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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A185422 Forests of k increasing plane unary-binary trees on n nodes. Generalized Stirling numbers of the second kind associated with A185415.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 9, 15, 6, 1, 39, 75, 45, 10, 1, 189, 459, 330, 105, 15, 1, 1107, 3087, 2709, 1050, 210, 21, 1, 7281, 23535, 23814, 11109, 2730, 378, 28, 1, 54351, 197235, 228285, 122850, 36099, 6174, 630, 36, 1
Offset: 1

Views

Author

Peter Bala, Jan 28 2011

Keywords

Comments

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing.
A080635 enumerates increasing plane (ordered) unary-binary trees with n nodes labeled from the set {1,2,...n}. The entry T(n,k) of the present table counts forests of k increasing plane unary-binary trees having n nodes in total. See below for an example.
The Stirling number of the second kind Stirling2(n,k) counts the partitions of the set [n] set into k blocks. Arranging the elements in each block in ascending numerical order provides an alternative combinatorial interpretation for Stirling2(n,k) as counting forests of k increasing unary trees on n nodes. Thus we may view the present array as generalized Stirling numbers of the second kind (associated with A080635 or with the polynomials P(n,x) of A185415 - see formulas (1) and (2) below).
For a table of ordered forests of increasing plane unary-binary trees see A185423. For the enumeration of forests and ordered forests in the non-plane case see A147315 and A185421.
The Bell transform of A080635(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......1
..3|....3......3......1
..4|....9.....15......6......1
..5|...39.....75.....45.....10......1
..6|..189....459....330....105.....15......1
..7|.1107...3087...2709...1050....210.....21......1
..
Examples of the recurrence:
T(5,1) = 39 = T(4,0)+1*T(4,1)+2*T(4,2) = 1*9+2*15;
T(6,3) = 330 = T(5,2)+3*T(5,3)+3*4*T(5,4) = 75+3*45+12*10.
Examples of forests:
T(4,1) = 9. The 9 plane increasing unary-binary trees on 4 nodes are shown in the example section of A080635.
T(4,2) = 15. The 15 forests consisting of two plane increasing unary-binary trees on 4 nodes consist of the 12 forests
......... ......... ...3.....
.2...3... .3...2... ...|.....
..\./.... ..\./.... ...2.....
...1...4. ...1...4. ...|.....
......... ......... ...1...4.
.
......... ......... ...4.....
.2...4... .4...2... ...|.....
..\./.... ..\./.... ...2.....
...1...3. ...1...3. ...|.....
......... ......... ...1...3.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...1...2. ...1...2. ...|.....
......... ......... ...1...2.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...2...1. ...2...1. ...|.....
......... ......... ...2...1.
.
and the three remaining forests
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... ..........
		

References

  • F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48.

Crossrefs

Programs

  • Maple
    #A185422
    P := proc(n,x)
    description 'polynomial sequence P(n,x) A185415'
    if n = 0
    return 1
    else
    return x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    with combinat:
    T:= (n,k) -> 1/k!*add ((-1)^(k-j)*binomial(k,j)*P(n,j),j = 0..k):
    for n from 1 to 10 do
    seq(T(n,k),k = 1..n);
    end do;
  • Mathematica
    t[n_, k_] := t[n, k] = t[n-1, k-1] + k*t[n-1, k] + k*(k+1)*t[n-1, k+1]; t[n_, n_] = 1; t[n_, k_] /; Not[1 <= k <= n] = 0; Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 22 2012, from given recurrence *)
  • PARI
    {T(n,k)=if(n<1||k<1||k>n,0,if(n==k,1,T(n-1,k-1)+k*T(n-1,k)+k*(k+1)*T(n-1,k+1)))}
    
  • PARI
    {T(n,k)=round(n!*polcoeff(polcoeff(exp(y*(-1/2+sqrt(3)/2*tan(sqrt(3)/2*x+Pi/6+x*O(x^n)))+y*O(y^k)),n,x),k,y))}
    
  • Sage
    # uses[bell_matrix from A264428]
    # Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: A080635(n+1), 10) # Peter Luschny, Jan 18 2016

Formula

TABLE ENTRIES
(1)... T(n,k) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*P(n,j),
where P(n,x) are the polynomials described in A185415.
Compare (1) with the formula for the Stirling numbers of the second kind
(2)... Stirling2(n,k) = 1/k!*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n.
RECURRENCE RELATION
(3)... T(n+1,k) = T(n,k-1) + k*T(n,k) + k*(k+1)*T(n,k+1).
GENERATING FUNCTION
Let E(t) = 1/2 + sqrt(3)/2*tan(sqrt(3)/2*t + Pi/6) be the e.g.f. for A080635.
The e.g.f. for the present triangle is
(4)... exp{x*(E(t)-1)} = Sum_{n>=0} R(n,x)*t^n/n!
= 1 + x*t + (x+x^2)*t^2/2! + (3*x+3*x^2+x^3)*t^3/3! + ....
ROW POLYNOMIALS
The row generating polynomials R(n,x) satisfy the recurrence
(5)... R(n+1,x) = x*{R(n,x)+R'(n,x)+R''(n,x)},
where the prime ' indicates differentiation with respect to x.
RELATIONS WITH OTHER SEQUENCES
Column 1 is A080635.
k!*T(n,k) counts ordered forests A185423(n,k).
The row polynomials R(n,x) are given by D^n(exp(x*t)) evaluated at t = 0, where D is the operator (1+t+t^2)*d/dt. Cf. A147315 and A008297. - Peter Bala, Nov 25 2011

A080635 Number of permutations on n letters without double falls and without initial falls.

