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 1-3 of 3 results.

A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 0, 10, 0, 1, 0, 0, 16, 0, 20, 0, 1, 0, 1, 0, 91, 0, 35, 0, 1, 0, 0, 64, 0, 336, 0, 56, 0, 1, 0, 1, 0, 820, 0, 966, 0, 84, 0, 1, 0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1, 0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1, 0, 0, 1024, 0, 87296, 0
Offset: 0

Views

Author

Paul D. Hanna, Jan 14 2008

Keywords

Comments

For partitions into blocks of even size see A156289.
Essentially the unsigned matrix inverse of triangle A121408.
From Peter Bala, Jul 28 2014: (Start)
Define a polynomial sequence x_(n) by setting x_(0) = 1 and for n = 1,2,... setting x_(n) = x*(x + n - 2)*(x + n - 4)*...*(x + n - 2*(n - 1)). Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k) for n = 0,1,2,.... An example is given below.
Let M denote the lower unit triangular array A119467 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle, omitting the first row and column, equals the infinite matrix product M(0)*M(1)*M(2)*.... (End)
Also the Bell transform of A000035(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle begins:
  1;
  0, 1;
  0, 0,   1;
  0, 1,   0,    1;
  0, 0,   4,    0,    1;
  0, 1,   0,   10,    0,     1;
  0, 0,  16,    0,   20,     0,    1;
  0, 1,   0,   91,    0,    35,    0,    1;
  0, 0,  64,    0,  336,     0,   56,    0,   1;
  0, 1,   0,  820,    0,   966,    0,   84,   0,   1;
  0, 0, 256,    0, 5440,     0, 2352,    0, 120,   0, 1;
  0, 1,   0, 7381,    0, 24970,    0, 5082,   0, 165, 0, 1;
T(5,3) = 10. The ten partitions of the set [5] into 3 odd-sized blocks are
(1)(2)(345), (1)(3)(245), (1)(4)(235), (1)(5)(234), (2)(3)(145),
(2)(4)(135), (2)(5)(134), (3)(4)(125), (3)(5)(124), (4)(5)(123).
Connection constants: Row 5 = [0,1,0,10,0,1]. Hence, with the polynomial sequence x_(n) as defined in the Comments section we have x^5 = x_(1) + 10*x_(3) + x_(5) = x + 10*x*(x+1)*(x-1) + x*(x+3)*(x+1)*(x-1)*(x-3).
		

References

  • L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225-226.

Crossrefs

Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.

Programs

  • Maple
    A136630 := proc (n, k) option remember; if k < 0 or n < k then 0 elif k = n then 1 else procname(n-2, k-2) + k^2*procname(n-2, k) end if end proc: seq(seq(A136630(n, k), k = 1 .. n), n = 1 .. 12); # Peter Bala, Jul 27 2014
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> (n+1) mod 2, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    t[n_, k_] := Coefficient[ x^k/Product[ 1 - (2*j + k - 2*Quotient[k, 2])^2*x^2, {j, 0, k/2}] + x*O[x]^n, x, n]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2013, after Pari *)
    BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
    rows = 13;
    M = BellMatrix[Mod[#+1, 2]&, rows];
    Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
  • PARI
    {T(n,k)=polcoeff(x^k/prod(j=0,k\2,1-(2*j+k-2*(k\2))^2*x^2 +x*O(x^n)),n)}

