Zagros Lalo - cp's OEIS frontend

A318777 Coefficients in expansion of 1/(1 - x - 2*x^5).

1, 1, 1, 1, 1, 3, 5, 7, 9, 11, 17, 27, 41, 59, 81, 115, 169, 251, 369, 531, 761, 1099, 1601, 2339, 3401, 4923, 7121, 10323, 15001, 21803, 31649, 45891, 66537, 96539, 140145, 203443, 295225, 428299, 621377, 901667, 1308553, 1899003, 2755601, 3998355, 5801689, 8418795

Offset: 0

Zagros Lalo, Sep 25 2018

The coefficients in the expansion of 1/(1 - x - 2*x^5) are given by the sequence generated by the row sums in triangle A318775.

Coefficients in expansion of 1/(1 - x - 2*x^5) are given by the sum of numbers along "fourth Layer" skew diagonals pointing top-right in triangle A013609 ((1+2x)^n) and by the sum of numbers along "fourth Layer" skew diagonals pointing top-left in triangle A038207 ((2+x)^n), see links.

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Zagros Lalo, Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n.
Zagros Lalo, Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2).

Cf. A318775, A318776.
Cf. A013609, A038207.
Essentially a duplicate of A143447.

a:=[1,1,1,1,1,3];; for n in [7..50] do a[n]:=a[n-1]+2*a[n-5]; od; a; # Muniru A Asiru, Sep 26 2018

seq(coeff(series((1-x-2*x^5)^(-1),x,n+1), x, n), n = 0 .. 50); # Muniru A Asiru, Sep 26 2018

a[0] = 1; a[n_] := a[n] = If[n < 0, 0, a[n - 1] + 2 * a[n - 5]];Table[a[n], {n, 0, 45}] // Flatten
LinearRecurrence[{1, 0, 0, 0, 2}, {1, 1, 1, 1, 1}, 46]
CoefficientList[Series[1/(1 - x - 2 x^5), {x, 0, 45}], x]

a(0)=1, a(n) = a(n-1) + 2 * a(n-5) for n = 0,1...; a(n)=0 for n < 0.

G.f.: -1/(2*x^5 + x - 1). - Chai Wah Wu, Aug 03 2020

A318776 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.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 1, 64, 4, 128, 12, 256, 32, 512, 80, 1024, 192, 1, 2048, 448, 6, 4096, 1024, 24, 8192, 2304, 80, 16384, 5120, 240, 32768, 11264, 672, 1, 65536, 24576, 1792, 8, 131072, 53248, 4608, 40, 262144, 114688, 11520, 160, 524288, 245760, 28160, 560, 1048576, 524288, 67584, 1792, 1, 2097152, 1114112, 159744, 5376, 10

Offset: 0

Zagros Lalo, Sep 04 2018

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+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)
The coefficients in the expansion of 1/(1-2*x-x^5) are given by the sequence generated by the row sums.
The row sums give A098588.
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.

			Triangle begins:
        1;
        2;
        4;
        8;
       16;
       32,       1;
       64,       4;
      128,      12;
      256,      32;
      512,      80;
     1024,     192,      1;
     2048,     448,      6;
     4096,    1024,     24;
     8192,    2304,     80;
    16384,    5120,    240;
    32768,   11264,    672,    1;
    65536,   24576,   1792,    8;
   131072,   53248,   4608,   40;
   262144,  114688,  11520,  160;
   524288,  245760,  28160,  560;
  1048576,  524288,  67584, 1792,  1;
  2097152, 1114112, 159744, 5376, 10;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Row sums give A098588.
Cf. A013609, A038207, A128099, A207538.
Cf. also A000079 (column 0), A001787 (column 1), A001788 (column 2), A001789 (column 3)

t[n_, k_] := t[n, k] = 2^(n - 5 k)/((n - 5 k)! k!) (n - 4 k)!; Table[t[n, k], {n, 0, 21}, {k, 0, Floor[n/5]} ] // Flatten
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, 21}, {k, 0, Floor[n/5]}] // Flatten

