

A136630


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


11



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
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OFFSET

0,13


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)


REFERENCES

L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 6162.
L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225226.


LINKS

Table of n, a(n) for n=0..83.
Ch. A. Charalambides, Central factorial numbers and related expansions, Fib. Quarterly, Vol. 19, No 5, Dec 1981, pp. 451456.


FORMULA

G.f. for column k: x^k/Product_{j=0..[k/2]} (1  (2*j + k2*[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)^(kj)*binomial(k,j)*(2*jk)^n,
Recurrence relation T(n+2,k) = T(n,k2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = sum {n = 0..inf} 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 oddsized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(PP^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! + ....
Hockeystick recurrence: T(n+1,k+1) = sum {i = 0..floor((nk)/2)} binomial(n,2*i)*T(n2*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(n2*k,t) with R(0,t) = 1. (End)


EXAMPLE

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 oddsized 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)*(x1) + x*(x+3)*(x+1)*(x1)*(x3).


MAPLE

A136630 := proc (n, k) option remember; if k < 0 or n < k then 0 elif k = n then 1 else procname(n2, k2) + k^2*procname(n2, k) end if end proc: seq(seq(A136630(n, k), k = 1 .. n), n = 1 .. 12);
# Peter Bala, Jul 27 2014


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 (* JeanFrançois Alcover, Nov 22 2013, after Pari *)


PROG

(PARI) {T(n, k)=polcoeff(x^k/prod(j=0, k\2, 1(2*j+k2*(k\2))^2*x^2 +x*O(x^n)), n)}


CROSSREFS

Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.
Sequence in context: A036859 A036861 A120324 * A111728 A250219 A143784
Adjacent sequences: A136627 A136628 A136629 * A136631 A136632 A136633


KEYWORD

nonn,tabl


AUTHOR

Paul D. Hanna, Jan 14 2008


STATUS

approved