Formula

G.f. for column k: x^k/Product_{j=0..floor(k/2)} (1 - (2*j + k-2*floor(k/2))^2 * x^2).
G.f. for column 2*k: x^(2*k)/Product_{j=0..k} (1 - (2*j)^2*x^2).
G.f. for column 2*k+1: x^(2*k+1)/Product_{j=0..k} (1 - (2*j+1)^2*x^2).
From Peter Bala, Feb 21 2011 (Start)
T(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j-k)^n,
Recurrence relation T(n+2,k) = T(n,k-2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = Sum_{n>=0} R(n,x)*z^n/n! = 1 + x*z + x^2*z^2/2! + (x+x^3)*z^3/3! + ....
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x^2
R(3,x) = x+x^3.
The e.g.f. F(x,z) satisfies the partial differential equation d^2/dz^2(F) = x^2*F + x*F' + x^2*F'' where ' denotes differentiation w.r.t. x.
Hence the row polynomials satisfy the recurrence relation R(n+2,x) = x^2*R(n,x) + x*R'(n,x) + x^2*R''(n,x) with R(0,x) = 1.
The recurrence relation for T(n,k) given above follows from this.
(End)
For the corresponding triangle of ordered partitions into odd-sized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of exp(t*M) lists the row polynomials for the present triangle. - Peter Bala, Oct 06 2011
Row generating polynomials equal D^n(exp(x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A196776. - Peter Bala, Dec 06 2011
From Peter Bala, Jul 28 2014: (Start)
E.g.f.: exp(t*sinh(x)) = 1 + t*x + t^2*x^2/2! + (t + t^3)*x^3/3! + ....
Hockey-stick recurrence: T(n+1,k+1) = Sum_{i = 0..floor((n-k)/2)} binomial(n,2*i)*T(n-2*i,k).
Recurrence equation for the row polynomials R(n,t):
R(n+1,t) = t*Sum_{k = 0..floor(n/2)} binomial(n,2*k)*R(n-2*k,t) with R(0,t) = 1. (End)

A196776 Triangle T(n,k) gives the number of ordered partitions of an n set into k odd-sized blocks.

Original entry on oeis.org

1, 0, 2, 1, 0, 6, 0, 8, 0, 24, 1, 0, 60, 0, 120, 0, 32, 0, 480, 0, 720, 1, 0, 546, 0, 4200, 0, 5040, 0, 128, 0, 8064, 0, 40320, 0, 40320, 1, 0, 4920, 0, 115920, 0, 423360, 0, 362880, 0, 512, 0, 130560, 0, 1693440, 0, 4838400, 0, 3628800
Offset: 1

Views

Author

Peter Bala, Oct 06 2011

Keywords

Comments

See A136630 for the case of unordered partitions into odd-sized blocks. See A193474 for this triangle in row reverse form (but with an offset of 0).

Examples

			Triangle begins
.n\k.|..1....2....3....4.....5....6.....7
= = = = = = = = = = = = = = = = = = = = =
..1..|..1
..2..|..0....2
..3..|..1....0....6
..4..|..0....8....0...24
..5..|..1....0...60....0...120
..6..|..0...32....0..480.....0..720
..7..|..1....0..546....0..4200....0..5040
...
T(4,2) = 8: The 8 ordered partitions of the set {1,2,3,4} into 2 odd-sized blocks are {1}{2,3,4}, {2,3,4}{1}, {2}{1,3,4}, {1,3,4}{2}, {3}{1,2,4}, {1,2,4}{3}, {4}{1,2,3} and {1,2,3}{4}.
Example of recurrence relation: T(7,3) = 3^2*T(5,3) + 3*(3-1)*T(5,1) = 9*60 + 6*1 = 546.
		

Crossrefs

Cf. A006154 (row sums), A136630, A162590, A193474 (row reverse).

Formula

T(n,k) = 1/(2^k)*sum {j = 0..k}(-1)^(k-j)*binomial(k,j)*(2*j-k)^n.
Recurrence: T(n+2,k) = k^2*T(n,k) + k*(k-1)*T(n,k-2).
E.g.f.: x*sinh(t)/(1-x*sinh(t)) = x*t + 2*x^2*t^2/2! + (x+6*x^3)*t^3/3! + (8*x^2+24*x^4)*t^4/4! + (x+60*x^3+120*x^5)*t^5/5! + ....
O.g.f. for column 2*k: (2*k)!*x^(2*k)/Product {j = 0..k} (1 - (2*j)^2*x^2).
O.g.f. for column 2*k+1: (2*k+1)!*x^(2*k+1)/Product {j = 0..k} (1 - (2*j+1)^2*x^2).
Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for the present triangle.
n-th row sum = A006154(n).
Row generating polynomials equal D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A136630. - Peter Bala, Dec 06 2011

A162590 Polynomials with e.g.f. exp(x*t)/csch(t), triangle of coefficients read by rows.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 1, 0, 3, 0, 0, 4, 0, 4, 0, 1, 0, 10, 0, 5, 0, 0, 6, 0, 20, 0, 6, 0, 1, 0, 21, 0, 35, 0, 7, 0, 0, 8, 0, 56, 0, 56, 0, 8, 0, 1, 0, 36, 0, 126, 0, 84, 0, 9, 0, 0, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 1, 0, 55, 0, 330, 0, 462, 0, 165, 0, 11, 0, 0, 12, 0, 220, 0, 792, 0, 792, 0
Offset: 0