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

A318775 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 2 * T(n-5,k-1) for k = 0..floor(n/5); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

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, 60, 1, 22, 84, 8, 1, 24, 112, 32, 1, 26, 144, 80, 1, 28, 180, 160, 1, 30, 220, 280, 1, 32, 264, 448, 16, 1, 34, 312, 672, 80, 1, 36, 364, 960, 240, 1, 38, 420, 1320, 560, 1, 40, 480, 1760, 1120

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "fourth layer" skew diagonals pointing top-right in center-justified triangle given in A013609 ((1+2*x)^n) and along a "fourth 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+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)
The coefficients in the expansion of 1/(1-x-2*x^5) are given by the sequence generated by the row sums.
The row sums give A318777.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.4510850920547191..., when n approaches infinity.

			Triangle begins:
  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,  60;
  1, 22,  84,    8;
  1, 24, 112,   32;
  1, 26, 144,   80;
  1, 28, 180,  160;
  1, 30, 220,  280;
  1, 32, 264,  448,  16;
  1, 34, 312,  672,  80;
  1, 36, 364,  960, 240;
  1, 38, 420, 1320, 560;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Row sums give A318777.

Cf. A013609, A038207, A128099, A207538.

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[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

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

A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 10, 13, 25, 46, 76, 115, 190, 328, 556, 901, 1471, 2455, 4123, 6826, 11239, 18604, 30973, 51451, 85168, 140980, 233899, 388252, 643756, 1066696, 1768393, 2933149, 4864417, 8064505, 13369684, 22169131, 36762382, 60955897, 101064949, 167572342, 277859488, 460727179, 763922026, 1266639052

Offset: 0

Zagros Lalo, Sep 04 2018

The coefficients in the expansion of 1/(1 - x - 3*x^4) are given by the sequence generated by the row sums in triangle A318772.

Coefficients in expansion of 1/(1 - x - 3*x^4) are given by the sum of numbers along "third Layer" skew diagonals pointing top-right in triangle A013610 ((1+3x)^n) and by the sum of numbers along "third Layer" skew diagonals pointing top-left in triangle A027465 ((3+x)^n), see links.

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3 x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n
Index entries for linear recurrences with constant coefficients, signature (1,0,0,3).

Cf. A013610, A027465, A318772, A318773.

Essentially a duplicate of A143454.

[n le 4 select 1 else Self(n-1) +3*Self(n-4): n in [1..51]]; // G. C. Greubel, May 08 2021

CoefficientList[Series[1/(1-x-3x^4), {x, 0, 50}], x]
a[n_]:= a[n]= If[n<4, 1, a[n-1] + 3*a[n-4]]; Table[a[n], {n,0,50}]
LinearRecurrence[{1,0,0,3}, {1,1,1,1}, 51]

my(p=Mod('x,x^4-'x^3-3)); a(n) = vecsum(Vec(lift(p^n))); \\ Kevin Ryde, May 11 2021

def a(n): return 1 if (n<4) else a(n-1) + 3*a(n-4)
[a(n) for n in (0..50)] # G. C. Greubel, May 08 2021

a(n) = a(n-1) + 3*a(n-4) for n >= 0, a(n)=0 for n < 0, with a(0) = a(1) = a(2) = a(3) = 1.

A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

Original entry on oeis.org

1, 3, 9, 27, 81, 1, 243, 6, 729, 27, 2187, 108, 6561, 405, 1, 19683, 1458, 9, 59049, 5103, 54, 177147, 17496, 270, 531441, 59049, 1215, 1, 1594323, 196830, 5103, 12, 4782969, 649539, 20412, 90, 14348907, 2125764, 78732, 540, 43046721, 6908733, 295245, 2835, 1, 129140163, 22320522, 1082565, 13608, 15

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)
The coefficients in the expansion of 1/(1-3*x-x^4) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.035744112294..., when n approaches infinity.

