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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A358374 Numbers k such that the k-th standard ordered rooted tree is an identity tree (counted by A032027).

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 13, 17, 19, 21, 33, 34, 38, 39, 42, 45, 49, 51, 53, 65, 66, 67, 81, 97, 130, 131, 133, 134, 135, 145, 161, 162, 177, 193, 195, 209, 259, 261, 262, 263, 266, 269, 289, 290, 305, 321, 322, 353, 387, 389, 401, 417, 513, 517, 518, 519, 522
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.
A rooted identity tree is an unlabeled rooted tree with no repeated branches directly under the same root.

Examples

			The terms together with their corresponding ordered rooted trees begin:
   1: o
   2: (o)
   3: ((o))
   5: (((o)))
   6: ((o)o)
   7: (o(o))
  10: (((o))o)
  13: (o((o)))
  17: ((((o))))
  19: (((o))(o))
  21: ((o)((o)))
  33: (((o)o))
  34: ((((o)))o)
  38: (((o))(o)o)
  39: (((o))o(o))
  42: ((o)((o))o)
  45: ((o)o((o)))
		

Crossrefs

These trees are counted by A032027.
The unordered version is A276625, counted by A004111.
A000081 counts unlabeled rooted trees, ranked by A358378.
A358371 and A358372 count leaves and nodes in standard ordered rooted trees.
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]];
    Select[Range[100],FreeQ[srt[#],[_]?(!UnsameQ@@#&)]&]

A001263 Triangle of Narayana numbers T(n,k) = C(n-1,k-1)*C(n,k-1)/k with 1 <= k <= n, read by rows. Also called the Catalan triangle.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50, 50, 15, 1, 1, 21, 105, 175, 105, 21, 1, 1, 28, 196, 490, 490, 196, 28, 1, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 1, 55, 825, 4950, 13860, 19404, 13860, 4950, 825
Offset: 1

