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%I #43 Aug 30 2020 03:38:01
%S 2,4,1,1,2,3,4,4,6,4,8,2,5,1,6,1,2,6,2,5,5,8,8,10,1,3,3,7,1,2,9,4,5,6,
%T 7,8,8,2,10,4,12,3,7,8,10,1,2,3,4,5,6,7,8,12,9,8,8,10,8,12,12,14,8,9,
%U 10,11,16,4,9,4,9,6,13,1,12,12,14,12
%N Square array read by descending antidiagonals T(n,k): In the binary expansion of n, reverse the order of the bits in the same position as the active bits in A057716(k).
%C T(n,k) is the swapping of the positions of the bits in n according to the active bits in K, where K = A057716(k). The bit in the same position as the first active bit in K switches positions with the bit in the same position as the last active bit in K, the bit in the same position as the second active bit in K switches with the one in the same as the second to last position, and so on until all have swapped (without repeating).
%C Any sequence, f, of the form "reverse the order of the a-th, b-th, ... and z-th bits in n" can be expressed as f(n) = T(n,k), where A057716(k) = 2^a + 2^b + ... 2^z. As a result, this operation combines 1 or more bit-swapping operations, which could be useful for bit-manipulation in computer programming.
%H Davis Smith, <a href="/A336434/b336434.txt">Table of n, a(n) for n = 1..5050; the first 100 antidiagonals of the array</a>
%H S. E. Anderson, <a href="https://graphics.stanford.edu/~seander/bithacks.html#SwappingValuesXOR">Bit Twiddling Hacks</a>.
%F T(n,k) = A003987(n, Sum_{m=1..floor(M/2)} A003987(A030308(n,A133457(K,m)), A030308(n,A133457(K,M - (m - 1))))* (2^A133457(K,m) + 2^A133457(K,M - (m - 1)))), where K = A057716(k) and M = A000120(A057716(k)).
%F When A057716(k) = 2^A070939(n) - 1, T(n,k) = A030101(n).
%F When A057716(k) = 2^(A070939(n) - 1) - 1, T(n,k) = A059893(n).
%e The binary expansion of 18 is 10010_2 and the active bits in the binary expansion of A057716(22) = 27 = 11011_2 are 0, 1, 3, and 4. So, to get T(18,22), we swap the 0th and 4th bits and then the 1st and 3rd bits, which gives us T(18,22) = 9.
%e Square array T(n,k) begins:
%e \k 1 2 3 4 5 6 7 8 9 10 ...
%e n\
%e 1| 2 4 1 4 8 1 8 1 8 1 ...
%e 2| 1 2 4 2 2 8 2 2 2 8 ...
%e 3| 3 6 5 6 10 9 10 3 10 9 ...
%e 4| 4 1 2 1 4 4 4 8 4 4 ...
%e 5| 6 5 3 5 12 5 12 9 12 5 ...
%e 6| 5 3 6 3 6 12 6 10 6 12 ...
%e 7| 7 7 7 7 14 13 14 11 14 13 ...
%e 8| 8 8 8 8 1 2 1 4 1 2 ...
%e 9| 10 12 9 12 9 3 9 5 9 3 ...
%e 10| 9 10 12 10 3 10 3 6 3 10 ...
%o (PARI) A336434(n,k)={my(K=k+#binary(k+#binary(k)), P=select(Z->bittest(K,Z),[0..#binary(K)-1]), Q1=P[1..floor(#P/2)],Q2=Vecrev(P)[1..floor(#P/2)], Sum=vecsum(apply(p->if(bittest(n,Q1[p])!=bittest(n,Q2[p]), bitor(shift(1,Q1[p]),shift(1,Q2[p]))), [1..floor(#P/2)])));bitxor(n,Sum)}
%Y Cf. A000120, A003987, A030101, A030308, A057716, A059893, A070939, A080412 (column k=1), A133457.
%K nonn,tabl,base
%O 1,1
%A _Davis Smith_, Jul 21 2020
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