OFFSET
1,3
COMMENTS
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
EXAMPLE
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
KEYWORD
AUTHOR
Peter Bala, Oct 06 2011
STATUS
approved