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 1-4 of 4 results.

A101280 Triangle T(n,k) (n >= 1, 0 <= k <= floor((n-1)/2)) read by rows, where T(n,k) = (k+1)T(n-1,k) + (2n-4k)T(n-1,k-1).

Original entry on oeis.org

1, 1, 1, 2, 1, 8, 1, 22, 16, 1, 52, 136, 1, 114, 720, 272, 1, 240, 3072, 3968, 1, 494, 11616, 34304, 7936, 1, 1004, 40776, 230144, 176896, 1, 2026, 136384, 1328336, 2265344, 353792, 1, 4072, 441568, 6949952, 21953408, 11184128, 1, 8166, 1398000, 33981760
Offset: 1

Views

Author

Don Knuth, Jan 28 2005

Keywords

Comments

Row n has ceiling(n/2) terms.
The paper by Shapiro et al. explains why T(n,k) is the number of permutations of n elements having k peaks and with the further property that every rise (ascent) is immediately followed by a peak. [That is, the permutation p_1 ... p_n has the further property that (j > 1 and p_{j-1} < p_j) implies (j < n and p_j > p_{j+1}).] For example, the T(4,1)=8 permutations in the case n=4, k=1 are 1423, 2143, 2431, 3142, 3241, 3421, 4231, 4132.
A more elegant way to state this property: T(n,k) is the number of permutations of n objects with k descents such that every descent is a peak. The eight examples for n=4 and k=1 are now 1243, 1324, 1342, 1423, 2314, 2341, 2413, 3412.
The rows of this triangle are the gamma vectors of the n-dimensional (type A) permutohedra (Postnikov et al., p. 31). Cf. A055151 and A089627. - Peter Bala, Oct 28 2008

Examples

			Triangle begins:
  1;
  1,
  1,   2;
  1,   8,
  1,  22,  16;
  1,  52, 136,
  1, 114, 720, 272;
  ...
From _Peter Bala_, Jun 26 2012: (Start)
n = 4: the 9 weighted plane increasing 0-1-2 trees on 4 vertices are
..................................................................
..4...............................................................
..|...............................................................
..3..........4...4...............4...4...............3...3........
..|........./.....\............./.....\............./.....\.......
..2....2...3.......3...2...3...2.......2...3...4...2.......2...4..
..|.....\./.........\./.....\./.........\./.....\./.........\./...
..1...(t)1........(t)1....(t)1........(t)1....(t)1........(t)1....
..................................................................
..3...4...4...3...................................................
...\./.....\./....................................................
.(t)2....(t)2.....................................................
....|.......|.....................................................
....1.......1.....................................................
Hence R(4,t) = 1 + 8*t.
(End)
		

References

  • D. Foata and V. Strehl, "Euler numbers and variations of permutations", in Colloquio Internazionale sulle Teorie Combinatorie, Rome, September 1973, (Atti dei Convegni Lincei 17, Rome, 1976), 129.
  • Guoniu Han, Frédéric Jouhet, Jiang Zeng, Two new triangles of q-integers via q-Eulerian polynomials of type A and B, Ramanujan J (2013) 31:115-127, DOI 10.1007/s11139-012-9389-3
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 4.

Crossrefs

The numbers 2^{n-1-k} T(n, k) form the array shown in A008303.
Cf. A055151, A089627. - Peter Bala, Oct 28 2008
Cf. A008292, A094503, A080635 (row sums).

Programs

  • Maple
    T:=proc(n,k) if k<0 then 0 elif n=1 and k=0 then 1 elif k>floor((n-1)/2) then 0 else (k+1)*T(n-1,k)+(2*n-4*k)*T(n-1,k-1) fi end: for n from 1 to 13 do seq(T(n,k),k=0..floor((n-1)/2)) od; # yields sequence in triangular form # Emeric Deutsch, Aug 06 2005
  • Mathematica
    t[, k?Negative] = 0; t[1, 0] = 1; t[n_, k_] /; k > (n-1)/2 = 0; t[n_, k_] := t[n, k] = (k+1)*t[n-1, k] + (2*n-4*k)*t[n-1, k-1]; Table[t[n, k], {n, 1, 13}, {k, 0, (n-1)/2}] // Flatten (* Jean-François Alcover, Nov 22 2012 *)

