A128099 -id:A128099 - cp's OEIS frontend

A317494 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.

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, 80, 1, 20, 112, 160, 16, 1, 22, 144, 280, 80, 1, 24, 180, 448, 240, 1, 26, 220, 672, 560, 32, 1, 28, 264, 960, 1120, 192, 1, 30, 312, 1320, 2016, 672, 1, 32, 364, 1760, 3360, 1792, 64

Offset: 0

Zagros Lalo, Jul 30 2018

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.)
The coefficients in the expansion of 1/(1-x-2*x^3) are given by the sequence generated by the row sums.
The row sums give A003229.
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.

			Triangle begins:
  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,   80;
  1, 20, 112,  160,   16;
  1, 22, 144,  280,   80;
  1, 24, 180,  448,  240;
  1, 26, 220,  672,  560,   32;
  1, 28, 264,  960, 1120,  192;
  1, 30, 312, 1320, 2016,  672;
  1, 32, 364, 1760, 3360, 1792, 64;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 358, 359

Row sums give A003229.

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

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

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[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

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

A317495 Triangle read by rows: T(0,0) = 1; T(n,k) =2 * T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 2, 4, 8, 1, 16, 4, 32, 12, 64, 32, 1, 128, 80, 6, 256, 192, 24, 512, 448, 80, 1, 1024, 1024, 240, 8, 2048, 2304, 672, 40, 4096, 5120, 1792, 160, 1, 8192, 11264, 4608, 560, 10, 16384, 24576, 11520, 1792, 60, 32768, 53248, 28160, 5376, 280, 1, 65536, 114688, 67584, 15360, 1120, 12

Offset: 0

Zagros Lalo, Jul 30 2018

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and along a "second layer" of 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/(1-2x-x^3) are given by the sequence generated by the row sums.
The row sums give A008998 and Pisot sequences E(4,9), P(4,9) when n > 1, see A020708.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.205569430400..., when n approaches infinity.

			Triangle begins:
       1;
       2;
       4;
       8,      1;
      16,      4;
      32,     12;
      64,     32,      1;
     128,     80,      6;
     256,    192,     24;
     512,    448,     80,      1;
    1024,   1024,    240,      8;
    2048,   2304,    672,     40;
    4096,   5120,   1792,    160,     1;
    8192,  11264,   4608,    560,    10;
   16384,  24576,  11520,   1792,    60;
   32768,  53248,  28160,   5376,   280,   1;
   65536, 114688,  67584,  15360,  1120,  12;
  131072, 245760, 159744,  42240,  4032,  84;
  262144, 524288, 372736, 112640, 13440, 448, 1;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 358, 359.

Row sums give A008998, A020708.
Cf. A013609
Cf. A038207
Cf. A317494
Cf. A128099, A207538.
Cf. A000079 (column 0), A001787 (column 1), A001788 (column 2), A001789 (column 3), A003472 (column 4).

Flat(List([0..20],n->List([0..Int(n/3)],k->2^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Jul 31 2018

/* As triangle */ [[2^(n-3*k)/(Factorial(n-3*k)*Factorial(k))* Factorial(n-2*k): k in [0..Floor(n/3)]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 05 2018

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

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

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).

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).

A109767 Triangle T(n,k), 0 <= k <= n, defined by T(n,k) = 2^k*A001497(n,k).

Original entry on oeis.org

1, 2, 2, 12, 12, 4, 120, 120, 48, 8, 1680, 1680, 720, 160, 16, 30240, 30240, 13440, 3360, 480, 32, 665280, 665280, 302400, 80640, 13440, 1344, 64, 17297280, 17297280, 7983360, 2217600, 403200, 48384, 3584, 128, 518918400, 518918400

Offset: 0

Philippe Deléham, Aug 12 2005

Also square array of unsigned coefficients of Hermite polynomials.

T[n,k]is A128099(2n,k)*A001813(n-k). - Richard Turk, Sep 26 2017

			Rows begin:
     1
     2,    2,
    12,   12,   4,
   120,  120,  48,   8,
  1680, 1680, 720, 160, 16,
Unsigned coefficients of Hermite polynomials:
     1,     2,      4,       8, ...
     2,    12,     48,     160, ...
    12,   120,    720,    3360, ...
   120,  1680,  13440,   80640, ...
  1680, 30240, 302400, 2217600, ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Cf. A001497.