Original entry on oeis.org

1, 1, 1, 3, 9, 39, 189, 1107, 7281, 54351, 448821, 4085883, 40533129, 435847959, 5045745069, 62594829027, 828229153761, 11644113200031, 173331882039141, 2723549731505163, 45047085512477049, 782326996336904679, 14233537708408467549, 270733989894887810547
Offset: 0

Views

Author

Emanuele Munarini, Feb 28 2003

Keywords

Comments

A permutation w has a double fall at k if w(k) > w(k+1) > w(k+2) and has an initial fall if w(1) > w(2).
exp(x*(1-y+y^2)*D_y)*f(y)|_{y=0} = f(1-E(-x)) for any function f with a Taylor series. D_y means differentiation with respect to y and E(x) is the e.g.f. given below. For a proof of exp(x*g(y)*D_y)*f(y) = f(F^{-1}(x+F(y))) with the compositional inverse F^{-1} of F(y)=int(1/g(y),y) with F(0)=0 see, e.g., the Datolli et al. reference.
Number of increasing ordered trees on vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2. - David Callan, Mar 30 2007
Number of increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2. - Wenjin Woan, May 21 2011

Examples

			E.g.f. = 1 + x + (1/2)*x^2 + (1/2)*x^3 + (3/8)*x^4 + (13/40)*x^5 + (21/80)*x^6 + ...
G.f. = 1 + x + x^2 + 3*x^3 + 9*x^4 + 39*x^5 + 189*x^6 + 1107*x^7 + ...
For n = 3: 123, 132, 231. For n = 4: 1234, 1243, 1324, 1342, 1423, 2314, 2341, 2413, 3412.
a(4)=9. The 9 plane (ordered) increasing unary-binary trees are
...................................................................
..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......................................................
...................................................................
		

Crossrefs

Programs

  • Maple
    a:= proc(n) if n < 2 then 1 else n! * sum((sqrt(3)/(2*Pi*(k+1/3)))^(n+1), k=-infinity..infinity) fi end: seq(a(n), n=0..30); # Richard Ehrenborg, Dec 09 2013
    a := proc(n) option remember; local k; if n < 3 then 1 else
    add(binomial(n-1, k)*a(k)*a(n-k-1), k = 0..n-2) fi end:
    seq(a(n), n = 0..23); # Peter Luschny, May 24 2024
  • Mathematica
    Table[n!, {n, 0, 40}]*CoefficientList[Series[ (1 + 1/Sqrt[3] Tan[Sqrt[3]/2 x])/(1 - 1/Sqrt[3] Tan[Sqrt[3]/2 x]), {x, 0, 40}], x]
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1/2 + Sqrt[3]/2 Tan[ Pi/6 + Sqrt[3] x/2], {x, 0, n}]]; (* Michael Somos, May 22 2011 *)
    Join[{1}, FullSimplify[Table[3^((n+1)/2) * n! * (Zeta[n+1, 1/3] - (-1)^n*Zeta[n+1, 2/3]) / (2*Pi)^(n+1), {n, 1, 20}]]] (* Vaclav Kotesovec, Aug 06 2021 *)
  • Maxima
    a(n):=if n=0 then 1 else sum((-3)^((n-k)/2)*((-1)^(n-k)+1)*sum(binomial(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^(j)*stirling2(n,j+k),j,0,n-k),k,1,n); /* Vladimir Kruchinin, Feb 13 2019 */
  • PARI
    {a(n) = my(A); if( n<1, n==0, A = O(x); for( k=1, n, A = intformal( 1 + A + A^2)); n! * polcoeff( A, n))}; /* Michael Somos, Oct 04 2003 */
    
  • PARI
    {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(2*cosh(x +x*O(x^n)) - 1) ) )), n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 22 2016
    
  • 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)+2*k*c(n-1,k+1)
    def A080635(n):
        return add(c(n,k) for k in (0..n))
    [A080635(n) for n in (0..23)] # Peter Luschny, Jun 10 2014
    