Views

Author

Peter Luschny, Jul 07 2009

Keywords

Comments

Comment from Peter Bala (Dec 06 2011): "Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for" A196776(n,k), which gives the number of ordered partitions of an n set into k odd-sized blocks. - Peter Luschny, Dec 06 2011
The n-th row of the triangle is formed by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probability of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms in the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k) as the (2k)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k). - Luca Onnis, Oct 29 2023

Examples

			Triangle begins:
  0
  1,  0
  0,  2,  0
  1,  0,  3,  0
  0,  4,  0,  4,  0
  1,  0, 10,  0,  5,  0
  0,  6,  0, 20,  0,  6,  0
  1,  0, 21,  0, 35,  0,  7,  0
  ...
  p[0](x) = 0;
  p[1](x) = 1
  p[2](x) = 2*x
  p[3](x) = 3*x^2 +  1
  p[4](x) = 4*x^3 +  4*x
  p[5](x) = 5*x^4 + 10*x^2 +  1
  p[6](x) = 6*x^5 + 20*x^3 +  6*x
  p[7](x) = 7*x^6 + 35*x^4 + 21*x^2 + 1
  p[8](x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x
.
Cf. the triangle of odd-numbered terms in rows of Pascal's triangle (A034867).
p[n] (k), n=0,1,...
k=0:  0, 1,  0,   1,    0,     1, ... A000035, (A059841)
k=1:  0, 1,  2,   4,    8,    16, ... A131577, (A000079)
k=2:  0, 1,  4,  13,   40,   121, ... A003462
k=3:  0, 1,  6,  28,  120,   496, ... A006516
k=4:  0, 1,  8,  49,  272,  1441, ... A005059
k=5:  0, 1, 10,  76,  520,  3376, ... A081199, (A016149)
k=6:  0, 1, 12, 109,  888,  6841, ... A081200, (A016161)
k=7:  0, 1, 14, 148, 1400, 12496, ... A081201, (A016170)
k=8:  0, 1, 16, 193, 2080, 21121, ... A081202, (A016178)
k=9:  0, 1, 18, 244, 2952, 33616, ... A081203, (A016186)
k=10: 0, 1, 20, 301, 4040, 51001, ... ......., (A016190)
.
p[n] (k), k=0,1,...
p[0]: 0,  0,   0,    0,    0,     0, ... A000004
p[1]: 1,  1,   1,    1,    1,     1, ... A000012
p[2]: 0,  2,   4,    6,    8,    10, ... A005843
p[3]: 1,  4,  13,   28,   49,    76, ... A056107
p[4]: 0,  8,  40,  120,  272,   520, ... A105374
p[5]: 1, 16, 121,  496, 1441,  3376, ...
p[6]: 0, 32, 364, 2016, 7448, 21280, ...
		

Crossrefs

Cf. A119467.

Programs

  • Maple
    # Polynomials: p_n(x)
    p := proc(n,x) local k;
    pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
    add((k mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
    # Coefficients: a(n)
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)/csch(t), t,16),t,i),x,n), n=0..i)), i=0..8);
  • Mathematica
    p[n_, x_] := Sum[Binomial[n, 2*k-1]*x^(n-2*k+1), {k, 0, n+2}]; row[n_] := CoefficientList[p[n, x], x] // Append[#, 0]&; Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 28 2013 *)
    n = 15; "n-th row"
    mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
    mat[[1, 2]] = 1;
    mat[[n + 1, n]] = 1;
    For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
    For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
    mat // MatrixForm;
    P2 = Dot[mat, mat];
    R1 = Simplify[
      Eigenvectors[Transpose[P2]][[1]]/
       Total[Eigenvectors[Transpose[P2]][[1]]]]
    R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
    even = R1*2^(n - 1) (* Luca Onnis, Oct 29 2023 *)

Formula

p_n(x) = Sum_{k=0..n} (k mod 2)*binomial(n,k)*x^(n-k).
E.g.f.: exp(x*t)/csch(t) = 0*(t^0/0!) + 1*(t^1/1!) + (2*x)*(t^2/2!) + (3*x^2+1)*(t^3/3!) + ...
The 'co'-polynomials with generating function exp(x*t)*sech(t) are the Swiss-Knife polynomials (A153641).
Showing 1-3 of 3 results.