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%I #31 Jan 30 2021 11:50:20
%S 1,1,2,1,1,4,1,0,3,8,1,-1,2,5,16,1,-2,1,2,11,32,1,-3,0,-1,6,21,64,1,
%T -4,-1,-4,1,10,43,128,1,-5,-2,-7,-4,-1,22,85,256,1,-6,-3,-10,-9,-12,1,
%U 42,171,512,1,-7,-4,-13,-14,-23,-20,-1,86,341,1024
%N A000079 is the first row. For the second row, subtract A001045. For the third row, subtract A001045 from the second one, etc. The resulting array is read by ascending antidiagonals.
%C Every row has the signature (1,2).
%C (Among consequences: a(n) read by antidiagonals is
%C 1,
%C 1, 2,
%C 1, 1, 4,
%C 1, 0, 3, 8,
%C 1, -1, 2, 5, 16
%C 1, -2, 1, 2, 11, 32,
%C 1, -3, 0, -1, 6, 21, 64,
%C ... .
%C The row sums and their first two difference table terms are
%C 1, 3, 6, 12, 23, 45, 88, ... = A086445(n+1) - 1
%C 2, 3, 6, 11, 22, 43, 86, ... = A005578(n+2)
%C 1, 3, 5, 11, 21, 43, 85, ... = A001045(n+2).
%C The antidiagonal sums are
%C b(n) = 1, 1, 3, 2, 5, 3, 9, 4, 15, 5, 27, 6, 49, 7, ... .)
%F A(n,k) = 2^k - n*round(2^k/3).
%e Square array:
%e 1, 2, 4, 8, 16, 32, 64, 128, ... = A000079(n)
%e 1, 1, 3, 5, 11, 21, 43, 85, ... = A001045(n+1)
%e 1, 0, 2, 2, 6, 10, 22, 42, ... = A078008(n)
%e 1, -1, 1, -1, 1, -1, 1, -1, ... = A033999(n)
%e 1, -2, 0, -4, -4, -12, -20, -44, ... = -A084247(n)
%e 1, -3, -1, -7, -9, -23, -41, -87, ... = (-1)^n*A140966(n+1)
%e 1, -4, -2, -10, -14, -34, -62, -130, ... = -A135440(n)
%e 1, -5, -3, -13, -19, -45, -83, -173, ... = -A155980(n+3) or -A171382(n+1)
%e ...
%p A:= (n, k)-> (<<0|1>, <2|1>>^k. <<1, 2-n>>)[1$2]:
%p seq(seq(A(d-k, k), k=0..d), d=0..12); # _Alois P. Heinz_, Jan 21 2021
%t A340660[m_, n_] := LinearRecurrence[{1, 2}, {1, m}, {n}]; Table[Reverse[Table[A340660[m, n + m - 2] // First, {m, 2, -n + 3, -1}]], {n, 1, 11}] // Flatten (* _Robert P. P. McKone_, Jan 28 2021 *)
%o (PARI) T(n, k) = 2^k - n*(2^k - (-1)^k)/3;
%o matrix(10,10,n,k,T(n-1,k-1)) \\ _Michel Marcus_, Jan 19 2021
%Y Cf. A000079, A001045, A033999, A078008, A084247, A135440, A140966, A155980, A171382.
%Y Cf. A005578, A086445.
%K sign,tabl,easy
%O 0,3
%A _Paul Curtz_, Jan 15 2021