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A304223 Triangle read by rows: T(0,0)=1; T(n,k) = T(n-1,k)-2*T(n-2,k-1)+2*T(n-3,k-2) for k = 0..floor(2*n/3); T(n,k)=0 for n or k < 0. 1

%I

%S 1,1,1,-2,1,-4,2,1,-6,8,1,-8,18,-8,1,-10,32,-32,4,1,-12,50,-80,36,1,

%T -14,72,-160,136,-24,1,-16,98,-280,360,-160,8,1,-18,128,-448,780,-592,

%U 128,1,-20,162,-672,1484,-1632,720,-64

%N Triangle read by rows: T(0,0)=1; T(n,k) = T(n-1,k)-2*T(n-2,k-1)+2*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-left in center-justified triangle given in A304209.

%C The coefficients in the expansion of 1/(1-x+2*x^2-2*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="/A304223/a304223.pdf">Left-justified Triangle</a>

%e Triangle begins:

%e 1;

%e 1;

%e 1, -2;

%e 1, -4, 2;

%e 1, -6, 8;

%e 1, -8, 18, -8;

%e 1, -10, 32, -32, 4;

%e 1, -12, 50, -80, 36;

%e 1, -14, 72, -160, 136, -24;

%e 1, -16, 98, -280, 360, -160, 8;

%e 1, -18, 128, -448, 780, -592, 128;

%e 1, -20, 162, -672, 1484, -1632, 720, -64;

%e 1, -22, 200, -960, 2576, -3752, 2624, -640, 16;

%e ...

%o (PARI) T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1,k)-2*T(n-2,k-1)+2*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 A077953.

%Y Cf. A304209.

%K tabf,easy,sign

%O 0,4

%A _Shara Lalo_, May 08 2018

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Last modified June 20 15:03 EDT 2021. Contains 345165 sequences. (Running on oeis4.)