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Square array A(n, k), n, k >= 0, read by antidiagonals; for any m > 0, the position of the m-th rightmost 0 in the binary expansion of A(n, k) is the least of the positions of the m-th rightmost 0 in the binary expansions of n and k (the least significant bit having position 0).
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%I #13 Jun 28 2022 11:00:27

%S 0,0,0,0,1,0,0,2,2,0,0,1,2,1,0,0,4,2,2,4,0,0,1,4,3,4,1,0,0,2,2,4,4,2,

%T 2,0,0,1,2,5,4,5,2,1,0,0,8,2,6,4,4,6,2,8,0,0,1,8,3,4,5,4,3,8,1,0,0,2,

%U 2,8,4,6,6,4,8,2,2,0,0,1,2,9,8,5,6,5,8,9,2,1,0

%N Square array A(n, k), n, k >= 0, read by antidiagonals; for any m > 0, the position of the m-th rightmost 0 in the binary expansion of A(n, k) is the least of the positions of the m-th rightmost 0 in the binary expansions of n and k (the least significant bit having position 0).

%C Leading 0's are taken into account.

%H Rémy Sigrist, <a href="/A355245/b355245.txt">Table of n, a(n) for n = 0..10010</a>

%F A(n, k) = A(k, n).

%F A(m, A(n, k)) = A(A(m, n), k).

%F A(n, n) = n.

%F A(n, 0) = 0.

%F A(n, 1) = A006519(n) for any n > 0.

%F A(n, k) < 2^m for any n < 2^m and k < 2^m.

%F A(m, A355246(n, k)) = A355246(A(m, n), A(m, k)).

%F A355246(m, A(n, k)) = A(A355246(m, n), A355246(m, k)).

%e Array A(n, k) begins:

%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

%e ---+-----------------------------------------------------------

%e 0| 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

%e 1| 0 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1

%e 2| 0 2 2 2 4 2 2 2 8 2 2 2 4 2 2 2

%e 3| 0 1 2 3 4 5 6 3 8 9 10 3 12 5 6 3

%e 4| 0 4 4 4 4 4 4 4 8 4 4 4 4 4 4 4

%e 5| 0 1 2 5 4 5 6 5 8 9 10 5 12 5 6 5

%e 6| 0 2 2 6 4 6 6 6 8 10 10 6 12 6 6 6

%e 7| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 7

%e 8| 0 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

%e 9| 0 1 2 9 4 9 10 9 8 9 10 9 12 9 10 9

%e 10| 0 2 2 10 4 10 10 10 8 10 10 10 12 10 10 10

%e .

%e For n = 876 and k = 425:

%e - the corresponding binary expansions and pairings of 0's are as follows (stars indicate least positions of 0's):

%e * * *

%e 876 ... 0 0 0 1 1 0 1 1 0 1 1 0 0

%e \ \ \ \ | / /

%e 425 ... 0 0 0 0 1 1 0 1 0 1 0 0 1

%e * * * * * *

%e -----------------------------

%e 428 ... 0 0 0 0 1 1 0 1 0 1 1 0 0

%e - so A(876, 425) = 428.

%o (PARI) A(n,k) = { my (v=0, zn=0, zk=0, w=1, b=1); while (n || k, if (n%2==0, zn++); if (k%2==0, zk++); if (max(zn, zk)==w, w++, v+=b); n\=2; k\=2; b*=2); v }

%Y See A355246 for a similar sequence.

%Y Cf. A006519.

%K nonn,base,tabl

%O 0,8

%A _Rémy Sigrist_, Jun 25 2022