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%I #13 May 31 2021 02:11:28
%S 0,0,0,0,1,0,0,2,2,0,0,2,2,2,0,0,4,2,2,4,0,0,4,4,3,4,4,0,0,4,4,4,4,4,
%T 4,0,0,4,4,4,4,4,4,4,0,0,8,4,6,4,4,6,4,8,0,0,8,8,6,4,5,4,6,8,8,0,0,8,
%U 8,8,4,4,4,4,8,8,8,0,0,8,8,8,8,5,6,5,8,8,8,8,0
%N Square array T(n, k), n, k >= 0, read by antidiagonals; T(n, k) = (n * 2^max(0, w(k)-w(n))) AND (k * 2^max(0, w(n)-w(k))) (where AND denotes the bitwise AND operator and w = A070939).
%C In other words, we right pad the binary expansion of the lesser of n and k with zeros (provided it is positive) so that both numbers have the same number of binary digits, and then apply the bitwise AND operator.
%H Rémy Sigrist, <a href="/A344834/b344834.txt">Table of n, a(n) for n = 0..10010</a>
%H Rémy Sigrist, <a href="/A344834/a344834.png">Colored representation of the table for n, k < 2^10</a>
%F T(n, k) = T(k, n).
%F T(m, T(n, k)) = T(T(m, n), k).
%F T(n, n) = n.
%F T(n, 0) = n.
%F T(n, 1) = A053644(n).
%e Array T(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 2 4 4 4 4 8 8 8 8 8 8 8 8
%e 2| 0 2 2 2 4 4 4 4 8 8 8 8 8 8 8 8
%e 3| 0 2 2 3 4 4 6 6 8 8 8 8 12 12 12 12
%e 4| 0 4 4 4 4 4 4 4 8 8 8 8 8 8 8 8
%e 5| 0 4 4 4 4 5 4 5 8 8 10 10 8 8 10 10
%e 6| 0 4 4 6 4 4 6 6 8 8 8 8 12 12 12 12
%e 7| 0 4 4 6 4 5 6 7 8 8 10 10 12 12 14 14
%e 8| 0 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8
%e 9| 0 8 8 8 8 8 8 8 8 9 8 9 8 9 8 9
%e 10| 0 8 8 8 8 10 8 10 8 8 10 10 8 8 10 10
%e 11| 0 8 8 8 8 10 8 10 8 9 10 11 8 9 10 11
%e 12| 0 8 8 12 8 8 12 12 8 8 8 8 12 12 12 12
%e 13| 0 8 8 12 8 8 12 12 8 9 8 9 12 13 12 13
%e 14| 0 8 8 12 8 10 12 14 8 8 10 10 12 12 14 14
%e 15| 0 8 8 12 8 10 12 14 8 9 10 11 12 13 14 15
%o (PARI) T(n,k,op=bitand,w=m->#binary(m)) = { op(n*2^max(0, w(k)-w(n)), k*2^max(0, w(n)-w(k))) }
%Y Cf. A004198, A053644, A070939.
%Y Cf. A344835 (OR), A344836 (XOR), A344837 (min), A344838 (max), A344839 (absolute difference).
%K nonn,base,tabl
%O 0,8
%A _Rémy Sigrist_, May 29 2021