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%I #11 Sep 25 2018 02:38:27
%S 1,2,4,8,16,1,32,4,64,12,128,32,256,80,1,512,192,6,1024,448,24,2048,
%T 1024,80,4096,2304,240,1,8192,5120,672,8,16384,11264,1792,40,32768,
%U 24576,4608,160,65536,53248,11520,560,1,131072,114688,28160,1792,10,262144,245760,67584,5376,60
%N Triangle read by rows: T(0,0) = 1; T(n,k) = 2 T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.
%C Unsigned version of the triangle in A317506.
%C The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A038207 ((2+x)^n), see links. (Note: First layer skew diagonals in center-justified triangles of coefficients in expansions of (1+2x)^n and (2+x)^n are given in A128099 and A207538 respectively.)
%C The coefficients in the expansion of 1/(1-2x-x^4) are given by the sequence generated by the row sums.
%C The row sums give A008999.
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.106919340376..., 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.
%H Zagros Lalo, <a href="/A317501/a317501.pdf">Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n</a>
%H Zagros Lalo, <a href="/A317501/a317501_1.pdf">Third layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n</a>
%F T(n,k) = 2^(n - 4*k) / ((n - 4*k)! k!) * (n - 3*k)! where n >= 0 and 0 <= k <= floor(n/4).
%e Triangle begins:
%e 1;
%e 2;
%e 4;
%e 8;
%e 16, 1;
%e 32, 4;
%e 64, 12;
%e 128, 32;
%e 256, 80, 1;
%e 512, 192, 6;
%e 1024, 448, 24;
%e 2048, 1024, 80;
%e 4096, 2304, 240, 1;
%e 8192, 5120, 672, 8;
%e 16384, 11264, 1792, 40;
%e 32768, 24576, 4608, 160;
%e 65536, 53248, 11520, 560, 1;
%e 131072, 114688, 28160, 1792, 10;
%e 262144, 245760, 67584, 5376, 60;
%t t[n_, k_] := t[n, k] = 2^(n - 4 k)/((n - 4 k)! k!) (n - 3 k)!; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]} ] // Flatten.
%t t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2 t[n - 1, k] + t[n - 4, k - 1]]; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]}] // Flatten.
%Y Row sums give A008999.
%Y Cf. A013609, A038207, A128099, A207538, A317506.
%K tabf,nonn,easy
%O 0,2
%A _Zagros Lalo_, Sep 03 2018
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