A327490 - OEIS

%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