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Triangle read by rows, derived from an array of sequences generated from (1 + x)/ (1 - r*x - r*x^2).
5

%I #45 Feb 01 2021 00:14:34

%S 1,1,2,1,3,3,1,4,8,5,1,5,15,22,8,1,6,24,57,60,13,1,7,35,116,216,164,

%T 21,1,8,48,205,560,819,448,34,1,9,63,330,1200,2704,3105,1224,55,1,10,

%U 80,497,2268,7025,13056,11772,3344,89,1,11,99,712,3920,15588,41125,63040,44631,9136,144

%N Triangle read by rows, derived from an array of sequences generated from (1 + x)/ (1 - r*x - r*x^2).

%C Row sums = A180166: (1, 3, 7, 18, 51, 161, 560, 2163, ...).

%C Rows of the array, with other offsets: (row 1 = A000045 starting with offset 2: (1, 2, 3, 5, 8, 13, ...); and for rows > 1, the entries: A028859, A125145, A086347, and A180033 start with offset 0; with the offset in the present array = 1.

%H Robert P. P. McKone, <a href="/A180165/b180165.txt">Antidiagonals n = 0..199, flattened</a>

%F Triangle read by rows, generated from an array of sequences generated from (1 + x)/(1 - r*x - r*x^2); r > 0.

%F Alternatively, given the array with offset 1, the sequence r-th sequence is generated from a(k) = r*a(k-1) + r*(k-2); a(1) = 1, a(2) = r+1.

%F With a matrix method, the array can be generated from a 2 X 2 matrix of the form [0,1; r,r] = M, such that M^n * [1,r+1] = [r,n+1; r,n+2].

%F Also, for r > 1, the (r+1)-th row of the array is the INVERT transform of the r-th row.

%e First few rows of the triangle:

%e 1;

%e 1, 2;

%e 1, 3, 3;

%e 1, 4, 8, 5;

%e 1, 5, 15, 22, 8;

%e 1, 6, 24, 57, 60, 13;

%e 1, 7, 35, 116, 216, 164, 21;

%e 1, 8, 48, 205, 560, 819, 448, 34;

%e 1, 9, 63, 330, 1200, 2704, 3105, 1224, 55;

%e 1, 10, 80, 497, 2268, 7025, 13056, 11772, 3344, 89;

%e 1, 11, 99, 712, 3920, 15588, 41125, 63040, 44631, 9136, 144;

%e 1, 12, 120, 981, 6336, 30919, 107136, 240750, 304384, 169209, 24960, 233;

%e ...

%e As an array A(r,k) by upwards antidiagonals:

%e k=1 k=2 k=3 k=4 k=5

%e r=1: 1, 2, 3, 5, 8, ...

%e r=2: 1, 3, 8, 22, 60, ...

%e r=3: 1, 4, 15, 57, 216, ...

%e r=4: 1, 5, 24, 116, 560, ...

%e r=5: 1, 6, 35, 205, 1200, ...

%e Row r=5 = A180033 = (1, 6, 35, 205,...) and is generated from (1+x)/(1-5*x-5*x^2); is the INVERT transform of row r=4; and the array term A(5,4) = 205 = 5*35 + 5*6.

%e Terms A(2,4) and A(2,5) = [22,60] = [0,1; 2,2]^3 * [1,3].

%t A180165[a_] := Reverse[Table[Table[CoefficientList[Series[(1 + x)/(1 - r*x - r*x^2), {x, 0, a - 2}], x], {r, 1, a + 1}][[k, n - k]], {n, 1, a}, {k, 1, n - 1}], 2] // Flatten;

%t A180165[12] (* _Robert P. P. McKone_, Jan 19 2021 *)

%Y Cf. A180166, A000045, A028859, A125145, A086347, A180033.

%Y Cf. A340156.

%K nonn,tabl

%O 1,3

%A _Gary W. Adamson_, Aug 14 2010

%E a(35) corrected by _Robert P. P. McKone_, Dec 31 2020