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%I #22 Mar 17 2021 15:30:14
%S 0,0,0,0,1,0,0,2,2,0,0,3,4,3,0,0,4,6,6,4,0,0,5,8,7,8,5,0,0,6,10,12,12,
%T 10,6,0,0,7,12,15,16,15,12,7,0,0,8,14,14,20,20,14,14,8,0,0,9,16,15,24,
%U 21,24,15,16,9,0,0,10,18,24,28,30,30,28,24,18,10,0,0,11,20,27,32,31,28
%N OR-numbral multiplication table, read by antidiagonals.
%C See A048888 for the definition of OR-numbral arithmetic
%H Antti Karttunen, <a href="/A067138/b067138.txt">Table of n, a(n) for n = 0..10439; the first 144 antidiagonals of square array</a>
%F From _Rémy Sigrist_, Mar 17 2021: (Start)
%F T(n, 0) = 0.
%F T(n, 1) = n.
%F T(n, 2^k) = n*2^k for any k >= 0.
%F T(n, n) = A067398(n).
%F (End)
%F For all n, k: A048720(n,k) <= A(n,k) <= A004247(n,k). - _Antti Karttunen_, Mar 17 2021
%e The top left 0..16 x 0..16 corner of the array:
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
%e 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
%e 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30,
%e 0, 3, 6, 7, 12, 15, 14, 15, 24, 27, 30, 31, 28, 31, 30, 31,
%e 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60,
%e 0, 5, 10, 15, 20, 21, 30, 31, 40, 45, 42, 47, 60, 61, 62, 63,
%e 0, 6, 12, 14, 24, 30, 28, 30, 48, 54, 60, 62, 56, 62, 60, 62,
%e 0, 7, 14, 15, 28, 31, 30, 31, 56, 63, 62, 63, 60, 63, 62, 63,
%e 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120,
%e 0, 9, 18, 27, 36, 45, 54, 63, 72, 73, 90, 91, 108, 109, 126, 127,
%e 0, 10, 20, 30, 40, 42, 60, 62, 80, 90, 84, 94, 120, 122, 124, 126,
%e 0, 11, 22, 31, 44, 47, 62, 63, 88, 91, 94, 95, 124, 127, 126, 127,
%e 0, 12, 24, 28, 48, 60, 56, 60, 96, 108, 120, 124, 112, 124, 120, 124,
%e 0, 13, 26, 31, 52, 61, 62, 63, 104, 109, 122, 127, 124, 125, 126, 127,
%e 0, 14, 28, 30, 56, 62, 60, 62, 112, 126, 124, 126, 120, 126, 124, 126,
%e 0, 15, 30, 31, 60, 63, 62, 63, 120, 127, 126, 127, 124, 127, 126, 127,
%e 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240,
%e .
%e Multiplying 3 ("11" in binary) with itself in this system means taking bitwise-or of "11" with itself, when shifted one bit-position left:
%e 11
%e 110
%e -------
%e OR: 111 = 7 in decimal = A(3,3).
%e .
%e Multiplying 10 (= "1010" in binary) and 11 (= "1011" in binary) in this system means taking bitwise-or of binary number 1011 when shifted once left with the same binary number when shifted three bit-positions left:
%e 10110
%e 1011000
%e -------
%e OR: 1011110 = 94 in decimal = A(10,11) = A(11,10).
%o (PARI) t(n, k) = {res = 0; for (i=0, length(binary(n))-1, if (bittest(n, i), res = bitor(res, shift(k, i)));); return (res);} \\ _Michel Marcus_, Apr 14 2013
%Y Cf. A003986, A067139, A048888, A007059, A067398 (main diagonal).
%Y Cf. also A004247, A048720 for analogous multiplication tables.
%K nonn,tabl,look
%O 0,8
%A _Jens Voß_, Jan 02 2002
%E Example-section rewritten by _Antti Karttunen_, Mar 17 2021
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