A013609 -id:A013609 - cp's OEIS frontend

A228484 a(n) = 2^n(3n)!/(n!(2n)!).

1, 6, 60, 672, 7920, 96096, 1188096, 14883840, 188280576, 2399654400, 30766095360, 396363202560, 5126871859200, 66538909237248, 866061993246720, 11300615801536512, 147773778404769792, 1936073567335219200, 25408660721789829120, 333963051307735449600

Offset: 0

Johannes W. Meijer, Aug 22 2013

Oblique diagonal of the Pell-Jacobsthal triangle A013609. Its mirror diagonal is A006588.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A005809, A013609, A006588, A137209.

[2^n*Factorial(3*n)/(Factorial(n)*Factorial(2*n)): n in [0..20]]; // Vincenzo Librandi, Aug 24 2013

a := n -> 2^n*binomial(3*n, n): seq(a(n), n=0..16);

Table[2^n (3 n)!/(n! (2 n)!), {n, 0, 20}] (* Vincenzo Librandi, Aug 24 2013 *)

a(n) = 2^n*binomial(3*n, 2*n); \\ Michel Marcus, Mar 06 2022

a(n) = 2^n*A005809(n).
a(n) = A013609(3*n, n).
a(n) = A006588(n)/2^n.
a(n) = (2*n+1)*A153231(n).
Asymptotic approximation of a(n) ~ C*(13.5)^n/sqrt(n) with C = (1/2)*sqrt(3/Pi) = A137209.
Sum_{n>=0} 1/a(n) = (11*Pi - 12*log(2) + 270)/250. - Amiram Eldar, Mar 06 2022
From Karol A. Penson, Feb 26 2025: (Start)
G.f.:   hypergeom([1/3,2/3],[1/2],27*z/2).
E.g.f.: hypergeom([1/3,2/3],[1/2,1],27*z/2). (End)

More terms from Vincenzo Librandi, Aug 24 2013

A249307 Triangle read by rows: A249095(n,k) * 2^k, k = 0 .. 2*n+1.

Original entry on oeis.org

1, 1, 2, 4, 1, 2, 8, 8, 16, 1, 2, 12, 16, 48, 32, 64, 1, 2, 16, 24, 96, 96, 256, 128, 256, 1, 2, 20, 32, 160, 192, 640, 512, 1280, 512, 1024, 1, 2, 24, 40, 240, 320, 1280, 1280, 3840, 2560, 6144, 2048, 4096, 1, 2, 28, 48, 336, 480, 2240, 2560, 8960, 7680

Offset: 0

Reinhard Zumkeller, Nov 14 2014

Length of row n = 2*n+1;
T(n,2*n+1) = 4^n = A000302(n);
for n > 0: sum of row n = 7*5^(n-1), cf. A005055.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A249095, A000079, A005408 (row lengths), A249308 (central terms), A000302 (right edge), A005055 (row sums, except for initial 1), A013609.

Cf. A013609.

a249307 n k = a249307_tabf !! n !! k
a249307_row n = a249307_tabf !! n
a249307_tabf = map (zipWith (*) a000079_list) a249095_tabf

A114192 Riordan array (1/(1-2x),x/(1-2x)^2).

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 24, 10, 1, 16, 80, 60, 14, 1, 32, 240, 280, 112, 18, 1, 64, 672, 1120, 672, 180, 22, 1, 128, 1792, 4032, 3360, 1320, 264, 26, 1, 256, 4608, 13440, 14784, 7920, 2288, 364, 30, 1, 512, 11520, 42240, 59136, 41184, 16016, 3640, 480, 34, 1

Offset: 0

Paul Barry, Nov 16 2005

Factors as (1/(1-x),x/(1-x))*(1/(1-x),x*(1+x)/(1-x)^2) or A007318 times A114188. Also (1/(1-2*x),x/(1-2*x))*(1,x*(1+2*x)). Inverse is A114193. Row sums are A007583. Diagonal sums are A007051.

			Triangle begins
1;
2, 1;
4, 6, 1;
8, 24, 10, 1;
16, 80, 60, 14, 1;
32, 240, 280, 112, 18, 1;

T(n,k) = sum{j=0..n, C(k, j)*C(n, k+j)}*2^(n-k).
T(n,k) = 2^(n-k)*binomial(n+k,2k) = 2^(n-k)*A085478(n,k). - Philippe Deléham, May 05 2006
T(n,k) = A013609(n+k, n-k). - Johannes W. Meijer, Sep 05 2013
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 4*T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014

A188440 Triangle T(n,k) read by rows: number of size-k antisymmetric subsets of {1,2,...,n}.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 4, 4, 1, 4, 4, 1, 6, 12, 8, 1, 6, 12, 8, 1, 8, 24, 32, 16, 1, 8, 24, 32, 16, 1, 10, 40, 80, 80, 32, 1, 10, 40, 80, 80, 32, 1, 12, 60, 160, 240, 192, 64, 1, 12, 60, 160, 240, 192, 64, 1, 14, 84, 280, 560, 672, 448, 128, 1, 14, 84, 280

