf. A121448. - Tom Copeland, Feb 07 2016
The bivariate o.g.f. in Copeland's Jan 29 2016 formulas can be related to conformal mappings of the complex plane and a solution of the dKP hierarchy. Cf. p. 24 of the Takebe et al. paper. - Tom Copeland, May 30 2018

Examples

			Triangle begins:
1;
0,  1;
1,  0,  1;
0,  3,  0,  1;
2,  0,  6,  0,  1;
0, 10,  0, 10,  0,  1;
5,  0, 30,  0, 15,  0,  1;
Row n has n+1 terms.
T(4,2)=6 because we have HHUD, HUDH, UDHH, HUHD, UHDH, UHHD, where U=(1,1), D=(1,-1) and H=(1,0).
		

References

  • G. J. Chaitin, Algorithmic Information Theory, Cambridge Univ. Press, 1987, page 169.

Crossrefs

Cf. A001006, A000108. A124027 is an essentially identical triangle.
Cf. A001263.

Programs

  • Maple
    G:=(1-t*z-sqrt(1-2*t*z+t^2*z^2-4*z^2))/2/z^2:
    Gser:=simplify(series(G,z=0,16)): P[0]:=1:
    for n from 1 to 13 do P[n]:=sort(coeff(Gser,z^n)) od:
    seq(seq(coeff(t*P[n],t^k),k=1..n+1),n=0..13);
    # Maple program for the triangular array:
    T:=proc(n,k) if n-k mod 2 = 0 and k<=n then n!/k!/((n-k)/2)!/((n-k)/2+1)! else 0 fi end: TT:=(n,k)->T(n-1,k-1): matrix(10,10,TT);
  • Mathematica
    T[n_,k_]:=If[n>=k&&EvenQ[n-k],n!/(k!((n-k)/2)!((n-k)/2+1)!),0];
    Flatten[Table[T[n,k],{n,0,20},{k,0,n}]] (* Peter J. C. Moses, Apr 06 2013 *)
    T[n_,k_] := If[OddQ[n - k], 0, Binomial[n, k] CatalanNumber[(n - k)/2]]; (* Peter Luschny, Jun 06 2018 *)

