A106340 Triangle T, read by rows, equal to the matrix inverse of the triangle defined by [T^-1](n,k) = (n-k)!*A008278(n+1,k+1), for n>=k>=0, where A008278 is a triangle of Stirling numbers of 2nd kind.
1, -1, 1, 1, -3, 1, -1, 9, -7, 1, 1, -45, 55, -15, 1, -1, 585, -835, 285, -31, 1, 1, -21105, 30835, -11025, 1351, -63, 1, -1, 1858185, -2719675, 977445, -121891, 6069, -127, 1, 1, -367958745, 538607755, -193649085, 24187051, -1213065, 26335, -255, 1, -1, 157169540745, -230061795355, 82717588485
Offset: 0
Examples
Triangle T begins: 1; -1,1; 1,-3,1; -1,9,-7,1; 1,-45,55,-15,1; -1,585,-835,285,-31,1; 1,-21105,30835,-11025,1351,-63,1; -1,1858185,-2719675,977445,-121891,6069,-127,1; 1,-367958745,538607755,-193649085,24187051,-1213065,26335,-255,1; ... Matrix inverse begins: 1; 1,1; 2,3,1; 6,12,7,1; 24,60,50,15,1; 120,360,390,180,31,1; ... where [T^-1](n,k) = (n-k)!*A008278(n+1,k+1).
Programs
-
Mathematica
rows = 10; M = Table[If[r >= c, (r-c)! Sum[(-1)^(r-c-m+1) m^r/m!/(r-c-m+1)!, {m, 0, r-c+1}], 0], {r, rows}, {c, rows}] // Inverse; T[n_, k_] := M[[n+1, k+1]]; Table[T[n, k], {n, 0, rows-1}, {k, 0, n}] (* Jean-François Alcover, Jun 27 2019, from PARI *)
-
PARI
{T(n,k)=(matrix(n+1,n+1,r,c,if(r>=c,(r-c)!* sum(m=0,r-c+1,(-1)^(r-c+1-m)*m^r/m!/(r-c+1-m)!)))^-1)[n+1,k+1]}
-
Sage
def A106340_matrix(d): def A130850(n, k): # EulerianNumber = A173018 return add(EulerianNumber(n,j)*binomial(n-j,k) for j in (0..n)) return matrix(ZZ, d, A130850).inverse() A106340_matrix(8) # Peter Luschny, May 21 2013
Formula
T(n, k) = A106338(n, k)/k!, for n>=k>=0.
Comments