Views

Author

Keywords

Comments

Number of antichains (or order ideals) in the poset 2*(k-1)*(n-k) or plane partitions with rows <= k-1, columns <= n-k and entries <= 2. - Mitch Harris, Jul 15 2000
T(n,k) is the number of Dyck n-paths with exactly k peaks. a(n,k) = number of pairs (P,Q) of lattice paths from (0,0) to (k,n+1-k), each consisting of unit steps East or North, such that P lies strictly above Q except at the endpoints. - David Callan, Mar 23 2004
Number of permutations of [n] which avoid-132 and have k-1 descents. - Mike Zabrocki, Aug 26 2004
T(n,k) is the number of paths through n panes of glass, entering and leaving from one side, of length 2n with k reflections (where traversing one pane of glass is the unit length). - Mitch Harris, Jul 06 2006
Antidiagonal sums given by A004148 (without first term).
T(n,k) is the number of full binary trees with n internal nodes and k-1 jumps. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007
From Gary W. Adamson, Oct 22 2007: (Start)
The n-th row can be generated by the following operation using an ascending row of (n-1) triangular terms, (A) and a descending row, (B); e.g., row 6:
A: 1....3....6....10....15
B: 15...10....6.....3.....1
C: 1...15...50....50....15....1 = row 6.
Leftmost column of A,B -> first two terms of C; then followed by the operation B*C/A of current column = next term of row C, (e.g., 10*15/3 = 50). Continuing with the operation, we get row 6: (1, 15, 50, 50, 15, 1). (End)
The previous comment can be upgraded to: The ConvOffsStoT transform of the triangular series; and by rows, row 6 is the ConvOffs transform of (1, 3, 6, 10, 15). Refer to triangle A117401 as another example of the ConvOffsStoT transform, and OEIS under Maple Transforms. - Gary W. Adamson, Jul 09 2012
For a connection to Lagrange inversion, see A134264. - Tom Copeland, Aug 15 2008
T(n,k) is also the number of order-decreasing and order-preserving mappings (of an n-element set) of height k (height of a mapping is the cardinal of its image set). - Abdullahi Umar, Aug 21 2008
Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p.60]. See A033282 for the corresponding array of f-vectors for associahedra of type A_n. See A008459 and A145903 for the h-vectors for associahedra of type B and type D respectively. The Hilbert transform of this triangle (see A145905 for the definition of this transform) is A145904. - Peter Bala, Oct 27 2008
T(n,k) is also the number of noncrossing set partitions of [n] into k blocks. Given a partition P of the set {1,2,...,n}, a crossing in P are four integers [a, b, c, d] with 1 <= a < b < c < d <= n for which a, c are together in a block, and b, d are together in a different block. A noncrossing partition is a partition with no crossings. - Peter Luschny, Apr 29 2011
Noncrossing set partitions are also called genus 0 partitions. In terms of genus-dependent Stirling numbers of the second kind S2(n,k,g) that count partitions of genus g of an n-set into k nonempty subsets, one has T(n,k) = S2(n,k,0). - Robert Coquereaux, Feb 15 2024
Diagonals of A089732 are rows of A001263. - Tom Copeland, May 14 2012
From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = 1/sqrt(y)*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang.
Generating function E(y)*E(x*y) = 1 + (1 + x)*y/(1!*2!) + (1 + 3*x + x^2)*y^2/(2!*3!) + (1 + 6*x + 6*x^2 + x^3)*y^3/(3!*4!) + .... Cf. A105278 with a generating function exp(y)*E(x*y).
The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse A103364 has a generating function E(x*y)/E(y). (End)
T(n,k) is the number of nonintersecting n arches above the x axis, starting and ending on vertices 1 to 2n, with k being the number of arches starting on an odd vertice and ending on a higher even vertice. Example: T(3,2)=3 [16,25,34] [14,23,56] [12,36,45]. - Roger Ford, Jun 14 2014
Fomin and Reading on p. 31 state that the rows of the Narayana matrix are the h-vectors of the associahedra as well as its dual. - Tom Copeland, Jun 27 2017
The row polynomials P(n, x) = Sum_{k=1..n} T(n, k)*x^(k-1), together with P(0, x) = 1, multiplied by (n+1) are the numerator polynomials of the o.g.f.s of the diagonal sequences of the triangle A103371: G(n, x) = (n+1)*P(n, x)/(1 - x)^{2*n+1}, for n >= 0. This is proved with Lagrange's theorem applied to the Riordan triangle A135278 = (1/(1 - x)^2, x/(1 - x)). See an example below. - Wolfdieter Lang, Jul 31 2017
T(n,k) is the number of Dyck paths of semilength n with k-1 uu-blocks (pairs of consecutive up-steps). - Alexander Burstein, Jun 22 2020
In case you were searching for Narayama numbers, the correct spelling is Narayana. - N. J. A. Sloane, Nov 11 2020
Named after the Canadian mathematician Tadepalli Venkata Narayana (1930-1987). They were also called "Runyon numbers" after John P. Runyon (1922-2013) of Bell Telephone Laboratories, who used them in a study of a telephone traffic system. - Amiram Eldar, Apr 15 2021 The Narayana numbers were first studied by Percy Alexander MacMahon (see reference, Article 495) as pointed out by Bóna and Sagan (see link). - Peter Luschny, Apr 28 2022
From Andrea Arlette España, Nov 14 2022: (Start)
T(n,k) is the degree distribution of the paths towards synchronization in the transition diagram associated with the Laplacian system over the complete graph K_n, corresponding to ordered initial conditions x_1 < x_2 < ... < x_n.
T(n,k) for n=2N+1 and k=N+1 is the number of states in the transition diagram associated with the Laplacian system over the complete bipartite graph K_{N,N}, corresponding to ordered (x_1 < x_2 < ... < x_N and x_{N+1} < x_{N+2} < ... < x_{2N}) and balanced (Sum_{i=1..N} x_i/N = Sum_{i=N+1..2N} x_i/N) initial conditions. (End)
From Gus Wiseman, Jan 23 2023: (Start)
Also the number of unlabeled ordered rooted trees with n nodes and k leaves. See the link by Marko Riedel. For example, row n = 5 counts the following trees:
((((o)))) (((o))o) ((o)oo) (oooo)
(((o)o)) ((oo)o)
(((oo))) ((ooo))
((o)(o)) (o(o)o)
((o(o))) (o(oo))
(o((o))) (oo(o))
The unordered version is A055277. Leaves in standard ordered trees are counted by A358371. (End)

Examples

			The initial rows of the triangle are:
  [1] 1
  [2] 1,  1
  [3] 1,  3,   1
  [4] 1,  6,   6,    1
  [5] 1, 10,  20,   10,    1
  [6] 1, 15,  50,   50,   15,    1
  [7] 1, 21, 105,  175,  105,   21,   1
  [8] 1, 28, 196,  490,  490,  196,  28,  1
  [9] 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
  ...
For all n, 12...n (1 block) and 1|2|3|...|n (n blocks) are noncrossing set partitions.
Example of umbral representation:
  A007318(5,k)=[1,5/1,5*4/(2*1),...,1]=(1,5,10,10,5,1),
  so A001263(5,k)={1,b(5)/b(1),b(5)*b(4)/[b(2)*b(1)],...,1}
  = [1,30/2,30*20/(6*2),...,1]=(1,15,50,50,15,1).
  First = last term = b.(5!)/[b.(0!)*b.(5!)]= 1. - _Tom Copeland_, Sep 21 2011
Row polynomials and diagonal sequences of A103371: n = 4,  P(4, x) = 1 + 6*x + 6*x^2 + x^3, and the o.g.f. of fifth diagonal is G(4, x) = 5* P(4, x)/(1 - x)^9, namely [5, 75, 525, ...]. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
		

References

  • Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), pp. 103-124.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 196.
  • P. A. MacMahon, Combinatory Analysis, Vols. 1 and 2, Cambridge University Press, 1915, 1916; reprinted by Chelsea, 1960, Sect. 495.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
  • A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.
  • T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 2.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 17.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.36(a) and (b).