Formula

G.f.: [1-tz-sqrt(1-2tz+t^2*z^2-4z^2)]/(2z^2).
T(n, k) = n!/[k!((n-k)/2)!((n-k)/2-1)! ] = A055151(n, (n-k)/2) if n-k is a nonnegative even number; otherwise T(n, k) = 0.
T(n, k) = C(n, k)*C((n-k)/2)*(1+(-1)^(n-k))/2 if k <= n, 0 otherwise. - Paul Barry, May 18 2005
T(n,k) = A121448(n,k)/2^k. - Philippe Deléham, Aug 17 2006
Sum_{k=0..n} T(n,k)*2^k = A000108(n+1). - Philippe Deléham, Aug 22 2006
Sum_{k=0..n} T(n,k)*3^k = A002212(n+1). - Philippe Deléham, Oct 03 2007
G.f.: 1/(1-x*y-x^2/(1-x*y-x^2/(1-x*y-x^2/.... (continued fraction). - Paul Barry, Dec 15 2008
Sum_{k=0..n} T(n,k)*4^k = A005572(n). - Philippe Deléham, Dec 03 2009
T(n,k) = A007318(n,k)*A126120(n-k). - Philippe Deléham, Dec 12 2009
From Tom Copeland, Jan 23 2016: (Start)
E.g.f.: M(x,t) = e^(xt) AC(t) = e^(xt) I_1(2t)/t = e(xt) * e.g.f.(A126120(t)) = e^(xt) Sum_{n>=0} t^(2n)/(n!(n+1)!) = exp[t P(.,x)].
The e.g.f. of this Appell sequence of polynomials P(n,x) is e^(xt) times the e.g.f. AC(t) of the aerated Catalan numbers A126120. AC(t) = I_1(2t)/t, where I_n(x) = T_n(d/dx) I_0(x) are the modified Bessel functions of the first kind and T_n, the Chebyshev polynomials of the first kind.
P(n,x) has the lowering and raising operators L = d/dx = D and R = d/dD log{M(x,D)} = x + d/dD log{AC(D)} = x + Sum_{n>=0} c(n) D^(2n+1)/(2n+1)! with c(n) = (-1)^n A180874(n+1), i.e., L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
(P(.,x) + y)^n = P(n,x+y) = Sum_{k=0..n} binomial(n,k) P(k,x) y^(n-k) = (b.+x+y)^n, where (b.)^k = b_k = A126120(k).
Exp(b.D) e^(xt) = exp[(x+b.)t] = exp[P(.,x)t] = e^(b.t) e^(xt) = e^(xt) AC(t).
See p. 12 of the Alexeev et al. link and A055151 for a refinement.
Shifted o.g.f: G(x,t) = [1-tx-sqrt[(1-tx)^2-4x^2]] / 2x = x + t x^2 + (1+t) x^3 + ... has the compositional inverse Ginv(x,t) = x / [1 + tx + x^2] = x - t x^2 +(-1+t^2) x^3 + (2t-t^3) x^4 + (1-3t^2+t^4) x^5 + ..., a shifted o.g.f. for the signed Chebyshev polynomials of the second kind of A049310 (cf. also the Fibonacci polynomials of A011973). Then the inversion formula of A134264, involving non-crossing partitions and free probability with their multitude of interpretations (cf. A125181 also), can be used with h_0 = 1, h_1 = t, and h_2 = 1 to interpret the coefficients of the Motzkin polynomials combinatorially.
(End)
From Tom Copeland, Jan 29 2016: (Start)
Provides coefficients of the inverse of f(x) = x / [1 + h1 x + h2 x^2], a bivariate generating function of A049310 (mod signs).
finv(x) = [(1-h1*x) - sqrt[(1-h1*x)^2-4h2*x^2]]/(2*h2*x) = x + h1 x^2 + (h2 + h1^2) x^3 + (3 h1 h2 + h1^3) x^4 + ... is a bivariate o.g.f. for this entry.
The infinitesimal generator for finv(x) is g(x) d/dx with g(x) = 1 /[df(x)/dx] = x^2 / [(f(x))^2 (1 - h2 x^2)] = (1 + h1 x + h2 x^2)^2 / (1 - h2 x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomials FI(n,h1,h2) of the compositional inverse of f(x), i.e., exp[x g(u)d/du] u |_(u=0) = finv(x) = 1 / [1 -x FI(.,h1,h2)]. Cf. A145271. E.g.,
FI(0,h1,h2) = 0
FI(1,h1,h2) = 1
FI(2,h1,h2) = 1 h1
FI(3,h1,h2) = 1 h2 + 1 h1^2
FI(4,h1,h2) = 3 h2 h1 + 1 h1^3
FI(5,h1,h2) = 2 h2^2 + 6 h2 h1^2 + 1 h1^4
FI(6,h1,h2) = 10 h2^2 h1 + 10 h2 h1^3 + 1 h1^5.
And with D = d/dh1, FI(n+1, h1,h2) = MT(n,h1,h2) = (b.y + h1)^n = Sum_{k=0..n} binomial(n,k) b(k) y^k h1^(n-k) = exp[(b.y D] (h1)^n = AC(y D) (h1)^n, where b(k) = A126120(k), y = sqrt(h2), and AC(t) is defined in my Jan 23 formulas above. Equivalently, AC(y D) e^(x h1) = exp[x MT(.,h1,h2)].
The MT polynomials comprise an Appell sequence in h1 with e.g.f. e^(h1*x) AC(xy) = exp[x MT(.,h1,h2)] with lowering operator L = d/dh1 = D, i.e., L MT(n,h1,h2) = dMT(n,h1,h2)/dh1 = n MT(n-1,h1,h2) and raising operator R = h1 + dlog{AC(y L)}/dL = h1 + Sum_{n>=0} c(n) h2^(n+1) D^(2n+1)/(2n+1)! = h1 + h2 d/dh1 - h2^2 (d/dh1)^3/3! + 5 h2^3 (d/dh1)^5/5! - ... with c(n) = (-1)^n A180874(n+1) (consistent with the raising operator in the Jan 23 formulas).
The compositional inverse finv(x) is also obtained from the non-crossing partitions of A134264 (or A125181) with h_0 = 1, h_1 = h1, h_2 = h2, and h_n = 0 for all other n.
See A238390 for the umbral compositional inverse in h1 of MT(n,h1,h2) and inverse matrix.
(End)
From Tom Copeland, Feb 13 2016: (Start)
z1(x,h1,h2) = finv(x), the bivariate o.g.f. above for this entry, is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,h1,h2),z2(x,h1,h2)) = (z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-h1*x)/(h2*x)] z + 1/h2, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,h1,h2) = (1 - h1*x)/(h2*x) - z1(x,h1,h2) = [1 - h1*x + sqrt[(1-h1*x)^2 - 4 h2*x^2]] / (2h2*x).
The two are the nontrivial zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2), (see Landweber et al., p. 14, Ellingsrud, and A121448), and the zeros for the Riccati equation z' = (z - z1)(z - z2), associated to soliton solutions of the KdV equation (see Copeland link).
(End)
Comparing the shifted o.g.f. S(x) = x / (1 - h_1 x + h_2 x^2) for the bivariate Chebyshev polynomials S_n(h_1,h_2) of A049310 with the shifted o.g.f. H(x) = x / ((1 - a x)(1 - b x)) for the complete homogeneous symmetric polynomials H_n(a,b) = (a^(n+1)-b^(n+1)) / (a - b) shows that S_n(h_1,h_2) = H_n(a,b) for h_1 = a + b and h_2 = ab and, conversely, a = (h_1 + sqrt(h_1^2 - 4 h_2)) / 2 and b = (h_1 - sqrt(h_1^2 - 4 h_2)) / 2. The compositional inverse about the origin of S(x) gives a bivariate o.g.f. for signed Motzkin polynomials M_n(h_1,h_2) of this entry, and that of H(x) gives one for signed Narayana polynomials N_n(a,b) of A001263, thereby relating the bivariate Motzkin and Narayana polynomials by the indeterminate transformations. E.g., M_2(h_1,h_2) = h_2 + h_1^2 = ab + (a + b)^2 = a^2 + 3 ab + b^2 = N_2(a,b). - Tom Copeland, Jan 27 2024

A055151 Triangular array of Motzkin polynomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 6, 2, 1, 10, 10, 1, 15, 30, 5, 1, 21, 70, 35, 1, 28, 140, 140, 14, 1, 36, 252, 420, 126, 1, 45, 420, 1050, 630, 42, 1, 55, 660, 2310, 2310, 462, 1, 66, 990, 4620, 6930, 2772, 132, 1, 78, 1430, 8580, 18018, 12012, 1716, 1, 91, 2002, 15015, 42042
Offset: 0

