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%I #10 Sep 28 2018 23:13:11
%S 1,2,-3,1,4,-12,13,-6,1,8,-36,66,-63,33,-9,1,16,-96,248,-360,321,-180,
%T 62,-12,1,32,-240,800,-1560,1970,-1683,985,-390,100,-15,1,64,-576,
%U 2352,-5760,9420,-10836,8989,-5418,2355,-720,147,-18,1,128,-1344,6496,-19152,38472,-55692,59906,-48639,29953,-13923,4809,-1197,203,-21,1
%N Triangle read by rows: T(0,0) = 1; T(n,k) = 2 T(n-1,k) - 3 T(n-1,k-1) + T(n-1,k-2) for k = 0..2n; T(n,k)=0 for n or k < 0.
%C Row n gives coefficients in expansion of (2 - 3*x + x^2)^n. Row sum s(n)= 1 when n = 0 and s(n)= 0 when n > 0, see link. In the center-justified triangle, the sum of numbers along "first layer" skew diagonals pointing top-right are the coefficients in expansion of 1/(1 - 2*x + 3*x^2 - x^3) and the sum of numbers along "first layer" skew diagonals pointing top-left are the coefficients in expansion of 1/(1-x+3*x^2-2x^3), see links. The generating function of the central terms is 1/sqrt(1 + 6*x + x^2), signed version of Central Delannoy numbers A001850.
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.
%H Shara Lalo, <a href="/A318685/a318685.pdf">Centre-justified triangle of coefficients in expansions of (2 - 3 x + x^2)^n</a>
%H Shara Lalo, <a href="/A318685/a318685_1.pdf">First layer skew diagonals in center-justified triangle of coefficients in expansion of (2 - 3 x + x^2)^n</a>
%F T(0,0) = 1; T(n,k) = 2 T(n-1,k) - 3 T(n-1,k-1) + T(n-1,k-2) for k = 0..2n; T(n,k)=0 for n or k < 0.
%F G.f.: 1/(1 - 2*t + 3*t*x - t*x^2).
%e Triangle begins:
%e 1;
%e 2, -3, 1;
%e 4, -12, 13, -6, 1;
%e 8, -36, 66, -63, 33, -9, 1;
%e 16, -96, 248, -360, 321, -180, 62, -12, 1;
%e 32, -240, 800, -1560, 1970, -1683, 985, -390, 100, -15, 1;
%e 64, -576, 2352, -5760, 9420, -10836, 8989, -5418, 2355, -720, 147, -18, 1;
%t t[n_, k_] := t[n, k] = Sum[(2^(n - k + i)/(n - k + i)!)*((-3)^(k - 2*i)/(k - 2*i)!)*(1/i!)*n!, {i, 0, k}];
%t Flatten[Table[t[n, k], {n, 0, 7}, {k, 0, 2*n}]]
%t t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2*t[n - 1, k] - 3*t[n - 1, k - 1] + t[n - 1, k - 2]];
%t Flatten[Table[t[n, k], {n, 0, 7}, {k, 0, 2*n}]]
%Y Cf. A001850.
%K tabf,sign,easy
%O 0,2
%A _Shara Lalo_, Sep 06 2018
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