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%I #21 Apr 28 2021 01:40:50
%S 0,2,2,5,3,5,9,9,9,9,14,12,10,12,14,20,20,16,16,20,20,27,25,27,21,27,
%T 25,27,35,35,35,35,35,35,35,35,44,42,40,42,36,42,40,42,44,54,54,50,50,
%U 46,46,50,50,54,54,65,63,65,59,57,55,57,59,65,63,65,77,77,77,77,69,69,69,69,77,77,77,77,90,88,86,88,90,80,78,80,90,88,86,88,90
%N Square array read by antidiagonals: A(n,k) = T(n XOR k, 2*(n AND k)), where T(n,k) is sequence A001477 considered as a two-dimensional table, AND is bitwise-and (A004198) and XOR is bitwise-xor (A003987).
%C The array is 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="/A286109/b286109.txt">Table of n, a(n) for n = 0..10584; the first 145 antidiagonals of array</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PairingFunction.html">Pairing Function</a>
%F A(n,k) = T(A003987(n,k), 2*A004198(n,k)), where T(n,k) is sequence A001477 considered as a two-dimensional table, that is, as a pairing function from [0, 1, 2, 3, ...] x [0, 1, 2, 3, ...] to [0, 1, 2, 3, ...].
%e The top left 0 .. 12 x 0 .. 12 corner of the array:
%e 0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90
%e 2, 3, 9, 12, 20, 25, 35, 42, 54, 63, 77, 88, 104
%e 5, 9, 10, 16, 27, 35, 40, 50, 65, 77, 86, 100, 119
%e 9, 12, 16, 21, 35, 42, 50, 59, 77, 88, 100, 113, 135
%e 14, 20, 27, 35, 36, 46, 57, 69, 90, 104, 119, 135, 144
%e 20, 25, 35, 42, 46, 55, 69, 80, 104, 117, 135, 150, 162
%e 27, 35, 40, 50, 57, 69, 78, 92, 119, 135, 148, 166, 181
%e 35, 42, 50, 59, 69, 80, 92, 105, 135, 150, 166, 183, 201
%e 44, 54, 65, 77, 90, 104, 119, 135, 136, 154, 173, 193, 214
%e 54, 63, 77, 88, 104, 117, 135, 150, 154, 171, 193, 212, 236
%e 65, 77, 86, 100, 119, 135, 148, 166, 173, 193, 210, 232, 259
%e 77, 88, 100, 113, 135, 150, 166, 183, 193, 212, 232, 253, 283
%e 90, 104, 119, 135, 144, 162, 181, 201, 214, 236, 259, 283, 300
%t T[a_, b_]:=((a + b)^2 + 3a + b)/2; A[n_, k_]:=T[BitXor[n, k], 2*BitAnd[n, k]]; Table[A[k, n - k ], {n, 0, 20}, {k, 0, n}] // Flatten (* _Indranil Ghosh_, May 20 2017 *)
%o (Scheme)
%o (define (A286109 n) (A286109bi (A002262 n) (A025581 n)))
%o (define (A286109bi row col) (let ((a (A003987bi row col)) (b (* 2 (A004198bi row col)))) (/ (+ (expt (+ a b) 2) (* 3 a) b) 2))) ;; Here A003987bi and A004198bi implement bitwise-xor (A003987) and bitwise-and (A004198).
%o (Python)
%o def T(a, b): return ((a + b)**2 + 3*a + b)//2
%o def A(n, k): return T(n^k, 2*(n&k))
%o for n in range(21): print([A(k, n - k) for k in range(n + 1)]) # _Indranil Ghosh_, May 20 2017
%Y Cf. A000096 (row 0 & column 0), A014105 (main diagonal).
%Y Cf. A003056, A003987, A004198.
%Y Cf. also arrays A286099, A286108, A286145, A286147, A286150, A286151.
%K nonn,tabl
%O 0,2
%A _Antti Karttunen_, May 03 2017
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