Formula

E.g.f.: (1 + 1/sqrt(3) * tan(sqrt(3)/2 * x)) / (1 - 1/sqrt(3) * tan( sqrt(3)/2 * x)).
Recurrence: a(n+1) = (Sum_{k=0..n} binomial(n, k) * a(k) * a(n-k)) - a(n) + 0^n.
E.g.f.: A(x) satisfies A' = 1 - A + A^2. - Michael Somos, Oct 04 2003
E.g.f.: E(x) = (3*cos((1/2)*3^(1/2)*x) + (3^(1/2))*sin((1/2)*3^(1/2)*x))/(3*cos((1/2)*3^(1/2)*x) - (3^(1/2))* sin((1/2)*3^(1/2)*x)). See the Michael Somos comment. - Wolfdieter Lang, Sep 12 2005
O.g.f.: A(x) = 1+x/(1-x-2*x^2/(1-2*x-2*3*x^2/(1-3*x-3*4*x^2/(1-... -n*x-n*(n+1)*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
From Peter Bala: (Start)
An alternative form of the e.g.f. for this sequence taken from [Bergeron et al.] is
(1)... (sqrt(3)/2)*tan((sqrt(3)/2)*x+Pi/6) [with constant term 1/2].
By comparing the egf for this sequence with the egf for the Eulerian numbers A008292 we can show that
(2)... a(n) = A(n,w)/(1+w)^(n-1) for n >= 1,
where w = exp(2*Pi*i/3) and {A(n,x),n>=1} = [1, 1+x, 1+4*x+x^2, 1+11*x+11*x^2+x^3,...] denotes the sequence of Eulerian polynomials. Equivalently,
(3)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k)*(-1/2 + sqrt(3)*i/6)^(k-1) for n >= 1, and
(4)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} (-1/2 + sqrt(3)*i/6)^(k-1)* Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n for n >= 1.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(5a)... a(p) == 1 (mod p) for prime p = 6*n+1,
(5b)... a(p) == -1 (mod p) for prime p = 6*n+5.
For the corresponding results for the case of non-plane unary-binary trees see A000111. For type B results see A001586. For a related sequence of polynomials see A185415. See also A185416 for a recursive method to compute this sequence. For forests of plane increasing unary binary trees see A185422 and A185423. (End)
O.g.f.: A(x) = x - (1/2)*x^2 + (1/2)*x^3 - (3/8)*x^4 + (13/40)*x^5 - (21/80)*x^6 + (123/560)*x^7 - (809/4480)*x^8 + (671/4480)*x^9 - (5541/44800)*x^10 + .... - Vladimir Kruchinin, Jan 18 2011
Let f(x) = 1+x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 0. - Peter Bala, Oct 06 2011
From Sergei N. Gladkovskii, May 06 2013 - Dec 24 2013: (Start)
Continued fractions:
G.f.: 1 + 1/Q(0), where Q(k) = 1/(x*(k+1)) - 1 - 1/Q(k+1).
E.g.f.: 1 + 2*x/(W(0)-x), where W(k) = 4*k + 2 - 3*x^2/W(k+1).
G.f.: 1 + x/Q(0), m=1, where Q(k) = 1 - m*x*(2*k+1) - m*x^2*(2*k+1)*(2*k+2)/( 1 - m*x*(2*k+2) - m*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - x*(k+1) - x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/( x^2*(k+1)*(k+2) - (1-x*(k+1))*(1-x*(k+2))/T(k+1) ).
G.f.: 1 + x/(G(0)-x), where G(k) = 1 + x*(k+1) - x*(k+1)/(1 - x*(k+2)/G(k+1) ). (End)
a(n) ~ 3^(3*(n+1)/2) * n^(n+1/2) / (exp(n)*(2*Pi)^(n+1/2)). - Vaclav Kotesovec, Oct 05 2013
a(n) = n! * Sum_{k=-oo..oo} (sqrt(3)/(2*Pi*(k+1/3)))^(n+1) for n >= 1. - Richard Ehrenborg, Dec 09 2013
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = (sqrt(3)/2)*tan((sqrt(3)/2)*x + Pi/6) satisfies the differential equation A"(x) = 2*A(x)*A'(x) with A(0) = 1/2 and A'(0) = 1, leading to the recurrence a(0) = 1/2, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the sequence [1/2, 1, 1, 3, 9, 39, 189, 1107, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 0 and a(1) = 1 produces A000182. Cf. A002105, A234797. (End)
E.g.f.: exp( Series_Reversion( Integral 1/(2*cosh(x) - 1) dx ) ). - Paul D. Hanna, Feb 22 2016
a(n) = Sum_{k=1..n} (-3)^((n-k)/2)*((-1)^(n-k)+1)*Sum_{j=0..n-k} C(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^j*Stirling2(n,j+k),n>0, a(0)=1. - Vladimir Kruchinin, Feb 13 2019
For n > 0, a(n) = 3^((n+1)/2) * n! * (zeta(n+1, 1/3) - (-1)^n*zeta(n+1, 2/3)) / (2*Pi)^(n+1). - Vaclav Kotesovec, Aug 06 2021

Extensions

Several typos corrected by Olivier Gérard, Mar 26 2011

A147309 Riordan array [sec(x), log(sec(x) + tan(x))].

