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%I #33 Apr 30 2020 13:36:05
%S 0,1,0,2,1,0,3,3,1,0,4,6,3,1,0,5,2,2,3,1,0,6,12,7,2,3,1,0,7,4,6,6,2,3,
%T 1,0,8,7,13,5,6,2,3,1,0,9,5,12,7,7,6,2,3,1,0,10,24,5,15,4,7,6,2,3,1,0,
%U 11,8,4,13,5,5,7,6,2,3,1,0,12,11,25,4,14,12,5,7,6,2,3,1,0,13,9,24,12,15,4,4,5,7,6,2,3,1,0,14,13,9,27,12,10,13,4,5,7,6,2,3,1,0
%N Square array A(r,c): A(0,c) = c, A(r,0) = 0, A(r>=1,c>=1) = A003188(1+A006068(A(r-1,c-1))) = A268717(1+A(r-1,c-1)), read by descending antidiagonals as A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ...
%H Antti Karttunen, <a href="/A268820/b268820.txt">Table of n, a(n) for n = 0..32895; the first 256 antidiagonals of array</a>
%F For row zero: A(0,k) = k, for column zero: A(n,0) = 0, and in other cases: A(n,k) = A003188(1+A006068(A(n-1,k-1)))
%F Other identities. For all n >= 0:
%F A(n,n) = A003188(n).
%F A(A006068(n),A006068(n)) = n.
%e The top left [0 .. 16] x [0 .. 19] section of the array:
%e 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
%e 0, 1, 3, 6, 2, 12, 4, 7, 5, 24, 8, 11, 9, 13, 15, 10, 14, 48, 16, 19
%e 0, 1, 3, 2, 7, 6, 13, 12, 5, 4, 25, 24, 9, 8, 15, 14, 11, 10, 49, 48
%e 0, 1, 3, 2, 6, 5, 7, 15, 13, 4, 12, 27, 25, 8, 24, 14, 10, 9, 11, 51
%e 0, 1, 3, 2, 6, 7, 4, 5, 14, 15, 12, 13, 26, 27, 24, 25, 10, 11, 8, 9
%e 0, 1, 3, 2, 6, 7, 5, 12, 4, 10, 14, 13, 15, 30, 26, 25, 27, 11, 9, 24
%e 0, 1, 3, 2, 6, 7, 5, 4, 13, 12, 11, 10, 15, 14, 31, 30, 27, 26, 9, 8
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 15, 13, 9, 11, 14, 10, 29, 31, 26, 30, 8
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 14, 15, 8, 9, 10, 11, 28, 29, 30, 31
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 10, 14, 24, 8, 11, 9, 20, 28, 31
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 11, 10, 25, 24, 9, 8, 21, 20
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 9, 11, 27, 25, 8, 24, 23
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 8, 9, 26, 27, 24, 25
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 24, 8, 30, 26, 25
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 25, 24, 31, 30
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 27, 25, 29
%e 0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 25, 26, 27
%t A003188[n_]:=BitXor[n, Floor[n/2]]; A006068[n_]:=If[n<2, n, Block[{m=A006068[Floor[n/2]]}, 2m + Mod[Mod[n,2] + Mod[m, 2], 2]]]; a[r_, 0]:= 0; a[0, c_]:=c; a[r_, c_]:= A003188[1 + A006068[a[r - 1, c - 1]]]; Table[a[c, r - c], {r, 0, 15}, {c, 0, r}] //Flatten (* _Indranil Ghosh_, Apr 02 2017 *)
%o (Scheme)
%o (define (A268820 n) (A268820bi (A002262 n) (A025581 n)))
%o (define (A268820bi row col) (cond ((zero? row) col) ((zero? col) 0) (else (A268717 (+ 1 (A268820bi (- row 1) (- col 1)))))))
%o (define (A268820bi row col) (cond ((zero? row) col) ((zero? col) 0) (else (A003188 (+ 1 (A006068 (A268820bi (- row 1) (- col 1))))))))
%o (PARI) A003188(n) = bitxor(n, n\2);
%o A006068(n) = if(n<2, n, {my(m = A006068(n\2)); 2*m + (n%2 + m%2)%2});
%o a(r, c) = if(r==0, c, if(c==0, 0, A003188(1 + A006068(a(r - 1, c - 1)))));
%o for(r=0, 15, for(c=0, r, print1(a(c, r - c),", "); ); print(); ); \\ _Indranil Ghosh_, Apr 02 2017
%o (Python)
%o def A003188(n): return n^(n//2)
%o def A006068(n):
%o if n<2: return n
%o else:
%o m=A006068(n//2)
%o return 2*m + (n%2 + m%2)%2
%o def a(r, c): return c if r<1 else 0 if c<1 else A003188(1 + A006068(a(r - 1, c - 1)))
%o for r in range(16):
%o print([a(c, r - c) for c in range(r + 1)]) # _Indranil Ghosh_, Apr 02 2017
%Y Cf. A003188, A006068.
%Y Inverses of these permutations can be found in table A268830.
%Y Row 0: A001477, Row 1: A268717, Row 2: A268821, Row 3: A268823, Row 4: A268825, Row 5: A268827, Row 6: A268831, Row 7: A268933.
%Y Rows converge towards A003188, which is also the main diagonal.
%Y Cf. array A268715 (can be extracted from this one).
%Y Cf. array A268833 (shows related Hamming distances with regular patterns).
%K nonn,tabl
%O 0,4
%A _Antti Karttunen_, Feb 14 2016
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