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

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

Author

Paul D. Hanna, Jan 14 2008

Keywords

• nonn
• tabl

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.

Links

• Ch. A. Charalambides, Central factorial numbers and related expansions, Fib. Quarterly, Vol. 19, No 5, Dec 1981, pp. 451-456.
• Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Appl. Anal. Disc. Math. (2024) Vol. 18, No. 1, 1-24. See p. 13.

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)