%I #8 Sep 22 2019 07:16:42
%S 1,1,1,2,2,2,1,1,1,1,4,2,4,2,4,3,3,3,3,3,3,2,4,2,4,2,4,2,1,1,1,1,1,1,
%T 1,1,8,2,4,2,8,2,4,2,8,7,7,3,3,7,7,3,3,7,7,6,8,6,4,6,8,6,4,6,8,6,5,5,
%U 5,5,5,5,5,5,5,5,5,5
%N T(n, k) = 1 + IFF(k - 1, n - k), where IFF is Boolean equality evaluated bitwise on the inputs, triangle read by rows, T(n, k) for n >= 1, 1 <= k <= n.
%C If row(n) has, seen as a set, only one element k then k is either 1 or 1 + 2^n and n has the form 2^n or 3*2^n.
%e 1
%e 1, 1
%e 2, 2, 2
%e 1, 1, 1, 1
%e 4, 2, 4, 2, 4
%e 3, 3, 3, 3, 3, 3
%e 2, 4, 2, 4, 2, 4, 2
%e 1, 1, 1, 1, 1, 1, 1, 1
%e 8, 2, 4, 2, 8, 2, 4, 2, 8
%e 7, 7, 3, 3, 7, 7, 3, 3, 7, 7
%e 6, 8, 6, 4, 6, 8, 6, 4, 6, 8, 6
%e 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
%p A327490 := (n, k) -> 1 + Bits:-Iff(k-1, n-k):
%p seq(seq(A327490(n, k), k=1..n), n=1..12);
%Y Cf. A327488 (Nand), A327489 (Nor), A280172 (Xor).
%K nonn,tabl
%O 1,4
%A _Peter Luschny_, Sep 22 2019