%I
%S 1,1,1,1,1,1,2,1,4,1,6,1,8,1,10,1,12,4,1,14,12,1,16,24,1,18,40,1,20,
%T 60,1,22,84,8,1,24,112,32,1,26,144,80,1,28,180,160,1,30,220,280,1,32,
%U 264,448,16,1,34,312,672,80,1,36,364,960,240,1,38,420,1320,560,1,40,480,1760,1120
%N Triangle read by rows: T(0,0) = 1; T(n,k) = T(n1,k) + 2 * T(n5,k1) for k = 0..floor(n/5); T(n,k)=0 for n or k < 0.
%C The numbers in rows of the triangle are along a "fourth layer" skew diagonals pointing topright in centerjustified triangle given in A013609 ((1+2*x)^n) and along a "fourth layer" skew diagonals pointing topleft in centerjustified triangle given in A038207 ((2+x)^n), see links. (Note: First layer skew diagonals in centerjustified 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/(1x2x^5) are given by the sequence generated by the row sums.
%C The row sums give A318777.
%C If s(n) is the row sum at n, then the ratio s(n)/s(n1) is approximately 1.4510850920547191..., when n approaches infinity.
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 9781999591403.
%H Zagros Lalo, <a href="/A318775/a318775.pdf">Fourth layer skew diagonals in centerjustified triangle of coefficients in expansion of (1 + 2x)^n</a>
%H Zagros Lalo, <a href="/A318775/a318775_1.pdf">Fourth layer skew diagonals in centerjustified triangle of coefficients in expansion of (2 + x)^n</a>
%F T(n,k) = 2^k / ((n  5*k)! k!) * (n  4*k)! where n >= 0 and 0 <= k <= floor(n/5).
%e Triangle begins:
%e 1;
%e 1;
%e 1;
%e 1;
%e 1;
%e 1, 2;
%e 1, 4;
%e 1, 6;
%e 1, 8;
%e 1, 10;
%e 1, 12, 4;
%e 1, 14, 12;
%e 1, 16, 24;
%e 1, 18, 40;
%e 1, 20, 60;
%e 1, 22, 84, 8;
%e 1, 24, 112, 32;
%e 1, 26, 144, 80;
%e 1, 28, 180, 160;
%e 1, 30, 220, 280;
%e 1, 32, 264, 448, 16;
%e 1, 34, 312, 672, 80;
%e 1, 36, 364, 960, 240;
%e 1, 38, 420, 1320, 560;
%e ...
%t t[n_, k_] := t[n, k] = 2^k/((n  5 k)! k!) (n  4 k)!; Table[t[n, k], {n, 0, 24}, {k, 0, Floor[n/5]} ] // Flatten.
%t t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0  k < 0, 0, t[n  1, k] + 2 t[n  5, k  1]]; Table[t[n, k], {n, 0, 24}, {k, 0, Floor[n/5]}] // Flatten.
%Y Row sums give A318777.
%Y Cf. A013609, A038207, A128099, A207538.
%K tabf,nonn,easy
%O 0,7
%A _Zagros Lalo_, Sep 04 2018
