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%I #26 Mar 05 2024 16:32:52
%S 0,1,1,2,4,2,3,3,3,3,4,6,8,6,4,5,5,9,9,5,5,6,16,6,12,6,16,6,7,7,7,7,7,
%T 7,7,7,8,18,12,18,16,18,12,18,8,9,9,13,13,17,17,13,13,9,9,10,12,10,24,
%U 18,20,18,24,10,12,10,11,11,11,11,19,19,19,19,11,11,11,11,12,14,32,14
%N Bit-interleaved number addition table; like binary addition but carries shift 2 instead of 1; addition base sqrt(2).
%H Reinhard Zumkeller, <a href="/A054240/b054240.txt">Antidiagonals n=0..125 of square array, flattened</a>
%F From _Peter Munn_, Dec 10 2019: (Start)
%F A(m,0) = A(0,m) = m.
%F A(n,k) = A(k,n).
%F A(n, A(m,k)) = A(A(n,m), k).
%F A(m,m) = 4*m.
%F A(2*n, 2*k) = 2*A(n,k).
%F A(A000695(n), A000695(k)) = A000695(n+k).
%F A(A000695(n), 2*A000695(k)) = A000695(n) + 2*A000695(k).
%F A(A000695(n) + 2*A000695(m), k) = A(A000695(n), k) + A(2*A000695(m), k) - k.
%F A(A057300(n), A057300(k)) = A057300(A(n,k)).
%F (End)
%e T(3,1)=6 because (0*2 + 1*sqrt(2) + 1*1) + (0*2 + 0*sqrt(2) + 1*1) = (1*2 + 1*sqrt(2) + 0*1) (i.e., base sqrt(2) addition).
%o (Haskell)
%o import Data.Bits (xor, (.&.), shift)
%o a054240 :: Integer -> Integer -> Integer
%o a054240 x 0 = x
%o a054240 x y = a054240 (x `xor` y) (shift (x .&. y) 2)
%o a054240_adiag n = map (\k -> a054240 (n - k) k) [0..n]
%o a054240_square = map a054240_adiag [0..]
%o -- _Reinhard Zumkeller_, Dec 03 2011
%Y Cf. A000695, A054239, A057300, A062880, A352909 (pairs (i,j) such that A(i,j) = i+j).
%Y Cf. A201651 (triangle read by rows).
%K nonn,base,tabl,easy,look
%O 0,4
%A _Marc LeBrun_, Feb 07 2000
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