Views

Author

Michael Somos, Jun 14 2000

Keywords

Comments

T(n,k) = number of Motzkin paths of length n with k up steps. T(n,k)=number of 0-1-2 trees with n edges and k+1 leaves, n>0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) E.g., T(4,1)=6 because we have UDHH, UHDH, UHHD, HHUD, HUHD, HUDH, where U=(1,1), D(1,-1), H(1,0). - Emeric Deutsch, Nov 30 2003
Coefficients in series reversion of x/(1+H*x+U*D*x^2) corresponding to Motzkin paths with H colors for H(1,0), U colors for U(1,1) and D colors for D(1,-1). - Paul Barry, May 16 2005
Eigenvector equals A119020, so that A119020(n) = Sum_{k=0..[n/2]} T(n,k)*A119020(k). - Paul D. Hanna, May 09 2006
Row reverse of A107131. - Peter Bala, May 07 2012
Also equals the number of 231-avoiding permutations of n+1 for which descents(w) = peaks(w) = k, where descents(w) is the number of positions i such that w[i]>w[i+1], and peaks(w) is the number of positions i such that w[i-1]w[i+1]. For example, T(4,1) = 6 because 13245, 12435, 14235, 12354, 12534, 15234 are the only 231-avoiding permutations of 5 elements with descents(w) = peaks(w) = 1. - Kyle Petersen, Aug 02 2013
Apparently, a refined irregular triangle related to this triangle (and A097610) is given in the Alexeev et al. link on p. 12. This entry's triangle is also related through Barry's comment to A125181 and A134264. The diagonals of this entry are the rows of A088617. - Tom Copeland, Jun 17 2015
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 24 2015

Examples

			The irregular triangle T(n,k) begins:
n\k 0  1   2    3   4  5 ...
0:  1
1:  1
2:  1  1
3:  1  3
4:  1  6   2
5:  1 10  10
6:  1 15  30    5
7:  1 21  70   35
8:  1 28 140  140  14
9:  1 36 252  420 126
10: 1 45 420 1050 630 42
... reformatted. - _Wolfdieter Lang_, Aug 24 2015
		

References

  • Miklos Bona, Handbook of Enumerative Combinatorics, CRC Press (2015), page 617, Corollary 10.8.2
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 4.3.

Crossrefs

A107131 (row reversed), A080159 (with trailing zeros), A001006 = row sums, A000108(n) = T(2n, n), A001700(n) = T(2n+1, n), A119020 (eigenvector), A001263 (Narayana numbers), A089627 (gamma vectors of type B associahedra), A101280 (gamma vectors of type A permutohedra).
Cf. A014531.

Programs

  • Maple
    b:= proc(x, y) option remember;
          `if`(y>x or y<0, 0, `if`(x=0, 1, expand(
           b(x-1, y) +b(x-1, y+1) +b(x-1, y-1)*t)))
        end:
    T:= n-> (p-> seq(coeff(p, t, i), i=0..degree(p)))(b(n, 0)):
    seq(T(n), n=0..20);  # Alois P. Heinz, Feb 05 2014
  • Mathematica
    m=(1-x-(1-2x+x^2-4x^2y)^(1/2))/(2x^2 y); Map[Select[#,#>0&]&, CoefficientList[ Series[m,{x,0,15}],{x,y}]]//Grid (* Geoffrey Critzer, Feb 05 2014 *)
    p[n_] := Hypergeometric2F1[(1-n)/2, -n/2, 2, 4 x]; Table[CoefficientList[p[n], x], {n, 0, 13}] // Flatten (* Peter Luschny, Jan 23 2018 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, n! / ((n-2*k)! * k! * (k+1)!))}
    
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, polcoeff( polcoeff( 2 / (1 - x + sqrt((1 - x)^2 - 4*y*x^2 + x * O(x^n))), n), k))} /* Michael Somos, Feb 14 2006 */
    
  • PARI
    {T(n, k) = n++; if( k<0 || 2*k>n, 0, polcoeff( polcoeff( serreverse( x / (1 + x + y*x^2) + x * O(x^n)), n), k))} /* Michael Somos, Feb 14 2006 */

Formula

T(n,k) = n!/((n-2k)! k! (k+1)!) = A007318(n, 2k)*A000108(k). - Henry Bottomley, Jan 31 2003
E.g.f. row polynomials R(n,y): exp(x)*BesselI(1, 2*x*sqrt(y))/(x*sqrt(y)). - Vladeta Jovovic, Aug 20 2003
G.f. row polynomials R(n,y): 2 / (1 - x + sqrt((1 - x)^2 - 4 *y * x^2)).
From Peter Bala, Oct 28 2008: (Start)
The rows of this triangle are the gamma vectors of the n-dimensional (type A) associahedra (Postnikov et al., p. 38). Cf. A089627 and A101280.
The row polynomials R(n,x) = Sum_{k = 0..n} T(n,k)*x^k begin R(0,x) = 1, R(1,x) = 1, R(2,x) = 1 + x, R(3,x) = 1 + 3*x. They are related to the Narayana polynomials N(n,x) := Sum_{k = 1..n} (1/n)*C(n,k)*C(n,k-1)*x^k through x*(1 + x)^n*R(n, x/(1 + x)^2) = N(n+1,x). For example, for n = 3, x*(1 + x)^3*(1 + 3*x/(1 + x)^2) = x + 6*x^2 + 6*x^3 + x^4, the 4th Narayana polynomial.
Recursion relation: (n + 2)*R(n,x) = (2*n + 1)*R(n-1,x) - (n - 1)*(1 - 4*x)*R(n-2,x), R(0,x) = 1, R(1,x) = 1. (End)
G.f.: M(x,y) satisfies: M(x,y)= 1 + x M(x,y) + y*x^2*M(x,y)^2. - Geoffrey Critzer, Feb 05 2014
T(n,k) = A161642(n,k)*A258820(n,k) = (binomial(n,k)/A003989(n+1, k+1))* A258820(n,k). - Tom Copeland, Jun 18 2015
Let T(n,k;q) = n!*(1+k)/((n-2*k)!*(1+k)!^2)*hypergeom([k,2*k-n],[k+2],q) then T(n,k;0) = A055151(n,k), T(n,k;1) = A008315(n,k) and T(n,k;-1) = A091156(n,k). - Peter Luschny, Oct 16 2015
From Tom Copeland, Jan 21 2016: (Start)
Reversed rows of A107131 are rows of this entry, and the diagonals of A107131 are the columns of this entry. The diagonals of this entry are the rows of A088617. The antidiagonals (bottom to top) of A088617 are the rows of this entry.
O.g.f.: [1-x-sqrt[(1-x)^2-4tx^2]]/(2tx^2), from the relation to A107131.
Re-indexed and signed, this triangle gives the row polynomials of the compositional inverse of the shifted o.g.f. for the Fibonacci polynomials of A011973, x / [1-x-tx^2] = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... . (End)
Row polynomials are P(n,x) = (1 + b.y)^n = Sum{k=0 to n} binomial(n,k) b(k) y^k = y^n M(n,1/y), where b(k) = A126120(k), y = sqrt(x), and M(n,y) are the Motzkin polynomials of A097610. - Tom Copeland, Jan 29 2016
Coefficients of the polynomials p(n,x) = hypergeom([(1-n)/2, -n/2], [2], 4x). - Peter Luschny, Jan 23 2018
Sum_{k=1..floor(n/2)} k * T(n,k) = A014531(n-1) for n>1. - Alois P. Heinz, Mar 29 2020