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 4, 0, 1, 5, 0, 10, 0, 1, 0, 40, 0, 20, 0, 1, 61, 0, 175, 0, 35, 0, 1, 0, 768, 0, 560, 0, 56, 0, 1, 1385, 0, 4996, 0, 1470, 0, 84, 0, 1, 0, 24320, 0, 22720, 0, 3360, 0, 120, 0, 1
Offset: 0

Views

Author

Paul Barry, Nov 05 2008

Keywords

Comments

Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
Unsigned version of A147308. - N. J. A. Sloane, Nov 07 2008
From Peter Bala, Jan 26 2011: (Start)
Define a polynomial sequence {Z(n,x)} n >= 0 by means of the recursion
(1)... Z(n+1,x) = 1/2*x*{Z(n,x-1)+Z(n,x+1)}
with starting condition Z(0,x) = 1. We call Z(n,x) the zigzag polynomial of degree n. This table lists the coefficients of these polynomials (for n >= 1) in ascending powers of x, row indices shifted by 1. The first few polynomials are
... Z(1,x) = x
... Z(2,x) = x^2
... Z(3,x) = x + x^3
... Z(4,x) = 4*x^2 + x^4
... Z(5,x) = 5*x + 10*x^3 + x^5.
The value Z(n,1) equals the zigzag number A000111(n). The polynomials Z(n,x) occur in formulas for the enumeration of permutations by alternating descents A145876 and in the enumeration of forests of non-plane unary binary labeled trees A147315.
{Z(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting Z(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[m]+...+(n-1)^[m] = (1/(m+1))*Sum_{k=0..m} (-1)^floor(k/2)*binomial(m+1,k)*B_k*n^[m+1-k],
where {B_k} k >= 0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For similarly defined polynomial sequences to Z(n,x) see A185415, A185417 and A185419. See also A185424.
(End)
[gd(x)^(-1)]^m = Sum_{n>=m} Tg(n,m)*(m!/n!)*x^n, where gd(x) is Gudermannian function, Tg(n+1,m+1)=T(n,m). - Vladimir Kruchinin, Dec 18 2011
The Bell transform of abs(E(n)), E(n) the Euler numbers. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins
   1;
   0,  1;
   1,  0,   1;
   0,  4,   0,  1;
   5,  0,  10,  0,  1;
   0, 40,   0, 20,  0, 1;
  61,  0, 175,  0, 35, 0, 1;
		

Crossrefs

Programs

  • Maple
    Z := proc(n, x) option remember;
    description 'zigzag polynomials Z(n, x)'
    if n = 0 return 1 else return 1/2*x*(Z(n-1, x-1)+Z(n-1, x+1)) end proc:
    with(PolynomialTools):
    for n from 1 to 10 CoefficientList(Z(n, x), x); end do; # Peter Bala, Jan 26 2011
  • Mathematica
    t[n_, k_] := SeriesCoefficient[ 2^k*ArcTan[(E^x - 1)/(E^x + 1)]^k*n!/k!, {x, 0, n}]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten // Abs (* Jean-François Alcover, Jan 23 2015 *)
  • PARI
    T(n, k)=local(X); if(k<1 || k>n, 0, X=x+x*O(x^n); n!*polcoeff(polcoeff((tan(X)+1/cos(X))^y, n), k)) \\ Paul D. Hanna, Feb 06 2011
    
  • Sage
    R = PolynomialRing(QQ, 'x')
    @CachedFunction
    def zzp(n, x) :
        return 1 if n == 0 else x*(zzp(n-1, x-1)+zzp(n-1, x+1))/2
    def A147309_row(n) :
        x = R.gen()
        L = list(R(zzp(n, x)))
        del L[0]
        return L
    for n in (1..10) : print(A147309_row(n)) # Peter Luschny, Jul 22 2012
    
  • Sage
    # uses[bell_matrix from A264428]
    # Alternative: Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: abs(euler_number(n)), 10) # Peter Luschny, Jan 18 2016

