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 A099514 Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + z + 2*z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2. 1
 1, 1, 1, 1, 2, 2, 1, 3, 5, 4, 1, 4, 9, 13, 4, 1, 5, 14, 28, 18, 12, 1, 6, 20, 50, 49, 56, 8, 1, 7, 27, 80, 105, 161, 56, 32, 1, 8, 35, 119, 195, 366, 210, 200, 16, 1, 9, 44, 168, 329, 721, 581, 732, 160, 80, 1, 10, 54, 228, 518, 1288, 1337, 2045, 780, 640, 32, 1, 11, 65, 300, 774, 2142, 2716, 4824, 2674, 2884, 432, 192 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Row sums form A099515. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2). LINKS FORMULA G.f.: (1-x+x*y-2*x^2*y^2)/((1-x)^2-4*x^2*y^2+3*x^3*y^2+4*x^4*y^4). EXAMPLE Rows begin: [1], [1,1], [1,2,2], [1,3,5,4], [1,4,9,13,4], [1,5,14,28,18,12], [1,6,20,50,49,56,8], [1,7,27,80,105,161,56,32], [1,8,35,119,195,366,210,200,16], [1,9,44,168,329,721,581,732,160,80],... and can be derived from coefficients of (1+z+2*z^2)^n: [1], [1,1,2], [1,2,5,4,4], [1,3,9,13,18,12,8], [1,4,14,28,49,56,56,32,16], [1,5,20,50,105,161,210,200,160,80,32],... by shifting each column k down by [k/2] rows. MATHEMATICA With[{m = 11}, CoefficientList[CoefficientList[Series[(1-x+x*y-2*x^2*y^2)/ ((1-x)^2-4*x^2*y^2+3*x^3*y^2+4*x^4*y^4), {x, 0 , m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 17 2020 *) PROG (PARI) T(n, k)=if(n

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Last modified November 28 13:18 EST 2020. Contains 338724 sequences. (Running on oeis4.)