Formula

G.f.: Sum_{n>=1, k>=0} T(n, k) t^k z^n/n! = C(t)(2-C(t))/(exp^(-z sqrt(1-4t)) + 1 - C(t)) - C(t), where the sum on k is actually finite, running up to ceiling(n/2) - 1; here C(t) = (1 - sqrt(1-4t))/2t is the generating function for the Catalan numbers (A000108).
Sum_{k} Eulerian(n, k) x^k = Sum_{k} T(n, k) x^k (1+x)^(n-1-2k). E.g., 1 + 11x + 11x^2 + x^3 = (1+x)^3 + 8x(1+x).
From Peter Bala, Jun 26 2012: (Start)
T(n,k) = 2^k*A094503(n,k+1).
Let r(t) = sqrt(1 - 2*t) and w(t) = (1 - r(t))/(1 + r(t)). Define F(t,z) = r(t)*(1 + w(t)*exp(r(t)*z))/(1 - w(t)*exp(r(t)*z)) = 1 + t*z + t*z^2/2! + (t+t^2)*z^3/3! + (t+4*t^2)*z^4/4! + ...; F(t,z) is the e.g.f. for A094503. The e.g.f. for the present table is A(t,z) := (F(2*t,z) - 1)/(2*t) = z + z^2/2! + (1+2*t)*z^3/3! + (1+8*t)*z^4/4! + ....
The e.g.f. A(t,z) satisfies the autonomous partial differential equation dA/dz = 1 + A + t*A^2 with A(t,0) = 0. It follows that the inverse function A(t,z)^(-1) may be expressed as an integral: A(t,z)^(-1) = int {x = 0..z} 1/(1+x+t*x^2) dx.
Applying [Dominici, Theorem 4.1] to invert the integral gives the following method for calculating the row polynomials R(n,t) of the table: let f(t,x) = 1+x+t*x^2 and let D be the operator f(t,x)*d/dx. Then R(n+1,t) = D^n(f(t,x)) evaluated at x = 0.
By Bergeron et al., Theorem 1, the row polynomial R(n,t) is the generating function for rooted plane increasing 0-1-2 trees on n vertices, where the vertices of outdegree 2 have weight t and all other vertices have weight 1. An example is given below.
Row sums A080635.
(End)

Extensions

More terms from Emeric Deutsch, Aug 06 2005

A234797 E.g.f. satisfies: A'(x) = 1 + A(x) + 2*A(x)^2, where A(0)=0.

Original entry on oeis.org

1, 1, 5, 17, 109, 649, 5285, 44513, 448861, 4836601, 58743125, 766520657, 10939702669, 167136559849, 2746173173765, 48016925121473, 893361709338301, 17582667488919001, 365487998075525045, 7994319232001122097, 183644125043688405229, 4418905413530661307849
Offset: 1

Views

Author

Paul D. Hanna, Jan 09 2014

Keywords

Comments

a(n) = number of increasing ordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2 and the vertices of degree 2 are colored either white or black. An example is given below. - Peter Bala, Sep 13 2015

Examples

			E.g.f.: A(x) = x + x^2/2! + 5*x^3/3! + 17*x^4/4! + 109*x^5/5! + 649*x^6/6! +...
Related series.
A(x)^2 = 2*x^2/2! + 6*x^3/3! + 46*x^4/4! + 270*x^5/5! + 2318*x^6/6! +...
a(4) = 17. The 17 plane (ordered) increasing unary-binary trees on 4 vertices are shown below. A * indicates the vertex of outdegree 2 may be colored either white or black.
...................................................................
..4................................................................
..|................................................................
..3..........4...4...............4...4...............3...3.........
..|........./.....\............./.....\............./.....\........
..2....2...3.......3...2...3...2.......2...3...4...2.......2...4...
..|.....\./.........\./.....\./.........\./.....\./.........\./....
..1......1*..........1*......1*..........1*......1*..........1*....
...................................................................
..3...4...4...3....................................................
...\./.....\./.....................................................
....2*......2*......................................................
....|.......|......................................................
....1.......1......................................................
...................................................................
- _Peter Bala_, Sep 13 2015
		

Crossrefs

Cf. A094503.

Programs

  • Mathematica
    Rest[FullSimplify[CoefficientList[Series[(Sqrt[7]*Tan[Sqrt[7]*x/2 + ArcTan[1/Sqrt[7]]]-1)/4, {x, 0, 20}], x] * Range[0, 20]!]] (* Vaclav Kotesovec, Jan 13 2014 *)
    nmax = 20; Clear[g]; g[nmax+1] = 1; g[k_] := g[k] = 1 - x*(2*k+1) - 4*x^2*(k+1)*(2*k+1)/( 1 - x*(2*k+2) - 4*x^2*(k+1)*(2*k+3)/g[k+1] ); CoefficientList[Series[1/g[0], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 15 2015, after Sergei N. Gladkovskii *)
  • PARI
    {a(n)=local(A=x);for(i=1,n,A=intformal(1+A+2*A^2 +x*O(x^n))); n!*polcoeff(A,n)}
    for(n=1,25,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=serreverse(intformal(1/(1+x+2*x^2 +x*O(x^n))))); n!*polcoeff(A,n)}
    for(n=1,25,print1(a(n),", "))
    
  • Sage
    @CachedFunction
    def c(n,k) :
        if n==k: return 1
        if k<1 or k>n: return 0
        return ((n-k)//2+1)*c(n-1,k-1)+k*c(n-1,k+1)
    def A234797(n):
        return add(c(n,k)*2^(n-k) for k in (0..n))
    [A234797(n) for n in (1..22)] # Peter Luschny, Jun 10 2014