Crossrefs

Other versions are in A090181 and A131198. - Philippe Deléham, Nov 18 2007
Cf. variants: A181143, A181144. - Paul D. Hanna, Oct 13 2010
Row sums give A000108 (Catalan numbers), n>0.
A008459 (h-vectors type B associahedra), A033282 (f-vectors type A associahedra), A145903 (h-vectors type D associahedra), A145904 (Hilbert transform). - Peter Bala, Oct 27 2008
Cf. A016098 and A189232 for numbers of crossing set partitions.
Cf. A243752.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • GAP
    Flat(List([1..11],n->List([1..n],k->Binomial(n-1,k-1)*Binomial(n,k-1)/k))); # Muniru A Asiru, Jul 12 2018
  • Haskell
    a001263 n k = a001263_tabl !! (n-1) !! (k-1)
    a001263_row n = a001263_tabl !! (n-1)
    a001263_tabl = zipWith dt a007318_tabl (tail a007318_tabl) where
       dt us vs = zipWith (-) (zipWith (*) us (tail vs))
                              (zipWith (*) (tail us ++ [0]) (init vs))
    -- Reinhard Zumkeller, Oct 10 2013
    
  • Magma
    /* triangle */ [[Binomial(n-1,k-1)*Binomial(n,k-1)/k : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 19 2014
    
  • Maple
    A001263 := (n,k)->binomial(n-1,k-1)*binomial(n,k-1)/k;
    a:=proc(n,k) option remember; local i; if k=1 or k=n then 1 else add(binomial(n+i-1, 2*k-2)*a(k-1,i),i=1..k-1); fi; end:
    # Alternatively, as a (0,0)-based triangle:
    R := n -> simplify(hypergeom([-n, -n-1], [2], x)): Trow := n -> seq(coeff(R(n,x),x,j), j=0..n): seq(Trow(n), n=0..9); # Peter Luschny, Mar 19 2018
  • Mathematica
    T[n_, k_] := If[k==0, 0, Binomial[n-1, k-1] Binomial[n, k-1] / k];
    Flatten[Table[Binomial[n-1,k-1] Binomial[n,k-1]/k,{n,15},{k,n}]] (* Harvey P. Dale, Feb 29 2012 *)
    TRow[n_] := CoefficientList[Hypergeometric2F1[1 - n, -n, 2, x], x];
    Table[TRow[n], {n, 1, 11}] // Flatten (* Peter Luschny, Mar 19 2018 *)
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],Length[Position[#,{}]]==k&]],{n,2,9},{k,1,n-1}] (* Gus Wiseman, Jan 23 2023 *)
    T[1, 1] := 1; T[n_, k_]/;1<=k<=n := T[n, k] = (2n/k-1) T[n-1,k-1] + T[n-1, k]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    {a(n, k) = if(k==0, 0, binomial(n-1, k-1) * binomial(n, k-1) / k)};
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^2*y^j)*x^m/m) +O(x^(n+1))),n,x),k,y)} \\ Paul D. Hanna, Oct 13 2010
    
  • Sage
    @CachedFunction
    def T(n, k):
        if k == n or k == 1: return 1
        if k <= 0 or k > n: return 0
        return binomial(n, 2) * (T(n-1, k)/((n-k)*(n-k+1)) + T(n-1, k-1)/(k*(k-1)))
    for n in (1..9): print([T(n, k) for k in (1..n)])  # Peter Luschny, Oct 28 2014
    

