A263985 Triangle of signed Eulerian numbers on involutions, read by rows.

1, -1, 1, -1, -2, 1, 1, -2, -2, 1, 1, 6, 0, -2, 1, -1, 3, 14, 2, -3, 1, -1, -12, -15, 12, -1, -4, 1, 1, -4, -51, -76, 4, -3, -4, 1, 1, 20, 67, -10, -80, 30, 3, -4, 1, -1, 5, 137, 517, 414, 66, 75, 7, -5, 1, -1, -30, -192, -140, 721, 588, -49, 44, 0, -6, 1

Offset: 1

Author

Michel Marcus, Oct 31 2015

Keywords

• sign
• tabl

Examples

			Triangle begins:
1;
-1, 1;
-1, -2, 1;
1, -2, -2, 1;
1, 6, 0, -2, 1;
-1, 3, 14, 2, -3, 1;
-1, -12, -15, 12, -1, -4, 1;
...
		

Links

• M. Barnabei, F. Bonetti, M. Silimbani, The signed Eulerian numbers on involutions, PU. M. A. Vol. 19 (2008) pp. 117-126.
• M. Barnabei, F. Bonetti, M. Silimbani, The signed Eulerian numbers on involutions, arXiv:0803.2126 [math.CO], 2008.
• J. Desarmenien and D. Foata, The signed Eulerian numbers, Discrete Math. 99 (1992), no. 1-3, 49-58.

Crossrefs

Cf. A049061.

Programs

Mathematica

T[n_, k_] := Sum[(-1)^(k-m+1) Binomial[n+1, k-m+1] Sum[(-1)^j Binomial[ Binomial[m+1, 2]+j-1, j] Binomial[m, n-2j], {j, 0, n/2}], {m, 0, k+1}];
Table[T[n, k], {n, 1, 11}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Sep 26 2018 *)

PARI

T(n, k) = sum(m=0, k+1, (-1)^(k-m+1)*binomial(n+1,k-m+1)*sum(j=0,n\2, (-1)^j*binomial(binomial(m+1,2)+j-1,j)*binomial(m,n-2*j)));

Formula

T(n, k) = Sum_{m=0..k+1} (-1)^(k-m+1)*C(n+1,k-m+1)*Sum_{j=0..floor(n/2)} (-1)^j*C(C(m+1,2)+j-1,j)*C(m,n-2*j);