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%I #24 Mar 22 2021 03:42:38
%S 0,1,1,2,3,2,3,6,6,3,4,2,5,2,4,5,12,7,7,12,5,6,4,15,6,15,4,6,7,7,13,
%T 13,13,13,7,7,8,5,4,12,9,12,4,5,8,9,24,12,5,11,11,5,12,24,9,10,8,27,4,
%U 14,10,14,4,27,8,10,11,11,25,25,10,15,15,10,25,25,11,11,12,9,8,24,29,14,12,14,29,24,8,9,12,13,13,24,9,31,31,13,13,31,31,9,24,13,13
%N Square array A(i,j) = A003188(A006068(i) + A006068(j)), read by antidiagonals as A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ...
%C Each row n is row A006068(n) of array A268820 without its A006068(n) initial terms.
%H Antti Karttunen, <a href="/A268715/b268715.txt">Table of n, a(n) for n = 0..15050; the first 173 antidiagonals of the array</a>
%F A(i,j) = A003188(A006068(i) + A006068(j)) = A003188(A268714(i,j)).
%F A(row,col) = A268820(A006068(row), (A006068(row)+col)).
%e The top left [0 .. 15] x [0 .. 15] section of the array:
%e 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
%e 1, 3, 6, 2, 12, 4, 7, 5, 24, 8, 11, 9, 13, 15, 10, 14
%e 2, 6, 5, 7, 15, 13, 4, 12, 27, 25, 8, 24, 14, 10, 9, 11
%e 3, 2, 7, 6, 13, 12, 5, 4, 25, 24, 9, 8, 15, 14, 11, 10
%e 4, 12, 15, 13, 9, 11, 14, 10, 29, 31, 26, 30, 8, 24, 27, 25
%e 5, 4, 13, 12, 11, 10, 15, 14, 31, 30, 27, 26, 9, 8, 25, 24
%e 6, 7, 4, 5, 14, 15, 12, 13, 26, 27, 24, 25, 10, 11, 8, 9
%e 7, 5, 12, 4, 10, 14, 13, 15, 30, 26, 25, 27, 11, 9, 24, 8
%e 8, 24, 27, 25, 29, 31, 26, 30, 17, 19, 22, 18, 28, 20, 23, 21
%e 9, 8, 25, 24, 31, 30, 27, 26, 19, 18, 23, 22, 29, 28, 21, 20
%e 10, 11, 8, 9, 26, 27, 24, 25, 22, 23, 20, 21, 30, 31, 28, 29
%e 11, 9, 24, 8, 30, 26, 25, 27, 18, 22, 21, 23, 31, 29, 20, 28
%e 12, 13, 14, 15, 8, 9, 10, 11, 28, 29, 30, 31, 24, 25, 26, 27
%e 13, 15, 10, 14, 24, 8, 11, 9, 20, 28, 31, 29, 25, 27, 30, 26
%e 14, 10, 9, 11, 27, 25, 8, 24, 23, 21, 28, 20, 26, 30, 29, 31
%e 15, 14, 11, 10, 25, 24, 9, 8, 21, 20, 29, 28, 27, 26, 31, 30
%t A003188[n_] := BitXor[n, Floor[n/2]]; A006068[n_] := BitXor @@ Table[Floor[ n/2^m], {m, 0, Log[2, n]}]; A006068[0]=0; A[i_, j_] := A003188[A006068[i] + A006068[j]]; Table[A[i-j, j], {i, 0, 13}, {j, 0, i}] // Flatten (* _Jean-François Alcover_, Feb 17 2016 *)
%o (Scheme)
%o (define (A268715 n) (A268715bi (A002262 n) (A025581 n)))
%o (define (A268715bi row col) (A003188 (+ (A006068 row) (A006068 col))))
%o ;; Alternatively, extracting data from array A268820:
%o (define (A268715bi row col) (A268820bi (A006068 row) (+ (A006068 row) col)))
%o (Python)
%o def a003188(n): return n^(n>>1)
%o def a006068(n):
%o s=1
%o while True:
%o ns=n>>s
%o if ns==0: break
%o n=n^ns
%o s<<=1
%o return n
%o def T(n, k): return a003188(a006068(n) + a006068(k))
%o for n in range(21): print([T(n - k, k) for k in range(n + 1)]) # _Indranil Ghosh_, Jun 07 2017
%Y Cf. A003188, A006068, A268714, A268820.
%Y Main diagonal: A001969.
%Y Row 0, column 0: A001477.
%Y Row 1, column 1: A268717.
%Y Antidiagonal sums: A268837.
%Y Cf. A268719 (the lower triangular section).
%Y Cf. also A268725.
%K nonn,tabl
%O 0,4
%A _Antti Karttunen_, Feb 12 2016
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