Formula

a(n, k) = C(n-1, k-1)*C(n, k-1)/k for k!=0; a(n, 0)=0.
Triangle equals [0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is Deléham's operator defined in A084938.
0Mike Zabrocki, Aug 26 2004
T(n, k) = C(n, k)*C(n-1, k-1) - C(n, k-1)*C(n-1, k) (determinant of a 2 X 2 subarray of Pascal's triangle A007318). - Gerald McGarvey, Feb 24 2005
T(n, k) = binomial(n-1, k-1)^2 - binomial(n-1, k)*binomial(n-1, k-2). - David Callan, Nov 02 2005
a(n,k) = C(n,2) (a(n-1,k)/((n-k)*(n-k+1)) + a(n-1,k-1)/(k*(k-1))) a(n,k) = C(n,k)*C(n,k-1)/n. - Mitch Harris, Jul 06 2006
Central column = A000891, (2n)!*(2n+1)! / (n!*(n+1)!)^2. - Zerinvary Lajos, Oct 29 2006
G.f.: (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) = Sum_{n>0, k>0} a(n, k)*x^n*y^k.
From Peter Bala, Oct 22 2008: (Start)
Relation with Jacobi polynomials of parameter (1,1):
Row n+1 generating polynomial equals 1/(n+1)*x*(1-x)^n*Jacobi_P(n,1,1,(1+x)/(1-x)). It follows that the zeros of the Narayana polynomials are all real and nonpositive, as noted above. O.g.f for column k+2: 1/(k+1) * y^(k+2)/(1-y)^(k+3) * Jacobi_P(k,1,1,(1+y)/(1-y)). Cf. A008459.
T(n+1,k) is the number of walks of n unit steps on the square lattice (i.e., each step in the direction either up (U), down (D), right (R) or left (L)) starting from the origin and finishing at lattice points on the x axis and which remain in the upper half-plane y >= 0 [Guy]. For example, T(4,3) = 6 counts the six walks RRL, LRR, RLR, UDL, URD and RUD, from the origin to the lattice point (1,0), each of 3 steps. Compare with tables A145596 - A145599.
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for this array is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... = 1/(1 - x*t/(1 - x/(1 - x*t/(1 - x/(1 - ...))))) (as a continued fraction). Cf. A108767, A132081 and A141618. (End)
G.f.: 1/(1-x-xy-x^2y/(1-x-xy-x^2y/(1-... (continued fraction). - Paul Barry, Sep 28 2010
E.g.f.: exp((1+y)x)*Bessel_I(1,2*sqrt(y)x)/(sqrt(y)*x). - Paul Barry, Sep 28 2010
G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n ). - Paul D. Hanna, Oct 13 2010
With F(x,t) = (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2))/(2*x) an o.g.f. in x for the Narayana polynomials in t, G(x,t) = x/(t+(1+t)*x+x^2) is the compositional inverse in x. Consequently, with H(x,t) = 1/ (dG(x,t)/dx) = (t+(1+t)*x+x^2)^2 / (t-x^2), the n-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*D_x)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*D_u)u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 04 2011
With offset 0, A001263 = Sum_{j>=0} A132710^j / A010790(j), a normalized Bessel fct. May be represented as the Pascal matrix A007318, n!/[(n-k)!*k!], umbralized with b(n)=A002378(n) for n>0 and b(0)=1: A001263(n,k)= b.(n!)/{b.[(n-k)!]*b.(k!)} where b.(n!) = b(n)*b(n-1)...*b(0), a generalized factorial (see example). - Tom Copeland, Sep 21 2011
With F(x,t) = {1-(1-t)*x-sqrt[1-2*(1+t)*x+[(t-1)*x]^2]}/2 a shifted o.g.f. in x for the Narayana polynomials in t, G(x,t)= x/[t-1+1/(1-x)] is the compositional inverse in x. Therefore, with H(x,t)=1/(dG(x,t)/dx)=[t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, (see A119900), the (n-1)-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/du) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 30 2011
T(n,k) = binomial(n-1,k-1)*binomial(n+1,k)-binomial(n,k-1)*binomial(n,k). - Philippe Deléham, Nov 05 2011
A166360(n-k) = T(n,k) mod 2. - Reinhard Zumkeller, Oct 10 2013
Damped sum of a column, in leading order: lim_{d->0} d^(2k-1) Sum_{N>=k} T(N,k)(1-d)^N=Catalan(n). - Joachim Wuttke, Sep 11 2014
Multiplying the n-th column by n! generates the revert of the unsigned Lah numbers, A089231. - Tom Copeland, Jan 07 2016
Row polynomials: (x - 1)^(n+1)*(P(n+1,(1 + x)/(x - 1)) - P(n-1,(1 + x)/(x - 1)))/((4*n + 2)), n = 1,2,... and where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Mar 03 2017
The coefficients of the row polynomials R(n, x) = hypergeom([-n,-n-1], [2], x) generate the triangle based in (0,0). - Peter Luschny, Mar 19 2018
Multiplying the n-th diagonal by n!, with the main diagonal n=1, generates the Lah matrix A105278. With G equal to the infinitesimal generator of A132710, the Narayana triangle equals Sum_{n >= 0} G^n/((n+1)!*n!) = (sqrt(G))^(-1) * I_1(2*sqrt(G)), where G^0 is the identity matrix and I_1(x) is the modified Bessel function of the first kind of order 1. (cf. Sep 21 2011 formula also.) - Tom Copeland, Sep 23 2020
T(n,k) = T(n,k-1)*C(n-k+2,2)/C(k,2). - Yuchun Ji, Dec 21 2020
From Sergii Voloshyn, Nov 25 2024: (Start)
G.f.: F(x,y) = (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) is the solution of the differential equation x^3 * d^2(x*F(x,y))/dx^2 = y * d^2(x*F(x,y))/dy^2.
Let E be the operator x*D*D, where D denotes the derivative operator d/dx. Then (1/(n! (1 + n)!)) * E^n(x/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} C(n-1, k-1)*C(n, k-1)/k*x^k. For example, when n = 4 we have (1/4!/5!)*E^3(x/(1 - x)) = x (1 + 6 x + 6 x^2 + x^3)/(1 - x)^9. (End)

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A126120 Catalan numbers (A000108) interpolated with 0's.