/* As triangle */ [[Factorial(2*n-k)*2^k/(Factorial(k)*Factorial(n-k)): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 14 2015

seq(seq((2*n-k)!*2^k/(k!*(n-k)!),k=0..n),n=0..10); # Robert Israel, Sep 26 2017

y[n_, x_] := Sqrt[2/(Pi*x)]*E^(1/x)*BesselK[-n-1/2, 1/x]; t[n_, k_] := 2^n*Coefficient[y[n, x], x, k]; Table[t[n, k], {n, 0, 8}, {k, n, 0, -1}] // Flatten (* or *) t[n_, k_] := (2*n - k)!*2^k/(k!*(n-k)!); Table[t[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 01 2013 *)
Table[((2n-k)!*2^k)/(k!(n-k)!),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Nov 23 2017 *)

T(n,k) = (2n-k)!*2^k/(k!*(n-k)!).

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).

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).

A191897 Coefficients of the Z(n,x) polynomials; Z(0,x) = 1, Z(1,x) = x and Z(n,x) = xZ(n-1,x) - 2Z(n-2,x), n >= 2.

Original entry on oeis.org

1, 1, 0, 1, 0, -2, 1, 0, -4, 0, 1, 0, -6, 0, 4, 1, 0, -8, 0, 12, 0, 1, 0, -10, 0, 24, 0, -8, 1, 0, -12, 0, 40, 0, -32, 0, 1, 0, -14, 0, 60, 0, -80, 0, 16, 1, 0, -16, 0, 84, 0, -160, 0, 80, 0, 1, 0, -18, 0, 112, 0, -280, 0, 240, 0, -32

Offset: 0

Paul Curtz, Jun 19 2011

The coefficients of the Z(n,x) polynomials by decreasing exponents, see the formulas, define this triangle.

			The first few rows of the coefficients of the Z(n,x) are
  1;
  1,    0;
  1,    0,   -2;
  1,    0,   -4,    0;
  1,    0,   -6,    0,    4;
  1,    0,   -8,    0,   12,    0;
  1,    0,  -10,    0,   24,    0,   -8;
  1,    0,  -12,    0,   40,    0,  -32,    0;
  1,    0,  -14,    0,   60,    0,  -80,    0,   16;
  1,    0,  -16,    0,   84,    0, -160,    0,   80,    0;

Row sums: A107920(n+1). Main diagonal: A077966(n).
Z(n,x=1) = A107920(n+1), Z(n,x=2)  = A009545(n+1),
Z(n,x=3) = A000225(n+1), Z(n,x=4)  = A007070(n),
Z(n,x=5) = A107839(n),   Z(n,x=6)  = A154244(n),
Z(n,x=7) = A186446(n),   Z(n,x=8)  = A190975(n+1),
Z(n,x=9) = A190979(n+1), Z(n,x=10) = A190869(n+1).
Row sum without sign: A113405(n+1).
Cf. A128099, A013609.

nmax:=10: Z(0, x):=1 : Z(1, x):=x: for n from 2 to nmax do Z(n, x) := x*Z(n-1, x) - 2*Z(n-2, x) od: for n from 0 to nmax do for k from 0 to n do T(n, k) := coeff(Z(n, x), x, n-k) od: od: seq(seq(T(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jun 27 2011, revised Nov 29 2012

a[n_, k_] := If[OddQ[k], 0, 2^(k/2)*Coefficient[ ChebyshevU[n, x/2], x, n-k]]; Flatten[ Table[ a[n, k], {n, 0, 10}, {k, 0, n}]] (* Jean-François Alcover, Aug 02 2012, from 2nd formula *)

Z(0,x) = 1, Z(1,x) = x and Z(n,x) = x*Z(n-1,x) - 2*Z(n-2,x), n >= 2.

a(n,k) = A077957(k) * A053119(n,k). - Paul Curtz, Sep 30 2011

Edited and information added by Johannes W. Meijer, Jun 27 2011

A246047 Irregular triangle read by rows, of partial serial probabilities T(n,k)_{2,3} (see "comments" for definitions and explanation).

Original entry on oeis.org

1, 0, 2, 4, 4, 16, 8, 16, 48, 16, 96, 128, 64, 32, 384, 320, 512, 64, 1280, 256, 768, 2560, 128, 3840, 2560, 1792, 10240, 1024, 256, 10752, 15360, 4096, 35840, 12288, 512, 28672, 71680, 4096, 9216, 114688, 86016, 1024, 73728, 286720, 57344