Formula

From Peter Bala, Jan 26 2011: (Start)
GENERATING FUNCTION
The e.g.f., upon including a constant term of '1', is given by:
(1) F(x,t) = (tan(t) + sec(t))^x = Sum_{n>=0} Z(n,x)*t^n/n! = 1 + x*t + x^2*t^2/2! + (x+x^3)*t^3/3! + ....
Other forms include
(2) F(x,t) = exp(x*arcsinh(tan(t))) = exp(2*x*arctanh(tan(t/2))).
(3) F(x,t) = exp(x*(t + t^3/3! + 5*t^5/5! + 61*t^7/7! + ...)),
where the coefficients [1,1,5,61,...] are the secant or zig numbers A000364.
ROW GENERATING POLYNOMIALS
One easily checks from (1) that
d/dt(F(x,t)) = 1/2*x*(F(x-1,t) + F(x+1,t))
and so the row generating polynomials Z(n,x) satisfy the recurrence relation
(4) Z(n+1,x) = 1/2*x*{Z(n,x-1) + Z(n,x+1)}.
The e.g.f. for the odd-indexed row polynomials is
(5) sinh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n+1,x)*t^(2n+1)/(2n+1)!.
The e.g.f. for the even-indexed row polynomials is
(6) cosh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n,x)*t^(2n)/(2n)!.
From sinh(2*x) = 2*sinh(x)*cosh(x) we obtain the identity
(7) Z(2n+1,2*x) = 2*Sum_{k=0..n} binomial(2n+1,2k)*Z(2k,x)*Z(2n-2k+1,x).
The zeros of Z(n,x) lie on the imaginary axis (use (4) and adapt the proof given in A185417 for the zeros of the polynomial S(n,x)).
BINOMIAL EXPANSION
The form of the e.g.f. shows that {Z(n,x)} n >= 0 is a sequence of polynomials of binomial type. In particular, we have the expansion
(8) Z(n,x+y) = Sum_{k=0..n} binomial(n,k)*Z(k,x)*Z(n-k,y).
The delta operator D* associated with this binomial type sequence is
(9) D* = D - D^3/3! + 5*D^5/5! - 61*D^7/7! + 1385*D^9/9! - ..., and satisfies
the relation
(10) tan(D*)+sec(D*) = exp(D).
The delta operator D* acts as a lowering operator on the zigzag polynomials:
(11) (D*)Z(n,x) = n*Z(n-1,x).
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(12) Z(p,x) = (-1)^((p-1)/2)*x (mod p).
More generally, for k = 1,2,3,...
(13) Z(p+k-1,x) = (-1)^((p-1)/2)*Z(k,x) (mod p).
RELATIONS WITH OTHER SEQUENCES
Row sums [1,1,2,5,16,61,...] are the zigzag numbers A000111(n) for n >= 1.
Column 1 (with 0's omitted) is the sequence of Euler numbers A000364.
A145876(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j).
A147315(n-1,k-1) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A185421(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A012123(n) = (-i)^n*Z(n,i) where i = sqrt(-1). A012259(n) = 2^n*Z(n,1/2).
(End)
T(n,m) = Sum(i=0..n-m, s(i)/(n-i)!*Sum(k=m..n-i, A147315(n-i,k)*Stirling1(k,m))), m>0, T(n,0) = s(n), s(n)=[1,0,1,0,5,0,61,0,1385,0,50521,...] (see A000364). - Vladimir Kruchinin, Mar 10 2011

A185416 Square array, read by antidiagonals, used to recursively calculate A080635.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 9, 6, 3, 1, 39, 24, 11, 4, 1, 189, 114, 51, 18, 5, 1, 1107, 648, 279, 96, 27, 6, 1, 7281, 4194, 1767, 594, 165, 38, 7, 1, 54351, 30816, 12699, 4176, 1143, 264, 51, 8, 1, 448821, 251586, 101979, 32922, 8865, 2034, 399, 66, 9, 1
Offset: 1

Views

Author

Peter Bala, Jan 28 2011

Keywords

Comments

The table entries T(n,k), n,k>=1, are defined by the recurrence relation
1)... T(n+1,k) = (k-1)*T(n,k-1)-k*T(n,k)+(k+1)*T(n,k+1) with boundary condition T(1,k)=1.
The first column of the table is A080635.
For similar tables to calculate the zigzag numbers, the Springer numbers and the number of minimax trees see A185414, A185418 and A185420, respectively.

Examples

			Triangle begins
n\k|....1......2......3......4......5.......6.......7
=====================================================
..1|....1......1......1......1......1.......1.......1
..2|....1......2......3......4......5.......6.......7
..3|....3......6.....11.....18.....27......38......51
..4|....9.....24.....51.....96....165.....264.....399
..5|...39....114....279....594...1143....2034....3399
..6|..189....648...1767...4176...8865...17304...31563
..7|.1107...4194..12699..32922..76203..161442..318339
..
Examples of the recurrence:
T(4,4) = 96 = 3*T(3,3)-4*T(3,4)+5*T(3,5) = 3*11-4*18+ 5*27;
T(5,1) = 39 = 0*T(4,0)-1*T(4,1)+2*T(4,2) = -1*9+2*24;
		

Crossrefs

Programs

  • Maple
    #A185416
    P := proc(n,x) description 'polynomial sequence P(n,x) A185415'
    if n = 0 return 1
    else return
    x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    for n from 1 to 10 do
    seq(P(n,k)/k,k = 1..10);
    end do;
  • PARI
    {T(n, k)=if(n==1, 1, (k-1)*T(n-1, k-1)-k*T(n-1,k)+(k+1)*T(n-1, k+1))}

Formula

(1)... T(n,k) = P(n,k)/k, where P(n,x) are the polynomials defined in A185415.

A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 11, 26, 12, 8, 57, 120, 136, 32, 16, 361, 970, 760, 560, 80, 32, 2763, 7052, 8860, 3680, 2000, 192, 64, 24611, 72530, 72884, 58520, 15120, 6496, 448, 128, 250737, 716528, 976464, 538048, 314720, 55552, 19712, 1024, 256
Offset: 1

Views

Author

Peter Bala, Jan 28 2011

Keywords

Comments

Define a polynomial sequence S(n,x) recursively by
(1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1.
This table lists the coefficients of these polynomials (for n>=1) in ascending powers of x.
The first few polynomials are
S(0,x) = 1
S(1,x) = 2*x+1
S(2,x) = 4*x^2+4*x+3
S(3,x) = 8*x^3+12*x^2+26*x+11.
The sequence [1,1,3,11,57,...] of constant terms of the polynomials is the sequence of Springer numbers A001586. The zeros of the polynomials S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
Compare the recurrence (1) with the recurrence relation satisfied by the coefficients T(n,k) of the polynomials of A104035, namely
(2)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1).