Original entry on oeis.org

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

Philippe Deléham, Mar 06 2007

Keywords

Comments

Inverse binomial transform of A001006.
The Hankel transform of this sequence gives A000012 = [1,1,1,1,1,...].
Counts returning walks (excursions) of length n on a 1-d integer lattice with step set {+1,-1} which stay in the chamber x >= 0. - Andrew V. Sutherland, Feb 29 2008
Moment sequence of the trace of a random matrix in G=USp(2)=SU(2). If X=tr(A) is a random variable (A distributed according to the Haar measure on G) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008
Essentially the same as A097331. - R. J. Mathar, Jun 15 2008
Number of distinct proper binary trees with n nodes. - Chris R. Sims (chris.r.sims(AT)gmail.com), Jun 30 2010
-a(n-1), with a(-1):=0, n>=0, is the Z-sequence for the Riordan array A049310 (Chebyshev S). For the definition see that triangle. - Wolfdieter Lang, Nov 04 2011
See A180874 (also A238390 and A097610) and A263916 for relations to the general Bell A036040, cycle index A036039, and cumulant expansion polynomials A127671 through the Faber polynomials. - Tom Copeland, Jan 26 2016
A signed version is generated by evaluating polynomials in A126216 that are essentially the face polynomials of the associahedra. This entry's sequence is related to an inversion relation on p. 34 of Mizera, related to Feynman diagrams. - Tom Copeland, Dec 09 2019

Examples

			G.f. = 1 + x^2 + 2*x^4 + 5*x^6 + 14*x^8 + 42*x^10 + 132*x^12 + 429*x^14 + ...
From _Gus Wiseman_, Nov 14 2022: (Start)
The a(0) = 1 through a(8) = 14 ordered binary rooted trees with n + 1 nodes (ranked by A358375):
  o  .  (oo)  .  ((oo)o)  .  (((oo)o)o)  .  ((((oo)o)o)o)
                 (o(oo))     ((o(oo))o)     (((o(oo))o)o)
                             ((oo)(oo))     (((oo)(oo))o)
                             (o((oo)o))     (((oo)o)(oo))
                             (o(o(oo)))     ((o((oo)o))o)
                                            ((o(o(oo)))o)
                                            ((o(oo))(oo))
                                            ((oo)((oo)o))
                                            ((oo)(o(oo)))
                                            (o(((oo)o)o))
                                            (o((o(oo))o))
                                            (o((oo)(oo)))
                                            (o(o((oo)o)))
                                            (o(o(o(oo))))
(End)
		

References

  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Ch. 49, Hemisphere Publishing Corp., 1987.

Crossrefs

Cf. A126216.
The unordered version is A001190, ranked by A111299.
These trees (ordered binary rooted) are ranked by A358375.

Programs

  • Magma
    &cat [[Catalan(n), 0]: n in [0..30]]; // Vincenzo Librandi, Jul 28 2016
    
  • Maple
    with(combstruct): grammar := { BB = Sequence(Prod(a,BB,b)), a = Atom, b = Atom }: seq(count([BB,grammar], size=n),n=0..47); # Zerinvary Lajos, Apr 25 2007
    BB := {E=Prod(Z,Z), S=Union(Epsilon,Prod(S,S,E))}: ZL:=[S,BB,unlabeled]: seq(count(ZL, size=n), n=0..45); # Zerinvary Lajos, Apr 22 2007
    BB := [T,{T=Prod(Z,Z,Z,F,F), F=Sequence(B), B=Prod(F,Z,Z)}, unlabeled]: seq(count(BB, size=n+1), n=0..45); # valid for n> 0. # Zerinvary Lajos, Apr 22 2007
    seq(n!*coeff(series(hypergeom([],[2],x^2),x,n+2),x,n),n=0..45); # Peter Luschny, Jan 31 2015
    # Using function CompInv from A357588.
    CompInv(48, n -> ifelse(irem(n, 2) = 0, 0, (-1)^iquo(n-1, 2))); # Peter Luschny, Oct 07 2022
  • Mathematica
    a[n_?EvenQ] := CatalanNumber[n/2]; a[n_] = 0; Table[a[n], {n, 0, 45}] (* Jean-François Alcover, Sep 10 2012 *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ BesselI[ 1, 2 x] / x, {x, 0, n}]]; (* Michael Somos, Mar 19 2014 *)
    bot[n_]:=If[n==1,{{}},Join@@Table[Tuples[bot/@c],{c,Table[{k,n-k-1},{k,n-1}]}]];
    Table[Length[bot[n]],{n,10}] (* Gus Wiseman, Nov 14 2022 *)
    Riffle[CatalanNumber[Range[0,50]],0,{2,-1,2}] (* Harvey P. Dale, May 28 2024 *)
  • Python
    from math import comb
    def A126120(n): return 0 if n&1 else comb(n,m:=n>>1)//(m+1) # Chai Wah Wu, Apr 22 2024
  • Sage
    def A126120_list(n) :
        D = [0]*(n+2); D[1] = 1
        b = True; h = 2; 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
    A126120_list(46) # Peter Luschny, Jun 03 2012
    

