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%I #14 Dec 08 2018 20:44:08
%S 0,1,1,2,3,2,3,12,12,3,4,7,12,7,4,5,24,56,56,24,5,6,51,24,15,24,51,6,
%T 7,28,3276,112,112,3276,28,7,8,15,28,455,48,455,28,15,8,9,48,240,120,
%U 25368,25368,120,240,48,9,10,99,48,31,56,51,56,31,48,99,10
%N Square array T(n, k) read by antidiagonals, n >= 0 and k >= 0: the lengths of runs in binary expansion of T(n, k) are obtained by adding those of n and of k (see Comments for precise definition).
%C For any n >= 0 and k >= 0:
%C - let r_n be the lengths of runs in binary expansion of n,
%C - for n = 0: we assume that r_0 = (0),
%C - let R_n be the #r_n-periodic sequence whose first #r_n terms match r_n,
%C - r_{T(n, k)} has lcm(#r_n, #r_k) terms and r_{T(n, k)}(i) = R_n(i) + R_k(i) for i = 1..lcm(#r_n, #r_k).
%F For any m >= 0, n >= 0 and k >= 0:
%F - T(n, k) = T(k, n) (T is commutative),
%F - T(m, T(n, k)) = T(T(m, n), k) (T is associative),
%F - T(n, 0) = n (0 is a neutral element for T),
%F - T(n, 1) = A175046(n),
%F - T(n, n) = A001196(n),
%F - A005811(T(n, k)) = max(A005811(n), A005811(k), lcm(A005811(n), A005811(k))),
%F - T(2^n - 1, 2^k - 1) = 2^(n+k) - 1,
%F - T(2^n, 2^k) = 3 * 2^(n+k) when n > 0 and k > 0,
%F - T(n, k) is odd iff both n and k are odd.
%e Array T(n, k) begins (in decimal):
%e n\k| 0 1 2 3 4 5 6 7 8 9 10
%e ---+------------------------------------------------------------------------
%e 0| 0 1 2 3 4 5 6 7 8 9 10
%e 1| 1 3 12 7 24 51 28 15 48 99 204
%e 2| 2 12 12 56 24 3276 28 240 48 12700 204
%e 3| 3 7 56 15 112 455 120 31 224 903 3640
%e 4| 4 24 24 112 48 25368 56 480 96 99896 792
%e 5| 5 51 3276 455 25368 51 29596 3855 199728 99 13421772
%e 6| 6 28 28 120 56 29596 60 496 112 116540 924
%e 7| 7 15 240 31 480 3855 496 63 960 7695 61680
%e 8| 8 48 48 224 96 199728 112 960 192 792688 3120
%e Array T(n, k) begins (in binary):
%e n\k| 0 1 10 11 100
%e ----+--------------------------------------------------------
%e 0| 0 1 10 11 100
%e 1| 1 11 1100 111 11000
%e 10| 10 1100 1100 111000 11000
%e 11| 11 111 111000 1111 1110000
%e 100| 100 11000 11000 1110000 110000
%e 101| 101 110011 110011001100 111000111 110001100011000
%e 110| 110 11100 11100 1111000 111000
%e 111| 111 1111 11110000 11111 111100000
%e 1000| 1000 110000 110000 11100000 1100000
%o (PARI) T(n,k) = my (v=0, p=1, rn=n, rk=k, b=if ((max(n,1)%2)&&(max(k,1)%2), 1, 0)); while (1, my (vn=if (rn==0, 0, valuation(rn+(rn%2), 2)), vk=if
%o (rk==0, 0, valuation(rk+(rk%2), 2)), w=vn+vk); v+=b*p*(2^w-1); rn\=2^vn; rk\=2^vk; if (rn==0 && rk==0, return (v), rn==0, rn=n, rk==0, rk=k); p*=2^w; b=1-b)
%Y See A322403 for the multiplicative variant.
%Y Cf. A001196, A005811, A175046.
%K nonn,base,tabl
%O 0,4
%A _Rémy Sigrist_, Dec 06 2018
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