A025230 a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ...+ a(n-1)*a(1) for n >= 3, with initial terms 3,1.

Original entry on oeis.org

3, 1, 6, 37, 234, 1514, 9996, 67181, 458562, 3172478, 22206420, 157027938, 1120292388, 8055001716, 58314533400, 424740506109, 3110401363122, 22888001498102, 169155516667524, 1255072594261142, 9345400450314924, 69812926066668044, 523072984217339304
Offset: 1

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Author

Keywords

Crossrefs

For Sum_{k = 0..n} m^(n-k)*binomial(n, k)*Catalan(k+1) see A126120 (m = -2), A001006 (m = -1), A000108 (m = 0), A002212 (m = 1), A005572 (m = 2), A182401 (m = 3), A025230 (m = 4).

Programs

  • Maple
    h := n -> simplify(4^n*hypergeom([3/2, -n], [3], -1)):
    a := n -> `if`(n=1, 3, h(n-2)):
    seq(a(n), n=1..21); # Peter Luschny, Feb 03 2015
  • Mathematica
    Rest[CoefficientList[Series[(1-Sqrt[1-12x+32x^2])/2,{x,0,30}],x]]  (* Harvey P. Dale, Feb 22 2011 *)
  • PARI
    a(n)=polcoeff((1-sqrt(1-12*x+32*x^2+x*O(x^n)))/2,n)
    
  • PARI
    {a(n)=if(n<2, 3*(n==1), n--; polcoeff( serreverse( x/(1+6*x+x^2) +x*O(x^n) ), n))} /* Michael Somos, Oct 14 2006 */

Formula

G.f.: (1-sqrt(1-12*x+32*x^2))/2. - Michael Somos, Jun 08 2000
D-finite with recurrence n*a(n) = (12*n-18)*a(n-1) - 32*(n-3)*a(n-2) - Richard Choulet, Dec 17 2009
a(n) ~ 2^(3*n-5/2)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 11 2013
a(n) = 4^(n-2)*hypergeom([3/2, -n+2], [3], -1) for n>1. - Peter Luschny, Feb 03 2015
a(n+1) = GegenbauerC(n-1, -n, -3)/n for n>=1. - Peter Luschny, May 09 2016
From Peter Bala, Feb 03 2024: (Start)
G.f.: 3*x + x^2/(1 - 4*x) * c(x/(1 - 4*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
a(n+2) = Sum_{k = 0..n} 4^(n-k)*binomial(n, k)*Catalan(k+1).
G.f.: 3*x + x^2/(1 - 8*x) * c(-x/(1 - 8*x))^2.
a(n+2) = 8^n * Sum_{k = 0..n} (-8)^(-k)*binomial(n, k)*Catalan(k+1).
a(n+2) = 8^n * hypergeom([-n, 3/2], [3], 1/2).
a(n) is odd iff n is a power of 2. (End)

Extensions

Name clarified by Robert C. Lyons, Feb 06 2025

A126216 Triangle read by rows: T(n,k) is the number of Schroeder paths of semilength n containing exactly k peaks but no peaks at level one (n >= 1; 0 <= k <= n-1).

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 14, 21, 9, 1, 42, 84, 56, 14, 1, 132, 330, 300, 120, 20, 1, 429, 1287, 1485, 825, 225, 27, 1, 1430, 5005, 7007, 5005, 1925, 385, 35, 1, 4862, 19448, 32032, 28028, 14014, 4004, 616, 44, 1, 16796, 75582, 143208, 148512, 91728, 34398, 7644, 936, 54, 1
Offset: 1

