%I #23 Mar 19 2023 05:02:19

%S 0,0,0,0,1,0,0,4,4,0,0,11,30,11,0,0,26,146,146,26,0,0,57,588,1198,588,

%T 57,0,0,120,2136,7792,7792,2136,120,0,0,247,7290,44089,78060,44089,

%U 7290,247,0,0,502,23902,227554,655114,655114,227554,23902,502,0

%N Triangle read by rows: T(n,k) = (Eulerian(n+1,k)-binomial(n,k))/2, for 0 <= k <= n.

%C The entries in the triangle of Eulerian numbers (A173018) and Pascal's triangle of binomial coefficients (A007318) have the same parity (see A047999) provided the final diagonal of zeros in the Eulerian triangle is removed and all its rows are moved up a notch. Compare

%C 1

%C 1 0

%C 1 1 0

%C 1 4 1 0 (A173018)

%C 1 11 11 1 0

%C 1 26 66 26 1 0

%C ...

%C and

%C 1

%C 1 1

%C 1 2 1 (A007318)

%C 1 3 3 1

%C 1 4 6 4 1

%C ...

%C The present triangle is the difference, halved.

%e The triangle begins:

%e [0]

%e [0, 0]

%e [0, 1, 0]

%e [0, 4, 4, 0]

%e [0, 11, 30, 11, 0]

%e [0, 26, 146, 146, 26, 0]

%e [0, 57, 588, 1198, 588, 57, 0]

%e [0, 120, 2136, 7792, 7792, 2136, 120, 0]

%e [0, 247, 7290, 44089, 78060, 44089, 7290, 247, 0]

%e ...

%p E:= proc(n, k) option remember;

%p if k=0 and n>=0 then 1

%p elif k<0 or k>n then 0

%p else (n-k) * E(n-1, k-1) + (k+1) * E(n-1, k)

%p fi

%p end:

%p T:=(n,k)->(E(n+1,k)-binomial(n,k))/2;

%p for n from 0 to 12 do lprint([seq(T(n,k),k=0..n)]); od:

%t Eu[n_, k_] := Eu[n, k] = Which[k == 0 && n >= 0, 1, k < 0 || k > n, 0, True, (n-k)*Eu[n-1, k-1] + (k+1)*Eu[n-1, k]];

%t T[n_, k_] := (Eu[n+1, k] - Binomial[n, k])/2;

%t Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Mar 19 2023, after Maple code *)

%Y Cf. A007318, A008292, A047999, A173018.

%K nonn,tabl

%O 0,8

%A _N. J. A. Sloane_, Aug 05 2017