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%I #20 Sep 05 2018 02:26:50
%S 1,4,16,1,64,8,256,48,1,1024,256,12,4096,1280,96,1,16384,6144,640,16,
%T 65536,28672,3840,160,1,262144,131072,21504,1280,20,1048576,589824,
%U 114688,8960,240,1,4194304,2621440,589824,57344,2240,24,16777216,11534336,2949120,344064,17920,336,1
%N Triangle read by rows: T(0,0)=1; T(n,k) = 4*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); 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 A013611 ((1+4*x)^n).
%C The coefficients in the expansion of 1/(1-4x-x^2) are given by the sequence generated by the row sums.
%C The row sums are A001076 (Denominators of continued fraction convergent to sqrt(5)).
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 4.236067977...; a metallic mean (see A098317), when n approaches infinity.
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 90, 373.
%H Shara Lalo, <a href="/A305833/a305833.pdf">Left justified triangle</a>
%H Shara Lalo, <a href="/A305833/a305833_1.pdf">Skew diagonals in triangle A013611</a>
%F G.f.: 1 / (1 - 4*t*x - t^2).
%e Triangle begins:
%e 1;
%e 4;
%e 16, 1;
%e 64, 8;
%e 256, 48, 1;
%e 1024, 256, 12;
%e 4096, 1280, 96, 1;
%e 16384, 6144, 640, 16;
%e 65536, 28672, 3840, 160, 1;
%e 262144, 131072, 21504, 1280, 20;
%e 1048576, 589824, 114688, 8960, 240, 1;
%e 4194304, 2621440, 589824, 57344, 2240, 24;
%e 16777216, 11534336, 2949120, 344064, 17920, 336, 1;
%e 67108864, 50331648, 14417920, 1966080, 129024, 3584, 28;
%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 4 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten
%Y Row sums give A001076.
%Y Cf. A000302 (column 0), A002697 (column 1), A038845 (column 2), A038846 (column 3), A040075 (column 4).
%Y Cf. A013611.
%Y Cf. A098317.
%K tabf,nonn,easy
%O 0,2
%A _Shara Lalo_, Jun 11 2018
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