Views

Author

Emeric Deutsch, Dec 20 2006

Keywords

Comments

A Schroeder path of semilength n is a lattice path in the first quadrant, from the origin to the point (2n,0) and consisting of steps U=(1,1), D=(1,-1) and H=(2,0).
Also number of Schroeder paths of semilength n containing exactly k doublerises but no (2,0) steps at level 0 (n >= 1; 0 <= k <= n-1). Also number of doublerise-bicolored Dyck paths (doublerises come in two colors; also called marked Dyck paths) of semilength n and having k doublerises of a given color (n >= 1; 0 <= k <= n-1). Also number of 12312- and 121323-avoiding matchings on [2n] with exactly k crossings.
Essentially the triangle given by [1,1,1,1,1,1,1,1,...] DELTA [0,1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 20 2007
Mirror image of triangle A033282. - Philippe Deléham, Oct 20 2007
For relation to Lagrange inversion, or series reversion and the geometry of associahedra, or Stasheff polytopes (and other combinatorial objects), see A133437. - Tom Copeland, Sep 29 2008
First column (k=0) gives the Catalan numbers (A000108). - Alexander Karpov, Jun 10 2018
T(n,k) is the multiplicity of the k-hook representation of the symmetric group in the (n-1)st parking space representation (see Pak and Postnikov, 1995). - Joshua Mundinger, Jul 18 2025

Examples

			T(3,1)=5 because we have HUUDD, UUDDH, UUUDDD, UHUDD and UUDHD.
Triangle starts:
   n\k  0      1      2      3      4     5    6   7  8
   1    1;
   2    2,     1;
   3    5,     5;     1;
   4   14,    21,     9,     1;
   5   42,    84,    56,    14,     1;
   6  132,   330,   300,   120,    20,    1;
   7  429,  1287,  1485,   825,   225,   27,   1;
   8 1430,  5005,  7007,  5005,  1925,  385,  35,  1;
   9 4862, 19448, 32032, 28028, 14014, 4004, 616, 44, 1;
  10 ...
Triangle [1,1,1,1,1,1,1,...] DELTA [0,1,0,1,0,1,0,1,...] begins:
   1;
   1,  0;
   2,  1,  0;
   5,  5,  1,  0;
  14, 21,  9,  1,  0;
  42, 84, 56, 14,  1,  0;
  ...
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->binomial(n,k)*binomial(2*n-k,n+1)/n: for n from 1 to 11 do seq(T(n,k),k=0..n-1) od; # yields sequence in triangular form
  • Mathematica
    Table[Binomial[n, k] Binomial[2 n - k, n + 1]/n, {n, 10}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Jan 09 2016 *)
  • PARI
    tabl(nn) = {mP = matrix(nn, nn, n, k, binomial(n-1, k-1)); mN = matrix(nn, nn, n, k, binomial(n-1, k-1) * binomial(n, k-1) / k); mprod = mN*mP; for (n=1, nn, for (k=1, n, print1(mprod[n, k], ", ");); print(););} \\ Michel Marcus, Apr 16 2015
    
  • PARI
    t(n,k) = binomial(n,k)*binomial(2*n-k,n+1)/n;
    concat(vector(10, n, vector(n, k, t(n,k-1))))  \\ Gheorghe Coserea, Apr 24 2016

Formula

T(n,k) = C(n,k)*C(2*n-k,n+1)/n (0 <= k <= n-1).
G.f.: G(t,z) = (1-2*z-t*z-sqrt(1-4*z-2*t*z+t^2*z^2))/(2*(1+t)*z).
Equals N * P, where N = the Narayana triangle (A001263) and P = Pascal's triangle, as infinite lower triangular matrices. A126182 = P * N. - Gary W. Adamson, Nov 30 2007
G.f.: 1/(1-x-(x+xy)/(1-xy/(1-(x+xy)/(1-xy/(1-(x+xy)/(1-xy/(1-.... (continued fraction). - Paul Barry, Feb 06 2009
Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A126216 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Oct 09 2011
From Tom Copeland, Oct 10 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = 1
P(3,t) = 2 + t
P(4,t) = 5 + 5 t + t^2
P(5,t) = 14 + 21 t + 9 t^2 + t^3
The o.g.f. A(x,t) = (1+x*t-sqrt((1-x*t)^2-4x))/(2(1+t)), and
B(x,t) = x - x^2/(1-t*x) = x - x^2 - ((t*x)^3 + (t*x)^4 + ...)/t^2 is the compositional inverse in x. [series corrected by Tom Copeland, Dec 10 2019]
Let h(x,t) = 1/(dB/dx) = (1-tx)^2/(1-(t+1)(2x-tx^2)) = 1/(1 - 2x - 3tx^2 + 4t^2x^3 + ...). Then P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(u,t)*d/du) u, evaluated at u=0, and dA/dx = h(A(x,t),t). (End)
From Tom Copeland, Dec 09 2019: (Start)
The polynomials in my 2011 formula entry above evaluate to an aerated, alternating sign sequence of the Catalan numbers A000108 with t = -2. The first few are P(2,-2) = 1, P(3,-2) = 0, P(4,t) = -1, P(5,-2) = 0, P(6,-2) = 2, P(7,-2) = 0, P(8,-2) = -5, P(9,-2) = 0, P(10,-2) = 14.
Generalizing the relations between w = theta and u = phi in Mizera on pp. 32-34, modify the inverse pair above to w = i * B(-i*u,t) = u + i * u^2/(1+i*t*u), where i is the imaginary number, and u = i*A(-i*w,t) = i*(1 - i*w*t - sqrt((1 + i*w*t)^2 + i*4*w))/(2(1+t)). Then the expression for V'(w) in Mizera generalizes to V'(w) = -i*(w - u) and reduces to V'(w) = (1 - sqrt(1-4 w^2))/2 when evaluated at t = -2, which is an o.g.f. for A126120. Cf. also A086810. (End)
Sum_{k = 0..n-1} (-1)^k*T(n,k)*binomial(x + 2*n - k, 2*n - k) = ( (x + 1) * ( Product_{k = 2..n} (x + k)^2 ) * (x + n + 1) )/(n!*(n + 1)!) for n >= 1. Cf. A243660 and A243661. - Peter Bala, Oct 08 2022

A358372 Number of nodes in the n-th standard ordered rooted tree.

