

A136630


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


9



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 A143784 A147311
Adjacent sequences: A136627 A136628 A136629 * A136631 A136632 A136633


KEYWORD

nonn,tabl,changed


AUTHOR

Paul D. Hanna, Jan 14 2008


STATUS

approved



