%I #11 Sep 22 2019 07:17:31

%S 1,1,1,2,1,2,1,1,1,1,4,1,2,1,4,3,3,1,1,3,3,2,3,2,1,2,3,2,1,1,1,1,1,1,

%T 1,1,8,1,2,1,4,1,2,1,8,7,7,1,1,3,3,1,1,7,7,6,7,6,1,2,3,2,1,6,7,6,5,5,

%U 5,5,1,1,1,1,5,5,5,5

%N T(n, k) = 1 + NOR(k - 1, n - k), where NOR is the Peirce arrow operating bitwise on the inputs, triangle read by rows, T(n, k) for n >= 1, 1 <= k <= n.

%e 1

%e 1, 1

%e 2, 1, 2

%e 1, 1, 1, 1

%e 4, 1, 2, 1, 4

%e 3, 3, 1, 1, 3, 3

%e 2, 3, 2, 1, 2, 3, 2

%e 1, 1, 1, 1, 1, 1, 1, 1

%e 8, 1, 2, 1, 4, 1, 2, 1, 8

%e 7, 7, 1, 1, 3, 3, 1, 1, 7, 7

%e 6, 7, 6, 1, 2, 3, 2, 1, 6, 7, 6

%e 5, 5, 5, 5, 1, 1, 1, 1, 5, 5, 5, 5

%e 4, 5, 6, 5, 4, 1, 2, 1, 4, 5, 6, 5, 4

%e 3, 3, 5, 5, 3, 3, 1, 1, 3, 3, 5, 5, 3, 3

%e 2, 3, 2, 5, 2, 3, 2, 1, 2, 3, 2, 5, 2, 3, 2

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

%p A327489 := (n, k) -> 1 + Bits:-Nor(k-1, n-k):

%p seq(seq(A327489(n, k), k=1..n), n=1..12);

%Y Cf. A327488 (Nand), A327490 (Iff), A280172 (Xor).

%Y T(2n+1,n+1) gives A080079.

%K nonn,tabl

%O 1,4

%A _Peter Luschny_, Sep 22 2019