Original entry on oeis.org

1, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 6, 7, 7, 7, 7, 6, 7, 7, 7, 7, 7, 7, 7, 6, 6, 7, 7, 7, 7, 7, 7, 6, 7, 7, 7, 7, 7, 7, 7, 5, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 7, 7, 8, 8, 8, 8, 8
Offset: 1

Views

Author

Gus Wiseman, Nov 14 2022

Keywords

Comments

We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

Examples

			The standard ordered rooted tree ranking begins:
  1: o        10: (((o))o)   19: (((o))(o))
  2: (o)      11: ((o)(o))   20: (((o))oo)
  3: ((o))    12: ((o)oo)    21: ((o)((o)))
  4: (oo)     13: (o((o)))   22: ((o)(o)o)
  5: (((o)))  14: (o(o)o)    23: ((o)o(o))
  6: ((o)o)   15: (oo(o))    24: ((o)ooo)
  7: (o(o))   16: (oooo)     25: (o(oo))
  8: (ooo)    17: ((((o))))  26: (o((o))o)
  9: ((oo))   18: ((oo)o)    27: (o(o)(o))
For example, the 25th ordered tree is (o,(o,o)) because the 24th composition is (1,4) and the 3rd composition is (1,1). Hence a(25) = 5.
		

Crossrefs

The triangle counting trees by leaves is A001263, unordered A055277.
The version for unordered trees is A061775, leaves A109129, edges A196050.
The leaves are counted by A358371.
A000081 counts unlabeled rooted trees, ranked by A358378.
A358374 ranks ordered identity trees, counted by A032027.
A358375 ranks ordered binary trees, counted by A126120

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    srt[n_]:=If[n==1,{},srt/@stc[n-1]];
    Table[Count[srt[n],_,{0,Infinity}],{n,100}]

A007853 Number of maximal antichains in rooted plane trees on n nodes.

Original entry on oeis.org

1, 2, 5, 15, 50, 178, 663, 2553, 10086, 40669, 166752, 693331, 2917088, 12398545, 53164201, 229729439, 999460624, 4374546305, 19250233408, 85120272755, 378021050306, 1685406494673, 7541226435054, 33852474532769, 152415463629568, 688099122024944
Offset: 1

Views

Author

Keywords

Comments

Also the number of initial subtrees (emanating from the root) of rooted plane trees on n vertices, where we require that an initial subtree contains either all or none of the branchings under any given node. The leaves of such a subtree comprise the roots of a corresponding antichain cover. Also, in the (non-commutative) multicategory of free pure multifunctions with one atom, a(n) is the number of composable pairs whose composite has n positions. - Gus Wiseman, Aug 13 2018
The g.f. is denoted by y_2 in Bacher 2004 Proposition 7.5 on page 20. - Michael Somos, Nov 07 2019

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 15*x^4 + 50*x^5 + 178*x^6 + 663*x^7 + 2553*x^8 + ... - _Michael Somos_, Nov 07 2019
		

Crossrefs

Programs

  • Mathematica
    ie[t_]:=If[Length[t]==0,1,1+Product[ie[b],{b,t}]];
    allplane[n_]:=If[n==1,{{}},Join@@Function[c,Tuples[allplane/@c]]/@Join@@Permutations/@IntegerPartitions[n-1]];
    Table[Sum[ie[t],{t,allplane[n]}],{n,9}] (* Gus Wiseman, Aug 13 2018 *)
  • Maxima
    a(n):=1/(n+1)*binomial(2*n,n)+sum((k+2)/(n+1)*binomial(2*n-k-1,n-k-1)*(sum(((binomial(2*i,i))*(binomial(k+i,3*i)))/(i+1),i,0,floor(k/2))),k,0,n-1); /* Vladimir Kruchinin, Apr 05 2019 */
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = sqrt(1 - 4*x + x * O(x^n)); polcoeff( (3 - 2*x - A - sqrt(2 - 16*x + 4*x^2 + (2 + 4*x) * A)) / 4, n))}; /* Michael Somos, Nov 07 2019 */

Formula

G.f.: (1/4) * (3 - 2*x - sqrt(1-4*x) - sqrt(2) * sqrt((1+2*x) * sqrt(1-4*x) + 1 - 8*x + 2*x^2)) [from Klazar]. - Sean A. Irvine, Feb 06 2018
a(n) = (1/(n+1))*C(2*n,n) + Sum_{k=0..n-1} ((k+2)/(n+1))*C(2*n-k-1,n-k-1)*Sum_{i=0..floor(k/2)} C(2*i,i)*C(k+i,3*i)/(i+1). - Vladimir Kruchinin, Apr 05 2019
Given the g.f. A(x) and the g.f. of A213705 B(x), then -x = A(-B(x)). - Michael Somos, Nov 07 2019

Extensions

More terms from Sean A. Irvine, Feb 06 2018

A126930 Inverse binomial transform of A005043.