			Triangle begins:
          1;
          3;
          9;
         27;
         81,        1;
        243,        6;
        729,       27;
       2187,      108;
       6561,      405,       1;
      19683,     1458,       9;
      59049,     5103,      54;
     177147,    17496,     270;
     531441,    59049,    1215,     1;
    1594323,   196830,    5103,    12;
    4782969,   649539,   20412,    90;
   14348907,  2125764,   78732,   540;
   43046721,  6908733,  295245,  2835,   1;
  129140163, 22320522, 1082565, 13608,  15;
  387420489, 71744535, 3897234, 61236, 135;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Row sums give A052917.
Cf. A013610, A027465.
Cf. A304236, A304249
Cf. A317496, A317497.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3).
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), A317497 (q=3), this sequence (q=4).

[3^(n-4*k)*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^(n-4k)*(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3T[n-1, k] + T[n-4, k-1]]; Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]}]//Flatten

flatten([[3^(n-4*k)*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^(n-4*k) * (n-3*k)!/(k! * (n-4*k)!) where n >= 0 and 0 <= k <= floor(n/4).

A318772 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 3 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 6, 1, 9, 1, 12, 1, 15, 9, 1, 18, 27, 1, 21, 54, 1, 24, 90, 1, 27, 135, 27, 1, 30, 189, 108, 1, 33, 252, 270, 1, 36, 324, 540, 1, 39, 405, 945, 81, 1, 42, 495, 1512, 405, 1, 45, 594, 2268, 1215, 1, 48, 702, 3240, 2835, 1, 51, 819, 4455, 5670, 243, 1, 54, 945, 5940, 10206, 1458

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)
The coefficients in the expansion of 1/(1-x-3*x^4) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.6580980673722..., when n approaches infinity.

			Triangle begins:
  1;
  1;
  1;
  1;
  1,  3;
  1,  6;
  1,  9;
  1, 12;
  1, 15,   9;
  1, 18,  27;
  1, 21,  54;
  1, 24,  90;
  1, 27, 135,   27;
  1, 30, 189,  108;
  1, 33, 252,  270;
  1, 36, 324,  540;
  1, 39, 405,  945,   81;
  1, 42, 495, 1512,  405;
  1, 45, 594, 2268, 1215;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Row sums give A318774.
Cf. A013610, A027465.
Cf. A304236, A304249.
Cf. A317496, A317497.

[3^k*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k]= 3^k(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 21}, {k, 0, Floor[n/4]} ] // Flatten
T[0, 0] = 1; T[n_, k_] := T[n, k] = If[n<0 || k<0, 0, T[n-1, k] + 3T[n-4, k-1]]; Table[T[n, k], {n, 0, 21}, {k, 0, Floor[n/4]}] // Flatten

flatten([[3^k*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^k * (n - 3*k)!/ ((n - 4*k)! k!), where n >= 0 and 0 <= k <= floor(n/4).

A317501 Triangle read by rows: T(0,0) = 1; T(n,k) = 2*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 2, 4, 8, 16, 1, 32, 4, 64, 12, 128, 32, 256, 80, 1, 512, 192, 6, 1024, 448, 24, 2048, 1024, 80, 4096, 2304, 240, 1, 8192, 5120, 672, 8, 16384, 11264, 1792, 40, 32768, 24576, 4608, 160, 65536, 53248, 11520, 560, 1, 131072, 114688, 28160, 1792, 10, 262144, 245760, 67584, 5376, 60

Offset: 0

Zagros Lalo, Sep 03 2018

Unsigned version of the triangle in A317506.
The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and along a "third 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+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)
The coefficients in the expansion of 1/(1-2*x-x^4) are given by the sequence generated by the row sums.
The row sums give A008999.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.106919340376..., when n approaches infinity.

			Triangle begins:
       1;
       2;
       4;
       8;
      16,      1;
      32,      4;
      64,     12;
     128,     32;
     256,     80,     1;
     512,    192,     6;
    1024,    448,    24;
    2048,   1024,    80;
    4096,   2304,   240,    1;
    8192,   5120,   672,    8;
   16384,  11264,  1792,   40;
   32768,  24576,  4608,  160;
   65536,  53248, 11520,  560,  1;
  131072, 114688, 28160, 1792, 10;
  262144, 245760, 67584, 5376, 60;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Row sums give A008999.