Examples

			Table begin
n\k|.....0.....1.....2.....3.....4.....5......6
===============================================
0..|.....1
1..|.....1.....2
2..|.....3.....4.....4
3..|....11....26....12.....8
4..|....57...120...136....32...16
5..|...361...970...760...560...80.....32
6..|..2763..7052..8860..3680..2000...192....64
...
		

Crossrefs

Cf A001586 (1st column and row sums), A104035, A126156, A147309, A185415, A185418, A185419

Programs

  • Maple
    #A185417
    S := proc(n,x) option remember;
    description 'polynomials S(n,x)'
    if n = 0 return 1 else return x*S(n-1,x-1)+(x+1)*S(n-1,x+1)
    end proc:
    with(PolynomialTools):
    for n from 1 to 10 CoefficientList(S(n,x),x); end do;
  • Mathematica
    S[0, ] = 1; S[n, x_] := S[n, x] = x*S[n-1, x-1] + (x+1)*S[n-1, x+1]; Table[ CoefficientList[S[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 15 2015 *)

Formula

E.g.f: F(x,t) = 1/(cos(t)-sin(t))*(tan(2*t)+sec(2*t))^x
= (cos(t)+sin(t))^x/(cos(t)-sin(t))^(x+1)
= 1 + (2*x+1)*t + (4*x^2+4*x+3)*t^2/2! + ....
Note that (tan(t)+sec(t))^x is the e.g.f for table A147309.
ROW POLYNOMIALS
The easily checked identity d/dt F(x,t) = x*F(x-1,t)+(x+1)*F(x+1,t) shows that the row generating polynomials of this table are the polynomials S(n,x) described in the Comments section above.
The polynomials S(n,-x) satisfy a Riemann hypothesis: that is, the zeros of S(n,-x) lie on the vertical line Re(x) = 1/2 in the complex plane - see the link.
RELATION WITH OTHER SEQUENCES
1st column [1,1,3,11,57,...] is A001586.
Row sums sequence [1,3,11,57,...] is also A001586.
For n>=1, the values 1/2^n*P(2*n,-1/2) = [1,7,139,5473,...] appear to be A126156.

A185419 Table of coefficients of a polynomial sequence of binomial type related to the enumeration of minimax trees A080795.

Original entry on oeis.org

1, 3, 1, 10, 9, 1, 42, 67, 18, 1, 248, 510, 235, 30, 1, 1992, 4378, 2835, 605, 45, 1, 19600, 44268, 34888, 10605, 1295, 63, 1, 222288, 524748, 461748, 178913, 31080, 2450, 84, 1, 2851712, 7103088, 6728428, 3069612, 690753, 77112, 4242, 108, 1
Offset: 1

Views

Author

Peter Bala, Feb 07 2011

Keywords

Comments

DEFINITION
Define a sequence of polynomials M(n,x) by means of the recurrence relation
(1)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)},
with starting value M(0,x) = 1. We call these the minimax polynomials.
The first few polynomials are
M(1,x) = x
M(2,x) = x*(x + 3)
M(3,x) = x*(x^2 + 9*x + 10)
M(4,x) = x*(x^3 + 18*x^2 + 67*x + 42)
M(5,x) = x*(x^4 + 30*x^3 + 235*x^2 + 510*x + 248).
This triangle lists the coefficients of these polynomials (apart from M(0,x)) in ascending powers of x.
RELATION TO MINIMAX TREES
The value M(n,1) equals the number of minimax trees on n nodes - A080795(n). This result can be used to recursively calculate the entries of A080795 - see A185420.
In addition, the minimax polynomials M(n,x) occur in the formula for the number T(n,k) of forests of k minimax trees on n nodes. ... T(n,k) = 1/k!*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*M(n,j).
ANALOGIES WITH THE MONOMIALS
{M(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting M(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[p]+...+(n-1)^[p] = -2*n^[p]+ 1/(p+1)*Sum_{k = 0..floor(p/2)} 8^k*binomial(p+1,2k)*B_(2k)*n^[p+1-2k], where {B_k}k>=0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For other polynomial sequences defined by recurrences similar to (1), and related to the zigzag numbers A000111 and the Springer numbers A001586, see A147309 and A185417, respectively. See also A185415.
The Bell transform of A143523(n). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins
n\k|.....1......2......3......4......5......6......7
====================================================
..1|.....1
..2|.....3......1
..3|....10......9......1
..4|....42.....67.....18......1
..5|...248....510....235.....30......1
..6|..1992...4378...2835....605.....45......1
..7|.19600..44268..34888..10605...1295.....63......1
..
Example of the generalized Bernoulli summation formula:
The second row of the triangle gives x^[2] = 3*x+x^2.
Then 1^[2]+2^[2]+...+(n-1)^[2] = (n^3+3*n^2-4*n)/3 = 1/3*(MB(3,n)-MB(3,0)).
From _R. J. Mathar_, Mar 15 2013: (Start)
The matrix inverse starts
       1;
      -3,       1;
      17,      -9,        1;
    -147,      95,      -18,      1;
    1697,   -1245,      305,    -30,      1;
  -24483,   19687,    -5670,    745,    -45,    1;
  423857, -365757,   118237, -18690,   1540,  -63,   1;
-8560947, 7819287, -2761122, 498197, -50190, 2842, -84, 1; (End)
		