Original entry on oeis.org

1, -1, 2, -3, 6, -10, 20, -35, 70, -126, 252, -462, 924, -1716, 3432, -6435, 12870, -24310, 48620, -92378, 184756, -352716, 705432, -1352078, 2704156, -5200300, 10400600, -20058300, 40116600, -77558760, 155117520, -300540195, 601080390, -1166803110
Offset: 0

Views

Author

Philippe Deléham, Mar 17 2007

Keywords

Comments

Successive binomial transforms are A005043, A000108, A007317, A064613, A104455. Hankel transform is A000012.
Moment sequence of the trace of the square of a random matrix in USp(2)=SU(2). If X=tr(A^2) is a random variable (a distributed with Haar measure) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008
From Tom Copeland, Nov 08 2014: (Start)
This array is one of a family of Catalan arrays related by compositions of the special fractional linear (Mobius) transformation P(x,t) = x/(1-t*x); its inverse Pinv(x,t) = P(x,-t); an o.g.f. of the Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2; and its inverse Cinv(x) = x*(1-x). The Motzkin sums, or Riordan numbers, A005043 are generated by Mot(x)=C[P(x,1)]. One could, of course, choose the Riordan numbers as the parent sequence.
O.g.f.: G(x) = C[P[P(x,1),1]1] = C[P(x,2)] = (1-sqrt(1-4*x/(1+2*x)))/2 = x - x^2 + 2 x^3 - ... = Mot[P(x,1)].
Ginv(x) = Pinv[Cinv(x),2] = P[Cinv(x),-2] = x(1-x)/[1-2x(1-x)] = (x-x^2)/[1-2(x-x^2)] = x*A146559(x).
Cf. A091867 and A210736 for an unsigned version with a leading 1. (End)

Crossrefs

Programs

  • Maple
    egf := BesselI(0,2*x) - BesselI(1,2*x):
    seq(n!*coeff(series(egf,x,34),x,n),n=0..33); # Peter Luschny, Dec 17 2014
  • Mathematica
    CoefficientList[Series[(1 + 2 x - Sqrt[1 - 4 x^2])/(2 x (1 + 2 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 23 2013 *)
    Table[2^n Hypergeometric2F1[3/2, -n, 2, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 02 2015 *)
  • PARI
    x='x+O('x^50); Vec((1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x))) \\ Altug Alkan, Nov 03 2015

Formula

a(n) = (-1)^n*C(n, floor(n/2)) = (-1)^n*A001405(n).
a(2*n) = A000984(n), a(2*n+1) = -A001700(n).
a(n) = (1/Pi)*Integral_{t=0..Pi}(2cos(2t))^n*2sin^2(t) dt. - Andrew V. Sutherland, Feb 29 2008, Mar 09 2008
a(n) = (-2)^n + Sum_{k=0..n-1} a(k)*a(n-1-k), a(0)=1. - Philippe Deléham, Dec 12 2009
G.f.: (1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x)). - Philippe Deléham, Mar 01 2013
O.g.f.: (1 + x*c(x^2))/(1 + 2*x), with the o.g.f. c(x) for the Catalan numbers A000108. From the o.g.f. of the Riordan type Catalan triangle A053121. This is the rewritten g.f. given in the previous formula. This is G(-x) with the o.g.f. G(x) of A001405. - Wolfdieter Lang, Sep 22 2013
D-finite with recurrence (n+1)*a(n) +2*a(n-1) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Dec 04 2013
Recurrence (an alternative): (n+1)*a(n) = 8*(n-2)*a(n-3) + 4*(n-2)*a(n-2) + 2*(-n-1)*a(n-1), n>=3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) ~ (-1)^n*2^(n+1/2)/sqrt(n*Pi). - Fung Lam, Mar 22 2014
E.g.f.: BesselI(0,2*x) - BesselI(1,2*x). - Peter Luschny, Dec 17 2014
a(n) = 2^n*hypergeom([3/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
G.f. A(x) satisfies: A(x) = 1/(1 + 2*x) + x*A(x)^2. - Ilya Gutkovskiy, Jul 10 2020

A097331 Expansion of 1 + 2x/(1 + sqrt(1 - 4x^2)).

Original entry on oeis.org

1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, 0, 132, 0, 429, 0, 1430, 0, 4862, 0, 16796, 0, 58786, 0, 208012, 0, 742900, 0, 2674440, 0, 9694845, 0, 35357670, 0, 129644790, 0, 477638700, 0, 1767263190, 0, 6564120420, 0, 24466267020, 0, 91482563640, 0, 343059613650, 0
Offset: 0

Views

Author

Paul Barry, Aug 05 2004

Keywords

Comments

Binomial transform is A097332. Second binomial transform is A014318.
Essentially the same as A126120. - R. J. Mathar, Jun 15 2008
Hankel transform is A087960(n) = (-1)^binomial(n+1,2). - Paul Barry, Aug 10 2009

Programs

  • Maple
    A097331_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w]:=a[w-1]-(-1)^w*add(a[j]*a[w-j-1],j=1..w-1) od; convert(a,list)end: A097331_list(48); # Peter Luschny, May 19 2011
  • Mathematica
    a[0] = 1; a[n_?OddQ] := CatalanNumber[(n-1)/2]; a[] = 0; Table[a[n], {n, 0, 48}] (* _Jean-François Alcover, Jul 24 2013 *)
  • Sage
    def A097331_list(n) :
        D = [0]*(n+2); D[1] = 1
        b = True; h = 1; R = []
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] -= D[k-1]
                h += 1; R.append(abs(D[1]))
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            b = not b
        return R
    A097331_list(49) # Peter Luschny, Jun 03 2012

