A211229 Matrix inverse of lower triangular array A211226.

1, -1, 1, 0, -1, 1, 0, 0, -1, 1, 1, 0, 0, -2, 1, -1, 1, 0, 0, -1, 1, 2, -3, 3, 0, 0, -3, 1, -2, 2, -3, 3, 0, 0, -1, 1, 9, -8, 8, -12, 6, 0, 0, -4, 1, -9, 9, -8, 8, -6, 6, 0, 0, -1, 1, 44, -45, 45, -40, 20, -30, 10, 0, 0, -5, 1, -44, 44, -45, 45, -20, 20, -10, 10, 0, 0, -1, 1

Offset: 0

Author

Peter Bala, Apr 05 2012

Keywords

• sign
• easy
• tabl

Comments

This triangle is related to the derangement numbers. The subtriangles (T(2*n,2*k))n,k>=0, -(T(2*n+1,2*k))n,k>=0, and (T(2*n+1,2*k+1))n,k>=0 are all equal to A008290, while the subtriangle (T(2*n,2*k+1))n,k>=0 equals -A180188 (with an extra initial row of zeros).

Examples

			Triangle begins:
   n\k |    0    1    2    3    4    5    6    7    8    9
  =====+==================================================
    0  |    1
    1  |   -1    1
    2  |    0   -1    1
    3  |    0    0   -1    1
    4  |    1    0    0   -2    1
    5  |   -1    1    0    0   -1    1
    6  |    2   -3    3    0    0   -3    1
    7  |   -2    2   -3    3    0    0   -1    1
    8  |    9   -8    8  -12    6    0    0   -4    1
    9  |   -9    9   -8    8   -6    6    0    0   -1    1
  ...
		

Crossrefs

Cf. A000166, A000240, A000387, A000449, A000475, A008290, A180188, A211226.

Programs

Mathematica

b[j_] = Floor[j/2]; h = If[EvenQ[n] && OddQ[k], 1, 0];
Table[(-1)^(n+k) (b[n]!/b[k]!) Sum[(-1)^i/i!, {i, 0, b[n-k]-h}], {n, 0, 31}, {k, 0, n}] //Flatten (* Manfred Boergens, Jan 10 2023 *)
(* Sum-free code *)
b[j_] = Floor[j/2]; h = If[EvenQ[n] && OddQ[k], 1, 0];
T[n_, k_] = (-1)^(n+k) (b[n]!/b[k]!) If[n-k<2, 1, Round[(b[n-k]-h)!/E]/(b[n-k]-h)!];
Table[T[n, k], {n, 0, 31}, {k, 0, n}] // Flatten
(* Manfred Boergens, Jan 10 2023 *)

PARI

f(n) = (n\2)!; \\ A081123
T(n,k) = f(n)/(f(k)*f(n-k)); \\ A211226
tabl(nn) = my(m=matrix(nn, nn, n, k, if (n>=k, T(n-1,k-1), 0))); 1/m; \\ Michel Marcus, Jan 10 2023

Formula

T(2*n,2*k) = T(2*n+1,2*k+1) = -T(2*n+1,2*k) = binomial(n,k)*A000166(n-k) = (n!/k!)*Sum_{i = 0..n-k} (-1)^i/i!;

T(2*n,2*k+1) = -n*binomial(n-1,k)*A000166(n-k-1) = -(n!/k!)*Sum_{i = 0..n-k-1} (-1)^i/i!.

T(n,k) = T(n-k,0)*A211226(n,k).

Column entries:

T(2*n,0) = A000166(n), T(2*n,2) = A000240(n), T(2*n,4) = A000387(n), T(2*n,6) = A000449(n), T(2*n,8) = A000475(n).

From Manfred Boergens, Jan 10 2023: (Start)

With b(j) = floor(j/2); h = 1 for n even and k odd, h = 0 else:

T(n,k) = (-1)^(n+k)*(b(n)!/b(k)!)*Sum_{i = 0..b(n-k)-h} (-1)^i/i!.

Sum-free formula:

T(n,k) = (-1)^(n+k)*(b(n)!/b(k)!) for n-k < 2.

T(n,k) = (-1)^(n+k)*(b(n)!/b(k)!)*round((b(n-k)-h)!/exp(1))/(b(n-k)-h)!) otherwise. (End)

Extensions

More terms from Manfred Boergens, Jan 10 2023