Crossrefs

Cf. A080253 (coeffs. of delta operator), A080795 (row sums), A143523 (column 1), A147309, A185415, A185417, A185420.

Programs

  • Maple
    #A185419
    M := proc(n,x) option remember;
    if n = 0 then
    return 1
    else return
    x*(2*M(n-1,x+1)-M(n-1,x-1))
    end if;
    end proc:
    with(PolynomialTools):
    for n from 1 to 10 do
    CoefficientList(M(n,x),x);
    end do;
  • Mathematica
    M[0, ] = 1; M[n, x_] := M[n, x] = x (2 M[n-1, x+1] - M[n-1, x-1]);
    Table[CoefficientList[M[n, x], x] // Rest, {n, 1, 10}] (* Jean-François Alcover, Jun 26 2019 *)
  • Sage
    # uses[bell_matrix from A264428]
    # Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: A143523(n), 10) # Peter Luschny, Jan 18 2016

Formula

GENERATING FUNCTION
Let a = 3-2*sqrt(2). Let f(t) = (1/2)*sqrt(2)*((1+a*exp(2*sqrt(2)*t))/ (1-a*exp(2*sqrt(2)*t))) = 1 + t + 4*t^2/2! + 20*t^3/3! + ... be the e.g.f. for A080795. Then the e.g.f. for the current table, including a constant 1, is
(1)... F(x,t) = f(t)^x = Sum_{n>=0} M(n,x)*t^n/n! = 1 + x*t + (3*x+x^2)*t^2/2! + (10*x+9*x^2+x^3)*t^3/3! + ....
ROW POLYNOMIALS
One easily checks that d/dt(F(x,t)) = x*(2*F(x+1,t)-F(x-1,t)) and hence the row generating polynomials M(n,x) satisfy the recurrence relation
(2)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)}.
The form of the e.g.f shows that the row polynomials are a polynomial sequence of binomial type. The associated delta operator D* is given by
(3)... D* = sqrt(2)/4*log((3+2*sqrt(2))*(sqrt(2)*exp(D)-1)/(sqrt(2)*exp(D)+1)),
where D is the derivative operator d/dx. This expands to
(4)... D* = D - 3*D^2/2! + 17*D^3/3! - 147*D^4/4! + ....
The sequence of coefficients [1,3,17,147,...] is A080253.
The delta operator D* acts as a lowering operator for the minimax polynomials
(5)...(D*) M(n,x) = n*M(n-1,x).
In what follows it will be convenient to denote M(n,x) by x^[n].
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(6)... x^[p] = (-1)^((p^2-1)/8)*x (mod p).
More generally, for k = 1,2,...
(7)... x^[p+k-1] = (-1)^((p^2-1)/8)*x^[k] (mod p).
GENERALIZED BERNOULLI POLYNOMIALS ASSOCIATED WITH THE MINIMAX POLYNOMIALS
The generalized Bernoulli polynomial MB(k,x) associated with the minimax polynomial x^[k] (= M(k,x)) may be defined as the result of applying the differential operator D*/(exp(D)-1) to the polynomial x^[k]:
(8)... MB(k,x) := {D*/(exp(D)-1)} x^[k].
The first few generalized Bernoulli polynomials are
MB(0,x) = 1,
MB(1,x) = x - 2,
MB(2,x) = x^2 - x + 4/3,
MB(3,x) = x^3 + 3*x^2 - 4*x,
MB(4,x) = x^4 + 10*x^3 + 3*x^2 - 14*x - 32/15.
Since exp(D)-1 is the forward difference operator it follows from (5) and (8) that
(9)... MB(k,x+1) - MB(k,x) = k*x^[k-1].
Summing (9) from x = 1 to x = n-1 and telescoping we find a closed form expression for the finite sums
(10)... 1^[p]+2^[p]+...+(n-1)^[p] = 1/(p+1)*{MB(p+1,n)-MB(p+1,1)}.
The generalized Bernoulli polynomials can be expanded in terms of the minimax polynomials x^[k]. Use (3) to express exp(D)-1 in terms of D*.
Substitute the resulting expression in (8) and expand as a power series in D* to arrive at the expansion:
(11)... MB(k,x) = -2*k*x^[k-1] + Sum_{j=0..floor(k/2)} 2^(3*j) * binomial(k,2j)*B_(2j)*x^[k-2j], where {B_j}j>=0 = [1,-1/2,1/6,0,-1/30,...] denotes the Bernoulli number sequence.
RELATION WITH OTHER SEQUENCES
Column 1 [1, 3, 10, 42, 248, ...] = A143523 with an offset of 1.
Row sums [1, 1, 4, 20, 128, 1024, ...] = A080795.

