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A279010 Alternating Jacobsthal triangle A_3(n,k) read by rows.

%I A279010 #15 Aug 13 2018 09:08:35
%S A279010 1,1,1,3,0,1,3,3,-1,1,9,0,4,-2,1,9,9,-4,6,-3,1,27,0,13,-10,9,-4,1,27,
%T A279010 27,-13,23,-19,13,-5,1,81,0,40,-36,42,-32,18,-6,1,81,81,-40,76,-78,74,
%U A279010 -50,24,-7,1,243,0,121,-116,154,-152,124,-74,31,-8,1
%N A279010 Alternating Jacobsthal triangle A_3(n,k) read by rows.
%H A279010 Kyu-Hwan Lee, Se-jin Oh, <a href="http://arxiv.org/abs/1601.06685">Catalan triangle numbers and binomial coefficients</a>, arXiv:1601.06685 [math.CO], 2016.
%e A279010 Triangle begins:
%e A279010     1;
%e A279010     1,  1;
%e A279010     3,  0,   1;
%e A279010     3,  3,  -1,    1;
%e A279010     9,  0,   4,   -2,   1;
%e A279010     9,  9,  -4,    6,  -3,    1;
%e A279010    27,  0,  13,  -10,   9,   -4,   1;
%e A279010    27, 27, -13,   23, -19,   13,  -5,   1;
%e A279010    81,  0,  40,  -36,  42,  -32,  18,  -6,  1;
%e A279010    81, 81, -40,   76, -78,   74, -50,  24, -7,  1;
%e A279010   243,  0, 121, -116, 154, -152, 124, -74, 31, -8, 1;
%e A279010   ...
%t A279010 A[n_, 0] := 3^Floor[n/2];
%t A279010 A[n_, k_] /; (k<0 || t>n) = 0;
%t A279010 A[n_, n_] = 1;
%t A279010 A[n_, k_] := A[n, k] = A[n-1, k-1] - A[n-1, k];
%t A279010 Table[A[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 12 2018 *)
%Y A279010 If initial column is omitted, this is very like the Riordan matrix A191582.
%Y A279010 Cf. A112468, A279006, A279009.
%K A279010 sign,tabl
%O A279010 0,4
%A A279010 _N. J. A. Sloane_, Dec 07 2016