Offset: 0

Dennis P. Walsh, Mar 31 2011

A subset S of {1,2,...,n} is antisymmetric if x is an element of S implies n+1-x is not an element of S. In other words, the sum of any two elements of S does not equal n+1. For example, {1,2,5} is an antisymmetric subset of {1,2,3,4,5,6,7}. If n is odd, (n+1)/2 cannot be an element of an antisymmetric subset of {1,2,...,n}. (Note that for n=0, we define {1,...,n} to be the empty set, and thus T(0,0)=1 since the empty set is vacuously antisymmetric.)
We note, for example, that T(100,k) provides the number of possible size-k committees of the U.S. Senate in which no two members are from the same state.
Triangle, read by rows, A013609 rows repeated. - Philippe Deléham, Apr 09 2012
Triangle, with zeros omitted, given by (1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 09 2012

			Triangle T(n,k) initial values 0 <= k <= floor(n/2), n=0..13:
  1
  1
  1   2
  1   2
  1   4   4
  1   4   4
  1   6  12   8
  1   6  12   8
  1   8  24  32  16
  1   8  24  32  16
  1  10  40  80  80  32
  1  10  40  80  80  32
  1  12  60 160 240 192  64
  1  12  60 160 240 192  64
  ...
For n=7 and k=2, T(7,2)=12 since there are 12 antisymmetric size-2 subsets of {1,2,...,7}:
  {1,2}, {1,3}, {1,5}, {1,6}, {2,3}, {2,5},
  {2,7}, {3,6}, {3,7}, {5,6}, {5,7}, and {6,7}.
(1, 0, -1, 0, 0, 0, 0, ...) DELTA (0, 2, -2, 0, 0, 0, 0, ...) begins:
  1
  1   0
  1   2   0
  1   2   0   0
  1   4   4   0   0
  1   4   4   0   0   0
  1   6  12   8   0   0   0
  1   6  12   8   0   0   0   0
  1   8  24  32  16   0   0   0   0
  1   8  24  32  16   0   0   0   0   0
  1  10  40  80  80  32   0   0   0   0   0
  1  10  40  80  80  32   0   0   0   0   0   0
  1  12  60 160 240 192  64   0   0   0   0   0   0
  1  12  60 160 240 192  64   0   0   0   0   0   0   0
- _Philippe Deléham_, Apr 09 2012

T. D. Noe, Rows n = 0..100, flattened
Dennis Walsh, Notes on antisymmetric subsets of {1,2,...,n}

Cf. A108411, row sums of triangle T(n,k).

Cf. A000079, A007318, A013109, A152198

seq(seq(binomial(floor(n/2),k)*2^k,k=0..floor(n/2)),n=0..22);

Table[ CoefficientList[(1 + 2*x)^n, x] , {n, 0, 7}, {2}] // Flatten (* Jean-François Alcover, Aug 19 2013, after Philippe Deléham *)

T(n,k) = 2^k*C(floor(n/2),k) where C(*,*) denotes a binomial coefficient.
Sum(T(n,k),k=0..floor(n/2)) = 3^floor(n/2) = A108411(n).
G.f. for columns(k fixed):(2t^2)^k/((1-t)*(1-t^2)^k).
T(n,k) = A152198(n,k)*2^k. - Philippe Deléham, Apr 09 2012
G.f.: (1+x)/(1-x^2-2*y*x^2). - Philippe Deléham, Apr 09 2012
T(n,k) = T(n-2,k) + 2*T(n-2,k-1), T(0,0) = T(1,0) = 1, T(1,1) = 0 and T(n,k) = 0 if k<0 or if k>n.- Philippe Deléham, Apr 09 2012

A277513 Irregular triangle read by rows: T(n,k) is the number of integers greater than 4 such that they have n trits and 2k+1 (k>=1) nonzero trits in their balanced ternary representation, with n>=3 and 1<=k<=(j-1)/2.

Original entry on oeis.org

4, 12, 24, 16, 40, 80, 60, 240, 64, 84, 560, 448, 112, 1120, 1792, 256, 144, 2016, 5376, 2304, 180, 3360, 13440, 11520, 1024, 220, 5280, 29568, 42240, 11264, 264, 7920, 59136, 126720, 67584, 4096, 312, 11440, 109824, 329472, 292864, 53248, 364, 16016, 192192

Offset: 3

Lei Zhou, Oct 18 2016

This is a subset of A013609 and A188440.
This sequence T(n,k) can be re-indexed into the form of b(m) where m is positive integer sequence that can be calculated by parametric function m = m(n,k): m(n=2j+1,k) = j^2-j+k, where n is odd and 1<=k<=j.
m(n=2j,k) = j^2-2j+1+k, where n is even and 1<=k<=j.
Here n is the number of digits of an odd number in balanced ternary representation, and 2k+1 is the number of nonzero trits (1 or T) of the same number in balanced ternary representation.