Offset: 0

Bob Selcoe, Aug 26 2014

Define "serial probability" as the probability that n will occur as a partial sum in an infinite sequence of numbers drawn randomly from set J = {j1,j2,..jz}, where 1 <= j1 < j2< ... < jz and z is the number of members in set J. Generally, serial probabilities are found by the recurrence equation: a(n) = (z^(j1-1)*a(n-j1) + z^(j2-1)*a(n-j2) + z^(j3-1)*a(n-j3) + ... + z^(jz-1)*a(n-jz))/z^n, where a(0)=1 and a(n)=0 when n < 0.
Denote the recurrence sequence for set J as S(n){J}, and denote serial probability (P) for set J as P(n){J}, such that P = S(n){J}/z^n. For example, S(n){2,3} = 2*a(n-2) + 4*a(n-3); therefore P(n)A176739(n)/2%5En;%20so%20for%20example,%20since%20A176739(9)%20=%20192,%20the%20probability%20that%209%20will%20occur%20as%20a%20partial%20sum%20in%20a%20randomly-generated%20infinite%20sequence%20of%202s%20and%203s%20is%20192/512%20=%203/8.%20That%20is,%20P(9)">{2,3} = (2*a(n-2) + 4*a(n-3))/2^n. This also is equivalent to A176739(n)/2^n; so for example, since A176739(9) = 192, the probability that 9 will occur as a partial sum in a randomly-generated infinite sequence of 2s and 3s is 192/512 = 3/8. That is, P(9){2,3} = 3/8.
Define "partial serial probability" (P'') as the probability that n would occur given the different ways to sort the compositions (ordered partitions) of n into j1's..jz's; and let T(n,k){J} be the triangle of partial serial probabilities for set J, such that P'' = T(n,k){J}/z^n. Denote these probabilities as P''(n,k)_{J}.
This triangle therefore is T(n,k){2,3}, and P''(n,k){2,3} = T(n,k)_{2,3}/2^n.
In general, row sums of T(n,k){J} are S(n){J}; thus, the row sums of T(n,k)A176739(n)%20and%20sums%20of%20P''(n,k)">{2,3} are A176739(n) and sums of P''(n,k){2,3} are A176739(n)/2^n.
For T(n,k)A176739(9)%20=%20192,%20the%20probability%20that%209%20will%20occur%20as%20a%20partial%20sum%20with%20three%20sorts%20of%202s%20and%20one%20sort%20of%203s%20is%20128/512%20=%201/4%20(n=9,%20k=0),%20and%20with%20zero%20sorts%20of%202s%20and%20three%20sorts%20of%203s%20is%2064/512%20=%201/8%20(n=9,%20k=1),%20totaling%20192/512%20=%203/8.%20That%20is,%20P''(9,0)">{2,3}: there are [(n-3*(n mod 2)-6k)/2] sorts of 2s, and [2k+(n mod 2)] sorts of 3s. So taking again example A176739(9) = 192, the probability that 9 will occur as a partial sum with three sorts of 2s and one sort of 3s is 128/512 = 1/4 (n=9, k=0), and with zero sorts of 2s and three sorts of 3s is 64/512 = 1/8 (n=9, k=1), totaling 192/512 = 3/8. That is, P''(9,0){2,3} = 1/4 and P''(9,1)_{2,3} = 1/8.
Given n, maximum k for T(n,k)_{2,3} is A103221(n)-1. That is, row lengths are floor(n/6)+1 unless n == 1 (mod 6); if n == 1 (mod 6), row length is floor(n/6).

			Triangle starts:
     1;
     0;
     2;
     4;
     4;
    16;
     8,    16;
    48;
    16,    96;
   128,    64;
    32,   384;
   320,   512;
    64,  1280,  256;
   768,  2560;
   128,  3840, 2560;
  1792, 10240, 1024;
E.g., T(13,0) = 768 because 768/2^13 (3/32) is the probability that 13 will occur as a partial sum in a randomly-generated infinite sequence of 2s and 3s, where the compositions of 13 are into five sorts of 2s and one sort of 3s. In other words, P''(13,0)_{2,3} = 3/32. The sorts are 5 and 1, respectively, because (13 - 3*(13 mod 2) - 6*0)/2 = 5 and 2*0 + (13 mod 2) = 1.

Cf. A176739, A007318 (binomial(n,k)), A103221, A128099 (related sequence).

tabf(nn) = {for (n=0, nn, for (k=0, max(0, (n+2)\2 - (n+2)\3 - 1), tnk = binomial((n - (n % 2) -2*k)/2, 2*k + (n % 2)) * 2^((n + (n % 2) + 2*k)/2); print1(tnk, ", ");); print(););} \\ Michel Marcus, Sep 26 2014

T(n,k) = binomial((n - (n mod 2) - 2*k)/2, 2*k + (n mod 2)) * 2^((n + (n mod 2) + 2*k)/2).

More terms from Michel Marcus, Sep 26 2014

cp's OEIS Frontend

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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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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.

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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.

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A191897 Coefficients of the Z(n,x) polynomials; Z(0,x) = 1, Z(1,x) = x and Z(n,x) = xZ(n-1,x) - 2Z(n-2,x), n >= 2.