|
%I #25 Sep 05 2018 02:29:07
%S 1,2,4,-2,8,-8,1,16,-24,8,32,-64,36,-4,64,-160,128,-32,1,128,-384,400,
%T -160,18,256,-896,1152,-640,136,-6,512,-2048,3136,-2240,720,-80,1,
%U 1024,-4608,8192,-7168,3120,-592,32,2048,-10240,20736,-21504,11872,-3264,360,-8,4096,-22528,51200,-61440,41216,-15008,2624,-160,1
%N Triangle read by rows: T(0,0) = 1; T(n,k) = 2*T(n-1,k) - 2*T(n-2,k-1) + T(n-3,k-2) for k = 0..floor(2*n/3); T(n,k)=0 for n or k < 0.
%C The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A304209.
%C The coefficients in the expansion of 1/(1-2*x+2*x^2-x^3) are given by the sequence generated by the row sums.
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 204, 205.
%H Shara Lalo, <a href="/A304213/a304213.pdf">Left-justified triangle</a>
%e Triangle begins:
%e 1;
%e 2;
%e 4, -2;
%e 8, -8, 1;
%e 16, -24, 8;
%e 32, -64, 36, -4;
%e 64, -160, 128, -32, 1;
%e 128, -384, 400, -160, 18;
%e 256, -896, 1152, -640, 136, -6;
%e 512, -2048, 3136, -2240, 720, -80, 1;
%e 1024, -4608, 8192, -7168, 3120, -592, 32;
%e 2048, -10240, 20736, -21504, 11872, -3264, 360, -8;
%e 4096, -22528, 51200, -61440, 41216, -15008, 2624, -160, 1;
%e 8192, -49152, 123904, -168960, 133632, -60928, 14896, -1632, 50;
%e 16384, -106496, 294912, -450560, 410880, -225792, 71680, -11776, 780, -10;
%e ...
%o (PARI) T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, 2*T(n-1,k)-2*T(n-2,k-1)+T(n-3,k-2)));
%o tabf(nn) = for (n=0, nn, for (k=0, 2*n\3, print1(T(n,k), ", ")); print); \\ _Michel Marcus_, May 10 2018
%Y Row sums is similar to A021823.
%Y Cf. A304209.
%K tabf,easy,sign
%O 0,2
%A _Shara Lalo_, May 08 2018
|
