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T(n, k) = Eulerian1(n - k, k), for n >= 0 and 0 <= k <= floor(n/2). Triangle read by rows.
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%I #9 May 18 2021 06:22:37

%S 1,1,1,0,1,1,1,4,0,1,11,1,1,26,11,0,1,57,66,1,1,120,302,26,0,1,247,

%T 1191,302,1,1,502,4293,2416,57,0,1,1013,14608,15619,1191,1,1,2036,

%U 47840,88234,15619,120,0,1,4083,152637,455192,156190,4293,1

%N T(n, k) = Eulerian1(n - k, k), for n >= 0 and 0 <= k <= floor(n/2). Triangle read by rows.

%C The antidiagonal representation of the first order Eulerian numbers (A173018).

%e Triangle starts:

%e [ 0] [1]

%e [ 1] [1]

%e [ 2] [1, 0]

%e [ 3] [1, 1]

%e [ 4] [1, 4, 0]

%e [ 5] [1, 11, 1]

%e [ 6] [1, 26, 11, 0]

%e [ 7] [1, 57, 66, 1]

%e [ 8] [1, 120, 302, 26, 0]

%e [ 9] [1, 247, 1191, 302, 1]

%e [10] [1, 502, 4293, 2416, 57, 0]

%e [11] [1, 1013, 14608, 15619, 1191, 1]

%p T := (n, k) -> combinat:-eulerian1(n - k, k):

%p seq(print(seq(T(n, k), k=0..n/2)), n = 0..11);

%Y Cf. A000800 (row sums).

%Y Cf. A173018.

%K nonn,tabf

%O 0,8

%A _Peter Luschny_, May 17 2021