			Odd numbers that can be expressed in 3 trits balanced ternary (bt) form are 5 = 1TT, 7 = 1T1, 9 = 100, 11 = 11T, 13 = 111. Among these five numbers, four have 3 nonzero digits, so a(1) = 4.
Odd numbers in 4 trits bt form are 15 = 1TT0, 17 = 1T0T, 19 = 1T01, 21 = 1T10, 23 = 10TT, 25 = 10T1, 27 = 1000, 29 = 101T, 31 = 1011, 33 = 11T0, 35 = 110T, 37 = 1101, 39 = 1110. Among these 13 numbers, 12 have 3 nonzero digits, so a(2) = 12.
The irregular triangle begins:
        k=1      2      3      4       5      6      7
n=3       4
n=4      12
n=5      24     16
n=6      40     80
n=7      60    240     64
n=8      84    560    448
n=9     112   1120   1792    256
n=10    144   2016   5376   2304
n=11    180   3360  13440  11520    1024
n=12    220   5280  29568  42240   11264
n=13    264   7920  59136 126720   67584   4096
n=14    312  11440 109824 329472  292864  53248
n=15    364  16016 192192 768768 1025024 372736  16384
These are the odd columns with the 1st column removed in the table in A013609.

Lei Zhou, Table of n, a(n) for n = 3..10002
Definition of Balanced Ternary.

Cf. A013609, A188440, A134021, A134022, A134023.

a = {}; Do[Do[ct = 2^(2k)*Binomial[n - 1, 2k]; AppendTo[a, ct], {k, 1, Floor[(n-1)/2]}], {n, 3, 15}]; a

T(n,k) = 2^(2k)*Binomial(n-1, 2k)

A300700 Triangle read by rows: T(n, n-k) = number of k-faces of the concertina n-cube.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 18, 42, 26, 1, 58, 252, 344, 150, 1, 190, 1420, 3380, 3230, 1082, 1, 614, 7770, 29200, 47130, 34452, 9366

Offset: 0

Tilman Piesk, Mar 11 2018

n-place formulas in first-order logic like Ax Ey P(x, y) can be ordered by implication. This Hasse diagram can be interpreted as an n-dimensional convex polytope with face dimensions ranging from 0 (the vertices) to n (the polytope itself).
The right diagonal (n-k = 0, number of vertices) is A000629, which is twice an ordered Bell number (A000670) for n>0.
The second right diagonal (n-k = 1, number of edges) is A300693.
The second left diagonal (k = 1, number of facets) is 2, 6, 18, 58, 190, 614, ... (not to be confused with A151282 or A193777).
The third left diagonal (k = 2, number of ridges) is 6, 42, 252, 1420, 7770, ...
The row sums are A300701. The central diagonal starts 1, 6, 252, 29200 and the row maxima start 1, 2, 6, 42, 344, 3380, 47130.
The corresponding triangle for hypercubes is A013609, and its row sums are A000244 (powers of 3). That for permutohedra is A019538, and its row sums are A000670 (ordered Bell numbers).

			First rows of the triangle:
  k      0     1     2     3     4     5     6         sums = A300701
n
0        1                                                1
1        1     2                                          3
2        1     6     6                                   13
3        1    18    42    26                             87
4        1    58   252   344   150                      805
5        1   190  1420  3380  3230  1082               9303
6        1   614  7770 29200 47130 34452  9366       128533
T(3, 3-1) = T(3, 2) = 42 is the number of 1-faces (edges) of the concertina 3-cube. It has 26 vertices, 42 edges, 18 faces and 1 cell.
In the reflected triangle the column number is the dimension of the counted faces:
  n-k    0     1     2     3     4     5     6
n
0        1
1        2     1
2        6     6     1
3       26    42    18     1
4      150   344   252    58     1
5     1082  3230  3380  1420   190     1
6     9366 34452 47130 29200  7770   614     1

Tilman Piesk, Formulas in predicate logic (Wikiversity)
Tilman Piesk, Skeleton and solid representation of a concertina cube
Tilman Piesk, SAGE code used to generate the sequence

Cf. A300701, A000629, A300693.
Cf. A013609, A000244 (for hypercubes).
Cf. A019538, A000670 (for permutohedra).

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.

Original entry on oeis.org

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).

cp's OEIS Frontend

A228484 a(n) = 2^n*(3*n)!/(n!*(2*n)!).

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A249307 Triangle read by rows: A249095(n,k) * 2^k, k = 0 .. 2*n+1.

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A188440 Triangle T(n,k) read by rows: number of size-k antisymmetric subsets of {1,2,...,n}.

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A300700 Triangle read by rows: T(n, n-k) = number of k-faces of the concertina n-cube.

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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.

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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.

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A228484 a(n) = 2^n(3n)!/(n!(2n)!).