Formula

a(2*n) = A000108(n), a(2*n+1) = 0.
a(n) = A053121(n,0).
(1/Pi) Integral_{0 .. Pi} (2*cos(x))^n *2*sin^2(x) dx. - Andrew V. Sutherland, Feb 29 2008
G.f.: (1 - sqrt(1 - 4*x^2)) / (2*x^2) = 1/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). - Philippe Deléham, Nov 24 2009
G.f. A(x) satisfies A(x) = 1 + x^2*A(x)^2. - Vladimir Kruchinin, Feb 18 2011
E.g.f.: I_1(2x)/x Where I_n(x) is the modified Bessel function. - Benjamin Phillabaum, Mar 07 2011
Apart from the first term the e.g.f. is given by x*HyperGeom([1/2],[3/2,2], x^2). - Benjamin Phillabaum, Mar 07 2011
a(n) = Integral_{x=-2..2} x^n*sqrt((2-x)*(2+x))/(2*Pi) dx. - Peter Luschny, Sep 11 2011
E.g.f.: E(0)/(1-x) where E(k) = 1-x/(1-x/(x-(k+1)*(k+2)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 05 2013
G.f.: 3/2- sqrt(1-4*x^2)/2 = 1/x^2 + R(0)/x^2, where R(k) = 2*k-1 - x^2*(2*k-1)*(2*k+1)/R(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 28 2013 (warning: this is not the g.f. of this sequence, R. J. Mathar, Sep 23 2021)
G.f.: 1/Q(0), where Q(k) = 2*k+1 + x^2*(1-4*(k+1)^2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 09 2014
a(n) = n!*[x^n]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 2^n*hypergeom([3/2,-n],[3],2). - Peter Luschny, Feb 03 2015
a(n) = ((-1)^n+1)*2^(2*floor(n/2)-1)*Gamma(floor(n/2)+1/2)/(sqrt(Pi)* Gamma(floor(n/2)+2)). - Ilya Gutkovskiy, Jul 23 2016
D-finite with recurrence (n+2)*a(n) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Mar 21 2021
From Peter Bala, Feb 03 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-2)^(-k)*binomial(n, k)*Catalan(k+1).
G.f.: 1/(1 + 2*x) * c(x/(1 + 2*x))^2 = 1/(1 - 2*x) * c(-x/(1 - 2*x))^2 = c(x^2), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. (End)

Extensions

An erroneous comment removed by Tom Copeland, Jul 23 2016

A080936 Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n and height k (1 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 18, 7, 1, 1, 31, 57, 33, 9, 1, 1, 63, 169, 132, 52, 11, 1, 1, 127, 482, 484, 247, 75, 13, 1, 1, 255, 1341, 1684, 1053, 410, 102, 15, 1, 1, 511, 3669, 5661, 4199, 1975, 629, 133, 17, 1, 1, 1023, 9922, 18579, 16017, 8778, 3366, 912, 168, 19, 1
Offset: 1

Views

Author

Henry Bottomley, Feb 25 2003

Keywords

Comments

Sum of entries in row n is A000108(n) (the Catalan numbers).
From Gus Wiseman, Nov 16 2022: (Start)
Also the number of unlabeled ordered rooted trees with n nodes and height k. For example, row n = 5 counts the following trees:
(oooo) ((o)oo) (((o))o) ((((o))))
((oo)o) (((o)o))
((ooo)) (((oo)))
(o(o)o) ((o(o)))
(o(oo)) (o((o)))
(oo(o))
((o)(o))
(End)

Examples

			T(3,2)=3 because we have UUDDUD, UDUUDD, and UUDUDD, where U=(1,1) and D=(1,-1). The other two Dyck paths of semilength 3, UDUDUD and UUUDDD, have heights 1 and 3, respectively. - _Emeric Deutsch_, Jun 08 2011
Triangle starts:
  1;
  1,  1;
  1,  3,   1;
  1,  7,   5,   1;
  1, 15,  18,   7,  1;
  1, 31,  57,  33,  9,  1;
  1, 63, 169, 132, 52, 11, 1;
		

References

  • N. G. de Bruijn, D. E. Knuth, and S. O. Rice, The average height of planted plane trees, in: Graph Theory and Computing (ed. T. C. Read), Academic Press, New York, 1972, pp. 15-22.