A185423 Ordered forests of k increasing plane unary-binary trees with n nodes.

Original entry on oeis.org

1, 1, 2, 3, 6, 6, 9, 30, 36, 24, 39, 150, 270, 240, 120, 189, 918, 1980, 2520, 1800, 720, 1107, 6174, 16254, 25200, 25200, 15120, 5040, 7281, 47070, 142884, 266616, 327600, 272160, 141120, 40320, 54351, 394470, 1369710, 2948400, 4331880
Offset: 1

Views

Author

Peter Bala, Jan 28 2011

Keywords

Comments

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A080635 enumerates increasing plane (ordered) unary-binary trees with n nodes with labels drawn from the set {1,2,...,n}. The entry T(n,k) of the present table counts ordered forests of k increasing plane unary-binary trees with n nodes. See below for an example. See A185421 for the corresponding table in the non-plane case. See A185422 for the table for unordered forests of increasing plane unary-binary trees.

Examples

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......2
..3|....3......6......6
..4|....9.....30.....36.....24
..5|...39....150....270....240....120
..6|..189....918...1980...2520...1800....720
..7|.1107...6174..16254..25200..25200..15120...5040
..
Examples of recurrence relation:
T(5,3) = 270 = 3*(T(4,2) + T(4,3) + T(4,4)) = 3*(30+36+24);
T(6,1) = 1*(T(5,0) + T(5,1) + T(5,2)) = 39+150.
Examples of ordered forests:
T(4,2) = 30. The 15 unordered forests consisting of two plane increasing unary-binary trees on 4 nodes are shown below. We can order the trees in a forest in two ways giving rise to 30 ordered forests.
......... ......... ...3.....
.2...3... .3...2... ...|.....
..\./.... ..\./.... ...2.....
...1...4. ...1...4. ...|.....
......... ......... ...1...4.
.
......... ......... ...4.....
.2...4... .4...2... ...|.....
..\./.... ..\./.... ...2.....
...1...3. ...1...3. ...|.....
......... ......... ...1...3.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...1...2. ...1...2. ...|.....
......... ......... ...1...2.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...2...1. ...2...1. ...|.....
......... ......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... ..........
		

Crossrefs

Programs

  • Maple
    #A185423
    P := proc(n,x)
    description 'polynomial sequence P(n,x) A185415'
    if n = 0 return 1
    else
    return x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    with combinat:
    T:=(n,k) -> add ((-1)^(k-j)*binomial(k,j)*P(n,j),j = 0..k):
    for n from 1 to 10 do
    seq(T(n,k),k = 1..n);
    end do;
  • PARI
    {T(n,k)=if(n<1||k<1||k>n,0,if(n==1,if(k==1,1,0),k*(T(n-1,k-1)+T(n-1,k)+T(n-1,k+1))))}

Formula

TABLE ENTRIES
(1)... T(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*P(n,j),
where P(n,x) are the polynomials described in A185415.
Compare (1) with the formula for the number of ordered set partitions
(2)... A019538(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*j^n.
RECURRENCE RELATION
(3)... T(n+1,k) = k*(T(n,k-1) + T(n,k) + T(n,k+1))
GENERATING FUNCTION
Let E(t) = 1/2 + (sqrt(3)/2)*tan(sqrt(3)/2*t + Pi/6) be the e.g.f. for A080635.
The e.g.f. for the present triangle is
(4)... 1/(1 + x*(1 - E(t))) = Sum {n >= 0} R(n,x)*t^n/n!
= 1 + x*t + (x + 2*x^2)*t^2/2! + (3*x + 6*x^2 + 6*x^3)*t^3/3! + ....
ROW POLYNOMIALS
The ordered Bell polynomials OB(n,x) are the row polynomials of A019538 given by
(5)... OB(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
By comparing the e.g.f.s for A019538 and the present table we obtain the surprising identity
(6)... (-i*sqrt(3))^(n-1)*OB(n,x)/x = R(n,y)/y,
where i = sqrt(-1) and x = (sqrt(3)*i/3)*y + (-1/2+sqrt(3)*i/6).
RELATIONS WITH OTHER SEQUENCES
Column 1: A080635.
A185422(n,k) = T(n,k)/k!.
Setting y = 0 in (6) gives
(7)... A080635(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k) * (-1/2+sqrt(3)*i/6)^(k-1).
The row polynomials have their zeros on the vertical line Re(z) = -1/2 in the complex plane (this follows from the above connection with the ordered Bell polynomials, which have negative real zeros). - Peter Bala, Oct 23 2023
Showing 1-7 of 7 results.