Formula

E.g.f.: Series_Reversion( Integral 1/(1 + x + 2*x^2) dx ).
E.g.f.: (sqrt(7)*tan(sqrt(7)*x/2 + arctan(1/sqrt(7)))-1)/4. - Vaclav Kotesovec, Jan 13 2014
a(n) ~ n! * 1/2*(sqrt(7)/(Pi - 2*arctan(1/sqrt(7))))^(n+1). - Vaclav Kotesovec, Jan 13 2014
O.g.f.: A(x) = x/(1-x - 2*1*2*x^2/(1-2*x - 2*2*3*x^2/(1-3*x - 2*3*4*x^2/(1-... -n*x - 2*n*(n+1)*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jun 12 2014
Let f(x) = 1 + x + 2*x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 0. - Peter Bala, Sep 13 2015
G.f.: 1/T(0), where m=4; u=x; T(k)= 1 - u*(2*k+1) - m*u^2*(k+1)*(2*k+1)/( 1 - u*(2*k+2) - m*u^2*(k+1)*(2*k+3)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2015
a(n) = 2^n*A(n, 1/2) where A(n,x) are the André polynomials. - Peter Luschny, May 05 2016

A113897 Triangle read by rows: number of simsun n-permutations with k descents.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 11, 4, 1, 26, 34, 1, 57, 180, 34, 1, 120, 768, 496, 1, 247, 2904, 4288, 496, 1, 502, 10194, 28768, 11056, 1, 1013, 34096, 166042, 141584, 11056, 1, 2036, 110392, 868744, 1372088, 349504, 1, 4083, 349500, 4247720, 11204160, 6213288, 349504
Offset: 1

Views

Author

Chak-On Chow (cchow(AT)alum.mit.edu), Jan 28 2006

Keywords

Comments

Is this A094503 after removal of the top row? - R. J. Mathar, Aug 13 2008
Yes. See formula of Peter Bala, Jun 26 2012 in A094503. - Stefano Spezia, Aug 09 2023

Examples

			Triangle begins
   1;
   1,   1;
   1,   4;
   1,  11,   4;
   1,  26,  34;
   1,  57, 180,  34;
   ...
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(2t-1)*(Sec[x*Sqrt[2t-1]/2]/(Sqrt[2t-1]- Tan[x*Sqrt[2t-1]/2]))^2,{x,0,n},{t,0,k}]n!,{n,11},{k,0,Floor[n/2]}]//Flatten (* Stefano Spezia, Aug 09 2023 *)

Formula

T(n, k) = (k+1)*T(n-1, k) + (n-2k+1)*T(n-1, k-1);
Row g.f.: T(n, t) = Sum_{k=0..floor(n/2)} T(n, k)*t^k,
T(n, t) = ((n-1)*t + 1)*T(n-1, t) + t*(1-2t)*T(n-1, t)'.
E.g.f.: Sum_{n>=1} T(n, t)*x^n/n! = (2t-1)*(sec(x*sqrt(2t-1)/2)/(sqrt(2t-1) - tan(x*sqrt(2t-1)/2)))^2.

Extensions

Corrected and extended by Vladeta Jovovic, Jan 30 2006

A131638 Increasing binary trees having exactly two vertices with outdegree 1.

Original entry on oeis.org

1, 11, 180, 4288, 141584, 6213288, 350400832, 24718075136, 2133652515072, 221311262045440, 27166907582280704, 3895974311462313984, 645512064907811491840, 122381396964887716078592, 26325690425815766552887296, 6377608610246241663568248832
Offset: 1

Views

Author

Wenjin Woan, Oct 03 2007

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n!*SeriesCoefficient[1/2*(-((x*Sec[x/Sqrt[2]]^2 *Tan[x/Sqrt[2]]) /Sqrt[2]) + 3*Sec[x/Sqrt[2]]^2 *Tan[x/Sqrt[2]]^2), {x, 0, n}], {n, 2, 40, 2}] (* Vaclav Kotesovec after Michel Marcus, Sep 25 2013 *)
  • PARI
    lista(m) = { default(realprecision, 30); x = y + O(y^m); egf = (3*tan(x/sqrt(2))^2/cos(x/sqrt(2))^2-x*tan(x/sqrt(2))/(sqrt(2)*cos(x/sqrt(2))^2))/2; forstep (n=2, m, 2, print1(round(n!*polcoeff(egf, n, y)), ", "));}  \\ Michel Marcus, Mar 03 2013

Formula

E.g.f.: (3*sec(x/sqrt(2))^2*tan(x/sqrt(2))^2-x*sec(x/sqrt(2))^2*tan(x/sqrt(2))/(sqrt(2)))/2. - Michel Marcus, Mar 03 2013
a(n) ~ (2*n)! * 2^(n+6)*n^3/Pi^(2*n+4). - Vaclav Kotesovec, Sep 25 2013
From Klaus K Haverkamp, Jul 02 2023: (Start)
a(n) = (A002105(n+2) - (n+1)*A002105(n+1))/2.
a(n) = A094503(2n+1,n). (End)

Extensions

More terms from Michel Marcus, Mar 03 2013
Showing 1-4 of 4 results.