Crossrefs

T(2n,n) gives A268316.
Counting by leaves instead of height gives A001263.
The unordered version is A034781.
The height statistic is ranked by A358379, unordered A109082.

Programs

  • Maple
    f := proc (k) options operator, arrow:
       sum(binomial(k-i, i)*(-z)^i, i = 0 .. floor((1/2)*k))
    end proc:
    h := proc (k) options operator, arrow:
       z^k/(f(k)*f(k+1))
    end proc:
    T := proc (n, k) options operator, arrow:
       coeff(series(h(k), z = 0, 25), z, n)
    end proc:
    for n to 11 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form Emeric Deutsch, Jun 08 2011
    # second Maple program:
    b:= proc(x, y, k) option remember; `if`(y>min(k, x) or y<0, 0,
          `if`(x=0, 1, b(x-1, y-1, k)+ b(x-1, y+1, k)))
        end:
    T:= (n, k)-> b(2*n, 0, k) -`if`(k=0, 0, b(2*n, 0, k-1)):
    seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Aug 06 2012
  • Mathematica
    b[x_, y_, k_] := b[x, y, k] = If[y > Min[k, x] || y<0, 0, If[x == 0, 1, b[x-1, y-1, k] + b[x-1, y+1, k]]]; T[n_, k_] := b[2*n, 0, k] - If[k == 0, 0, b[2*n, 0, k-1] ]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 14}] // Flatten (* Jean-François Alcover, Feb 26 2015, after Alois P. Heinz *)
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],Depth[#]-2==k&]],{n,1,9},{k,1,n-1}] (* Gus Wiseman, Nov 16 2022 *)

Formula

T(n, k) = A080934(n, k) - A080934(n, k-1).
The g.f. for Dyck paths of height k is h(k) = z^k/(f(k)*f(k+1)), where f(k) are Fibonacci type polynomials defined by f(0)=f(1)=1, f(k)=f(k-1)-z*f(k-2) or by f(k) = Sum_{i=0..floor(k/2)} binomial(k-i,i)*(-z)^i. Incidentally, the g.f. for Dyck paths of height at most k is H(k) = f(k)/f(k+1). - Emeric Deutsch, Jun 08 2011
For all n >= 1 and floor((n+1)/2) <= k <= n we have: T(n,k) = 2*(2*k+3)*(2*k^2+6*k+1-3*n)*(2*n)!/((n-k)!*(n+k+3)!). - Gheorghe Coserea, Dec 06 2015
T(n, k) = Sum_{i=1..k-1} (-1)^(i+1) * (Sum_{j=1..n} (Sum_{x=0..n} (-1)^(j+x) * binomial(x+2n-2j+1,x))) * a(k-i); a(1)=1, a(0)=0. - Tim C. Flowers, May 14 2018

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}]

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}]

A358377 Numbers k such that the k-th standard ordered rooted tree is a generalized Bethe tree (counted by A003238).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 11, 16, 17, 32, 37, 43, 64, 128, 129, 137, 171, 256, 257, 293, 512, 529, 683, 1024, 1025, 2048, 2185, 2341, 2731, 4096, 8192, 10923, 16384, 16913, 18725, 32768, 32769, 32897, 34953, 43691, 65536, 65537, 131072, 131329, 149797, 174763
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.
A generalized Bethe tree is an unlabeled rooted tree where all branches directly under the same root are equal.

Examples

			The terms together with their corresponding ordered rooted trees begin:
    1: o
    2: (o)
    3: ((o))
    4: (oo)
    5: (((o)))
    8: (ooo)
    9: ((oo))
   11: ((o)(o))
   16: (oooo)
   17: ((((o))))
   32: (ooooo)
   37: (((o))((o)))
   43: ((o)(o)(o))
   64: (oooooo)
  128: (ooooooo)
  129: ((ooo))
  137: ((oo)(oo))
  171: ((o)(o)(o)(o))
		

Crossrefs

These trees are counted by A003238.
The unordered version is A214577, also counted by A003238.
A000081 counts unlabeled rooted trees, ranked by A358378.
A358371 and A358372 count leaves and nodes in standard ordered rooted trees.
A358374 ranks ordered identity trees, counted by A032027.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    srt[n_]:=If[n==1,{},srt/@stc[n-1]];
    Select[Range[1000],FreeQ[srt[#],[_]?(!SameQ@@#&)]&]