Formula

a(n) = 0^n + Catalan((n-1)/2)(1-(-1)^n)/2.
Unsigned version of A090192, A105523. - Philippe Deléham, Sep 29 2006
From Paul Barry, Aug 10 2009: (Start)
G.f.: 1+xc(x^2), c(x) the g.f. of A000108;
G.f.: 1/(1-x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+... (continued fraction);
G.f.: 1+x/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). (End)
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+2*x) (continued fraction); more generally g.f. C(x/(1+2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
Conjecture: (n+1)*a(n) + n*a(n-1) + 4*(-n+2)*a(n-2) + 4*(-n+3)*a(n-3)=0. - R. J. Mathar, Dec 02 2012
Recurrence: (n+3)*a(n+2) = 4*n*a(n), a(0)=a(1)=1. For nonzero terms, a(n) ~ 2^(n+1)/((n+1)^(3/2)*sqrt(2*Pi)). - Fung Lam, Mar 17 2014

A100754 Triangle read by rows: T(n, k) = number of hill-free Dyck paths (i.e., no peaks at height 1) of semilength n and having k peaks.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 8, 8, 1, 1, 13, 29, 13, 1, 1, 19, 73, 73, 19, 1, 1, 26, 151, 266, 151, 26, 1, 1, 34, 276, 749, 749, 276, 34, 1, 1, 43, 463, 1781, 2762, 1781, 463, 43, 1, 1, 53, 729, 3758, 8321, 8321, 3758, 729, 53, 1, 1, 64, 1093, 7253, 21659, 31004, 21659, 7253, 1093, 64, 1
Offset: 2

Views

Author

Emeric Deutsch, Jan 14 2005

Keywords

Comments

Row n has n - 1 terms. Row sums yield the Fine numbers (A000957).
Related to the number of certain sets of non-crossing partitions for the root system A_n (p. 11, Athanasiadis and Savvidou). - Tom Copeland, Oct 19 2014
T(n,k) is the number of permutations pi of [n-1] with k - 1 descents such that s(pi) avoids the patterns 132, 231, and 312, where s is West's stack-sorting map. - Colin Defant, Sep 16 2018
The absolute values of the polynomials at -1 and j (cube root of 1) seem to be given by A126120 and A005043. - F. Chapoton, Nov 16 2021
Don Knuth observes that this sequence also arrises from the enumeration of restricted max-and-min-closed relations, only there it appears as an array read by antidiagonals: see the Knuth "Notes" link and A372068. Knuth also gives a formula expressing the array A372368 in terms of this array. He also reports that there is strong experimental evidence that the n-th term of row m in this array is a polynomial of degree 2*m-2 in n. - N. J. A. Sloane, May 12 2024

Examples

			T(4, 2) = 4 because we have UU*DDUU*DD, UU*DUU*DDD, UUU*DDU*DD and UUU*DU*DDD, where U = (1, 1), D = (1,-1) and * indicates the peaks.
Triangle starts:
   1;
   1,  1;
   1,  4,   1;
   1,  8,   8,    1;
   1, 13,  29,   13,    1;
   1, 19,  73,   73,   19,    1;
   1, 26, 151,  266,  151,   26,    1;
   1, 34, 276,  749,  749,  276,   34,   1;
   1, 43, 463, 1781, 2762, 1781,  463,  43,  1;
   1, 53, 729, 3758, 8321, 8321, 3758, 729, 53, 1;
   ...
As an array (for which the rows of the preceding triangle are the antidiagonals):
   1,  1,    1,     1,      1,      1,       1,        1,        1, ...
   1,  4,    8,    13,     19,     26,      34,       43,       53, ...
   1,  8,   29,    73,    151,    276,     463,      729,     1093, ...
   1, 13,   73,   266,    749,   1781,    3758,     7253,    13061, ...
   1, 19,  151,   749,   2762,   8321,   21659,    50471,   107833, ...
   1, 26,  276,  1781,   8321,  31004,   97754,   271125,   679355, ...
   1, 34,  463,  3758,  21659,  97754,  367285,  1196665,  3478915, ...
   1, 43,  729,  7253,  50471, 271125, 1196665,  4526470, 15118415, ...
   1, 53, 1093, 13061, 107833, 679355, 3478915, 15118415, 57500480, ...
   ...
		

Crossrefs

Programs

  • Maple
    T := (n, k) -> add((j/(n-j))*binomial(n-j, k-j)*binomial(n-j,k), j=0..min(k,n-k)): for n from 2 to 13 do seq(T(n, k), k = 1..n-1) od; # yields the sequence in triangular form
  • Mathematica
    T[n_, k_] := Sum[(j/(n-j))*Binomial[n-j, k-j]*Binomial[n-j, k], {j, 0, Min[k, n-k]}]; Table[T[n, k], {n, 2, 13}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Feb 19 2017, translated from Maple *)

Formula

T(n, k) = Sum_{j=0..min(k, n-k)} (j/(n-j)) * C(n-j, k-j) * C(n-j, k), n >= 2.
G.f.: t*z*r/(1 - t*z*r), where r = r(t, z) is the Narayana function defined by r = z*(1 + r)*(1 + t*r).
From Tom Copeland, Oct 19 2014: (Start)
With offset 0 for A108263 and offset 1 for A132081, row polynomials of this entry P(n, x) = Sum_{i} A108263(n, i)*x^i*(1 + x)^(n - 2*i) = Sum_{i} A132081(n - 2, i)*x^i*(1 + x)^(n - 2*i).
E.g., P(4, x) = 1*x*(1 + x)^(4 - 2*1) + 2*x^2*(1 + x)^(4 - 2*2) = x + 4*x^2 + x^3.
Equivalently, let Q(n, x) be the row polynomials of A108263. Then P(n, x) = (1 + x)^n * Q(n, x/(1 + x)^2).
E.g., P(4, x) = (1 + x)^4 * (x/(1 + x)^2 + 2 * (x/(1 + x)^2)^2).
See Athanasiadis and Savvidou (p. 7). (End)
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