Cf. A013609, A038207, A128099, A207538, A317506.

t[n_, k_] := t[n, k] = 2^(n - 4 k)/((n - 4 k)! k!) (n - 3 k)!; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]} ] // Flatten
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2 t[n - 1, k] + t[n - 4, k - 1]]; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]}] // Flatten

T(n,k) = 2^(n - 4*k) / ((n - 4*k)! k!) * (n - 3*k)! where n >= 0 and 0 <= k <= floor(n/4).

A317500 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.

Original entry on oeis.org

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, 20, 84, 32, 1, 22, 112, 80, 1, 24, 144, 160, 1, 26, 180, 280, 16, 1, 28, 220, 448, 80, 1, 30, 264, 672, 240, 1, 32, 312, 960, 560, 1, 34, 364, 1320, 1120, 32

Offset: 0

Zagros Lalo, Sep 03 2018

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+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)
The coefficients in the expansion of 1/(1-x-2*x^4) are given by the sequence generated by the row sums.
The row sums give A052942.
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.

			Triangle begins:
  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, 20,  84,  32;
  1, 22, 112,  80;
  1, 24, 144, 160;
  1, 26, 180, 280,  16;
  1, 28, 220, 448,  80;
  1, 30, 264, 672, 240;
...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Row sums give A052942.

Cf. A013609, A038207, A128099, A207538, A256099.

t[n_, k_] := t[n, k] = 2^k/((n - 4 k)! k!) (n - 3 k)!; Table[t[n, k], {n, 0, 20}, {k, 0, Floor[n/4]} ] // Flatten
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, 20}, {k, 0, Floor[n/4]}] // Flatten

T(n,k) = 2^k / ((n - 4*k)! k!) * (n - 3*k)! where n >= 0 and 0 <= k <= floor(n/4).

A317499 Coefficients in expansion of 1/(1 + 2x - 3x^3).

Original entry on oeis.org

1, -2, 4, -5, 4, 4, -23, 58, -104, 139, -104, -104, 625, -1562, 2812, -3749, 2812, 2812, -16871, 42178, -75920, 101227, -75920, -75920, 455521, -1138802, 2049844, -2733125, 2049844, 2049844, -12299063, 30747658, -55345784, 73794379, -55345784, -55345784

Offset: 0

Zagros Lalo, Jul 31 2018

The coefficients in the expansion of 1/(1 + 2*x - 3*x^3) are given by the sequence generated by the row sums in triangle A317503.

Coefficients in expansion of 1/(1 + 2*x - 3*x^3) are given by the sum of numbers along second Layer skew diagonals pointing top-left in triangle A303901 ((3-2*x)^n) and by the sum of numbers along second Layer skew diagonals pointing top-right in triangle A317498 ((-2+3*x)^n), see links.

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 396, 397.

Colin Barker, Table of n, a(n) for n = 0..1000
Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (3 - 2x)^n
Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (-2 + 3x)^n
Index entries for linear recurrences with constant coefficients, signature (-2,0,3).

Cf. A317502, A317503.

Cf. A303901, A317498.

a:=[1,-2,4];; for n in [4..40] do a[n]:=-2*a[n-1]+3*a[n-3]; od; a; # Muniru A Asiru, Aug 01 2018

seq(coeff(series(1/(1+2*x-3*x^3), x,n+1),x,n),n=0..40); # Muniru A Asiru, Aug 01 2018

CoefficientList[Series[1/(1 + 2 x - 3 x^3), {x, 0, 40}], x]
a[0] = 1; a[n_] := a[n] = If[n < 0, 0, -2 * a[n - 1] + 3 * a[n - 3]]; Table[a[n], {n, 0, 40}] // Flatten
LinearRecurrence[{-2, 0, 3}, {1, -2, 4}, 41]

Vec(1 / ((1 - x)*(1 + 3*x + 3*x^2)) + O(x^40)) \\ Colin Barker, Aug 02 2018

a(0)=1, a(n) = -2*a(n-1) + 3*a(n-3) for n = 0,1...; a(n)=0 for n < 0.

a(n) = (2^(-n)*(2^n + (-3-i*sqrt(3))^n*(3-2*i*sqrt(3)) + (-3+i*sqrt(3))^n*(3+2*i*sqrt(3)))) / 7 where i=sqrt(-1). - Colin Barker, Aug 02 2018

A317498 Triangle read by rows of coefficients in expansions of (-2 + 3*x)^n, where n is nonnegative integer.