A358378 Numbers k such that the k-th standard ordered rooted tree is fully canonically ordered (counted by A000081).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 15, 16, 17, 21, 25, 27, 29, 31, 32, 37, 41, 43, 49, 53, 57, 59, 61, 63, 64, 65, 73, 81, 85, 101, 105, 107, 113, 117, 121, 123, 125, 127, 128, 129, 137, 145, 165, 169, 171, 193, 201, 209, 213, 229, 233, 235, 241, 245, 249, 251
Offset: 1

Views

Author

Gus Wiseman, Nov 14 2022

Keywords

Comments

The ordering of finitary multisets is first by length and then lexicographically. This is also the ordering used for Mathematica expressions.
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 terms together with their corresponding ordered rooted trees begin:
   1: o
   2: (o)
   3: ((o))
   4: (oo)
   5: (((o)))
   7: (o(o))
   8: (ooo)
   9: ((oo))
  11: ((o)(o))
  13: (o((o)))
  15: (oo(o))
  16: (oooo)
  17: ((((o))))
  21: ((o)((o)))
		

Crossrefs

These trees are counted by A000081.
A358371 and A358372 count leaves and nodes in standard ordered rooted trees.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    srt[n_]:=If[n==1,{},srt/@stc[n-1]];
    Select[Range[1000],FreeQ[srt[#],[_]?(!OrderedQ[#]&)]&]

A358376 Numbers k such that the k-th standard ordered rooted tree is lone-child-avoiding (counted by A005043).

Original entry on oeis.org

1, 4, 8, 16, 18, 25, 32, 36, 50, 57, 64, 72, 100, 114, 121, 128, 137, 144, 200, 228, 242, 249, 256, 258, 274, 281, 288, 385, 393, 400, 456, 484, 498, 505, 512, 516, 548, 562, 569, 576, 770, 786, 793, 800, 897, 905, 912, 968, 996, 1010, 1017, 1024, 1032, 1096
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 initial terms and their corresponding trees:
    1: o
    4: (oo)
    8: (ooo)
   16: (oooo)
   18: ((oo)o)
   25: (o(oo))
   32: (ooooo)
   36: ((oo)oo)
   50: (o(oo)o)
   57: (oo(oo))
   64: (oooooo)
   72: ((oo)ooo)
  100: (o(oo)oo)
  114: (oo(oo)o)
  121: (ooo(oo))
  128: (ooooooo)
  137: ((oo)(oo))
  144: ((oo)oooo)
  200: (o(oo)ooo)
		

Crossrefs

These trees are counted by A005043.
The series-reduced case appears to be counted by A284778.
The unordered version is A291636, counted by A001678.
A000081 counts unlabeled rooted trees, ranked by A358378.
A358371 and A358372 count leaves and nodes in standard 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]];
    Select[Range[100],FreeQ[srt[#],[_]?(Length[#]==1&)]&]

A358508 Least Matula-Goebel number of a tree with exactly n permutations.

Original entry on oeis.org

1, 6, 12, 24, 48, 30, 192, 104, 148, 72, 3072, 60, 12288, 832, 144, 712, 196608, 222, 786432, 120, 288, 13312
Offset: 1

Views

Author

Gus Wiseman, Nov 20 2022

Keywords

Comments

The Matula-Goebel number of a rooted tree is the product of primes indexed by the Matula-Goebel numbers of the branches of its root, which gives a bijective correspondence between positive integers and unlabeled rooted trees.
To get a permutation of a tree, we choose a permutation of the multiset of branches of each node.

Examples

			The terms together with their corresponding trees begin:
      1: o
      6: (o(o))
     12: (oo(o))
     24: (ooo(o))
     48: (oooo(o))
     30: (o(o)((o)))
    192: (oooooo(o))
    104: (ooo(o(o)))
    148: (oo(oo(o)))
     72: (ooo(o)(o))
   3072: (oooooooooo(o))
     60: (oo(o)((o)))
  12288: (oooooooooooo(o))
    832: (oooooo(o(o)))
    144: (oooo(o)(o))
    712: (ooo(ooo(o)))
		

Crossrefs

Position of first appearance of n in A206487.
The sorted version is A358507.
A000081 counts rooted trees, ordered A000108.
A214577 and A358377 rank trees with no permutations.

Programs

  • Mathematica
    primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]
    MGTree[n_Integer]:=If[n===1,{},MGTree/@primeMS[n]]
    treeperms[t_]:=Times @@ Cases[t,b:{}:>Length[Permutations[b]],{0,Infinity}];
    uv=Table[treeperms[MGTree[n]],{n,100000}];
    Table[Position[uv,k][[1,1]],{k,Min@@Complement[Range[Max@@uv],uv]-1}]
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