|
%I #21 Sep 28 2018 22:56:30
%S 1,1,1,1,1,2,1,4,1,6,1,8,1,10,4,1,12,12,1,14,24,1,16,40,1,18,60,8,1,
%T 20,84,32,1,22,112,80,1,24,144,160,1,26,180,280,16,1,28,220,448,80,1,
%U 30,264,672,240,1,32,312,960,560,1,34,364,1320,1120,32
%N Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 2 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.
%C The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A013609 ((1+2*x)^n) and along a "third layer" skew diagonals pointing top-left 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-x-2x^4) are given by the sequence generated by the row sums.
%C The row sums give A052942.
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.543689012692076... (A256099: Decimal expansion of the real root of a cubic used by Omar Khayyám in a geometrical problem), 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="/A317500/a317500.pdf">Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n</a>
%H Zagros Lalo, <a href="/A317500/a317500_1.pdf">Third layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n</a>
%F T(n,k) = 2^k / ((n - 4*k)! k!) * (n - 3*k)! where n >= 0 and 0 <= k <= floor(n/4).
%e Triangle begins:
%e 1;
%e 1;
%e 1;
%e 1;
%e 1, 2;
%e 1, 4;
%e 1, 6;
%e 1, 8;
%e 1, 10, 4;
%e 1, 12, 12;
%e 1, 14, 24;
%e 1, 16, 40;
%e 1, 18, 60, 8;
%e 1, 20, 84, 32;
%e 1, 22, 112, 80;
%e 1, 24, 144, 160;
%e 1, 26, 180, 280, 16;
%e 1, 28, 220, 448, 80;
%e 1, 30, 264, 672, 240;
%e ...
%t t[n_, k_] := t[n, k] = 2^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, t[n - 1, k] + 2 t[n - 4, k - 1]]; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]}] // Flatten.
%Y Row sums give A052942.
%Y Cf. A013609, A038207, A128099, A207538, A256099.
%K tabf,nonn,easy
%O 0,6
%A _Zagros Lalo_, Sep 03 2018
|
