A317494 - OEIS

%I #17 Aug 19 2018 15:48:52

%S 1,1,1,1,2,1,4,1,6,1,8,4,1,10,12,1,12,24,1,14,40,8,1,16,60,32,1,18,84,

%T 80,1,20,112,160,16,1,22,144,280,80,1,24,180,448,240,1,26,220,672,560,

%U 32,1,28,264,960,1120,192,1,30,312,1320,2016,672,1,32,364,1760,3360,1792,64

%N Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 2 * T(n-3,k-1) for k = 0..floor(n/3); T(n,k)=0 for n or k < 0.

%C The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A013609 ((1+2*x)^n) and along a "second layer" of 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+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)

%C The coefficients in the expansion of 1/(1-x-2x^3) are given by the sequence generated by the row sums.

%C The row sums give A003229.

%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.695620769559862... (see A289265), 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. 358, 359

%H Zagros Lalo, <a href="/A317494/a317494.pdf">Second layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n</a>

%H Zagros Lalo, <a href="/A317494/a317494_1.pdf">Second layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n</a>

%F T(n,k) = 2^k / ((n - 3k)! k!) * (n - 2k)! where n is a nonnegative integer and k = 0..floor(n/3).

%e Triangle begins:

%e   1;

%e   1;

%e   1;

%e   1,  2;

%e   1,  4;

%e   1,  6;

%e   1,  8,   4;

%e   1, 10,  12;

%e   1, 12,  24;

%e   1, 14,  40,    8;

%e   1, 16,  60,   32;

%e   1, 18,  84,   80;

%e   1, 20, 112,  160,   16;

%e   1, 22, 144,  280,   80;

%e   1, 24, 180,  448,  240;

%e   1, 26, 220,  672,  560,   32;

%e   1, 28, 264,  960, 1120,  192;

%e   1, 30, 312, 1320, 2016,  672;

%e   1, 32, 364, 1760, 3360, 1792, 64;

%t t[n_, k_] := t[n, k] = 2^k/((n - 3 k)! k!) (n - 2 k)!; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/3]} ]  // 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 - 3, k - 1]]; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/3]}] // Flatten.

%o (GAP) Flat(List([0..20],n->List([0..Int(n/3)],k->2^k/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # _Muniru A Asiru_, Jul 31 2018

%Y Row sums give A003229.

%Y Cf. A013609, A038207, A289265, A317495, A128099, A207538.

%K tabf,nonn,easy

%O 0,5

%A _Zagros Lalo_, Jul 30 2018