Original entry on oeis.org

1, -2, 3, 4, -12, 9, -8, 36, -54, 27, 16, -96, 216, -216, 81, -32, 240, -720, 1080, -810, 243, 64, -576, 2160, -4320, 4860, -2916, 729, -128, 1344, -6048, 15120, -22680, 20412, -10206, 2187, 256, -3072, 16128, -48384, 90720, -108864, 81648, -34992, 6561, -512, 6912, -41472, 145152, -326592, 489888, -489888, 314928, -118098, 19683

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Zagros Lalo, Jul 31 2018

Row n gives coefficients in expansion of (-2 + 3*x)^n.
This is a signed version of A013620.
The coefficients in the expansion of 1/(1-x) are given by the sequence generated by the row sums.
The row sums give A000012 (The simplest sequence of positive numbers: the all 1's sequence).
The numbers in rows of triangles in A302747 and A303941 (Triangle of coefficients of Fermat polynomials) are along first layer skew diagonals pointing top-right and top-left in center-justified triangle of coefficients in expansions of (-2 + 3*x)^n, see links.

			Triangle begins:
     1;
    -2,     3;
     4,   -12,      9;
    -8,    36,    -54,     27;
    16,   -96,    216,   -216,      81;
   -32,   240,   -720,   1080,    -810,     243;
    64,  -576,   2160,  -4320,    4860,   -2916,     729;
  -128,  1344,  -6048,  15120,  -22680,   20412,  -10206,   2187;
   256, -3072,  16128, -48384,   90720, -108864,   81648, -34992,    6561;
  -512,  6912, -41472, 145152, -326592,  489888, -489888, 314928, -118098, 19683;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, Pages 394-396.

Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (-2 + 3 x)^n.
Eric Weisstein's World of Mathematics, Fermat Polynomial.

Row sums give A000012.
Cf. A013620 ((2+3*x)^n).
Cf. A302747, A303941.

Flat(List([0..8],n->List([0..n],k->(-2)^(n-k)*3^k/(Factorial(n-k)*Factorial(k))*Factorial(n)))); # Muniru A Asiru, Aug 01 2018

t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, -2 t[n - 1, k] + 3 t[n - 1, k - 1]]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
t[n_, k_] := t[n, k] = ((-2)^(n - k) 3^k)/((n - k)! k!) n!;Table[t[n, k], {n, 0, 9}, {k, 0, n} ] // Flatten
Table[CoefficientList[(-2 + 3 x)^n, x], {n, 0, 9}] // Flatten

trianglerows(n) = my(v=[]); for(k=0, n-1, v=Vec((-2+3*x)^k + O(x^(k+1))); print(v))
/* Print initial 10 rows of triangle as follows */
trianglerows(10) \\ Felix Fröhlich, Jul 31 2018

T(0,0) = 1; T(n,k) = -2 * T(n-1,k) + 3 * T(n-1,k-1) for k = 0,1,...,n and T(n,k)=0 for n or k < 0.
T(n, k) = ((-2)^(n - k) 3^k)/((n - k)! k!) n! for k = 0,1..n.
G.f.: 1 / (1 + 2*x - 3*x*t).

cp's OEIS Frontend

User: Zagros Lalo

A318777 Coefficients in expansion of 1/(1 - x - 2*x^5).

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A318776 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.

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A318775 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 2 * T(n-5,k-1) for k = 0..floor(n/5); T(n,k)=0 for n or k < 0.

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A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

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A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

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A318772 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 3 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

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A317499 Coefficients in expansion of 1/(1 + 2x - 3x^3).