A318776 - OEIS

%I #14 Sep 28 2018 23:02:38

%S 1,2,4,8,16,32,1,64,4,128,12,256,32,512,80,1024,192,1,2048,448,6,4096,

%T 1024,24,8192,2304,80,16384,5120,240,32768,11264,672,1,65536,24576,

%U 1792,8,131072,53248,4608,40,262144,114688,11520,160,524288,245760,28160,560,1048576,524288,67584,1792,1,2097152,1114112,159744,5376,10

%N Triangle read by rows: T(0,0) = 1; T(n,k) = 2 T(n-1,k) + T(n-5,k-1) 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 top-left in center-justified triangle given in A013609 ((1+2*x)^n) and along a "fourth 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^5) are given by the sequence generated by the row sums.

%C The row sums give A098588.

%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.0559673967128..., 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="/A318776/a318776.pdf">Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n</a>

%H Zagros Lalo, <a href="/A318776/a318776_1.pdf">Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n</a>

%F T(n,k) = 2^(n - 5*k) / ((n - 5*k)! k!) * (n - 4*k)! where n >= 0 and 0 <= k <= floor(n/5).

%e Triangle begins:

%e         1;

%e         2;

%e         4;

%e         8;

%e        16;

%e        32,       1;

%e        64,       4;

%e       128,      12;

%e       256,      32;

%e       512,      80;

%e      1024,     192,      1;

%e      2048,     448,      6;

%e      4096,    1024,     24;

%e      8192,    2304,     80;

%e     16384,    5120,    240;

%e     32768,   11264,    672,    1;

%e     65536,   24576,   1792,    8;

%e    131072,   53248,   4608,   40;

%e    262144,  114688,  11520,  160;

%e    524288,  245760,  28160,  560;

%e   1048576,  524288,  67584, 1792,  1;

%e   2097152, 1114112, 159744, 5376, 10;

%e   ...

%t t[n_, k_] := t[n, k] = 2^(n - 5 k)/((n - 5 k)! k!) (n - 4 k)!; Table[t[n, k], {n, 0, 22}, {k, 0, Floor[n/5]} ]  // 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 - 5, k - 1]]; Table[t[n, k], {n, 0, 22}, {k, 0, Floor[n/5]}] // Flatten.

%Y Row sums give A098588.

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

%Y Cf. also A000079 (column 0), A001787 (column 1), A001788 (column 2), A001789 (column 3)

%K tabf,nonn,easy

%O 0,2

%A _Zagros Lalo_, Sep 04 2018