A050228 -id:A050228 - cp's OEIS frontend

A344003 Erroneous version of A050228 (if initial 0 is ignored).

0, 1, 3, 6, 11, 19, 31, 49, 76, 106, 155, 232, 350

Offset: 0

N. J. A. Sloane, Jun 02 2021

dead

Included in accordance with OEIS rule of including published but erroneous sequences in order to serve as pointers to the correct versions.

Chu, Hung Viet. Various Sequences from Counting Subsets, Fib. Quart., 59:2 (May 2021), 152-157. [But beware errors.] [Note: there is a different paper on the arXiv with the same author and tile, but it lacks the sequences. Do not replace this reference with a link to the arXiv version.]

A077868 Expansion of 1/((1-x)*(1-x-x^3)).

Original entry on oeis.org

1, 2, 3, 5, 8, 12, 18, 27, 40, 59, 87, 128, 188, 276, 405, 594, 871, 1277, 1872, 2744, 4022, 5895, 8640, 12663, 18559, 27200, 39864, 58424, 85625, 125490, 183915, 269541, 395032, 578948, 848490, 1243523, 1822472, 2670963, 3914487, 5736960, 8407924, 12322412

Offset: 0

N. J. A. Sloane, Nov 17 2002

Row sums of Riordan array (1/(1-x), x*(1+x^2)). - Paul Barry, Feb 16 2005

a(n) is the number of partitions of {1, ..., n+3} into two blocks in which only 1- or 3-strings of consecutive integers can appear in a block and there is at least one 3-string. E.g., a(3)=5 because the enumerated partitions of {1,2,3,4,5,6} are 1235/46, 1345/26, 15/2346, 13/2456, 123/456. - Augustine O. Munagi, Apr 11 2005

Chu, Hung Viet. "Various Sequences from Counting Subsets." Fib. Quart., 59:2 (May 2021), 150-157.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Kassie Archer and Aaron Geary, Powers of permutations that avoid chains of patterns, arXiv:2312.14351 [math.CO], 2023. See p. 15.
Hung Viet Chu, Various sequences from counting subsets, arXiv:2005.10081 [math.CO], 2020-2021.
A. O. Munagi, Set Partitions with Successions and Separations, IJMMS 2005:3 (2005), 451-463.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-1).

Cf. A000071, A077888, A077941, A105489.
Cf. A000930, A050228, A144898, A144899, A144900, A144901, A144902, A144903, A144904, A226405.
Cf. A078012, A099567, A135851.

A077868:= func< n | n eq 0 select 0 else (&+[Binomial(n-2*j+, j+1): j in [0..Floor((n+1)/3)]]) >;
[A077868(n): n in [0..40]]; // G. C. Greubel, Jul 27 2022

a:= n-> (Matrix(4, (i,j)-> if i=j-1 then 1 elif j=1 then [2,-1,1,-1][i] else 0 fi)^n)[1,1]: seq(a(n), n=0..41); # Alois P. Heinz, Sep 05 2008
g:=(1+z+z^2)/(1-z-z^3): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-1, n=1..42); # Zerinvary Lajos, Jan 09 2009

LinearRecurrence[{1,1,0,0,-1}, {1,2,3,5,8,12}, 42] (* or *)
CoefficientList[Series[1/((1-x)(1-x-x^3)), {x, 0, 41}], x] (* Michael De Vlieger, Jun 06 2018 *)

Vec(1/(1-x)/(1-x-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

{a = vector(50);
a[1] = 1; a[2] = 2; a[3] = 3;
for(n=4,50,
a[n] = 1 + a[n-1] + a[n-3];
); a} \\ Gerry Martens, Jun 03 2018

{a(n) = if( n<0, n=-4-n; polcoeff( -1 / (1 - x) / (1 + x^2 - x^3) + x * O(x^n), n), polcoeff( 1 / (1 - x) / (1 - x - x^3) + x * O(x^n), n))}; /* Michael Somos, Jun 17 2018 */

def A077868(n): return sum(binomial(n-2*j+1, j+1) for j in (0..((n+1)//3)))
[A077868(n) for n in (0..40)] # G. C. Greubel, Jul 27 2022

Partial sums of A000930. a(n-1) = Sum_{k=0..floor(n/2)} binomial(n-2*k, k+1). - Paul Barry, Jul 07 2004
a(n-3) = Sum(binomial(n-r, r)), r=1, 2, ... which is the case t=3 and k=2 in the general case of t-strings and k blocks: a(n-3, k, t) = Sum(binomial(n-r*(t-1), r)*S2(n-r*(t-1)-1, k-1)), r=1, 2, ... - Augustine O. Munagi, Apr 11 2005
From Paul Weisenhorn, Oct 28 2011: (Start)
a(n) = a(n-1) + a(n-2) - a(n-5) for n > 4.
a(n) = a(n-2) + a(n-3) + a(n-4) + 2 for n > 3.
G.f.: 1/((1-x)*(1-x-x^3)). (End)
a(n) = 1 + a(n-1) + a(n-3), a(1)=1, a(2)=2, a(3)=3. - Gerry Martens, Jun 10 2018
a(n) = -A077888(-4-n) for all n in Z. - Michael Somos, Jun 17 2018
a(n) = A000930(n+3) - 1. - Greg Dresden, Jun 20 2021
a(n) = A099567(n+3, 4). - G. C. Greubel, Jul 27 2022

More terms from Augustine O. Munagi, Apr 11 2005

A099567 Riordan array (1/(1-x-x^3), 1/(1-x)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 3, 5, 6, 4, 1, 4, 8, 11, 10, 5, 1, 6, 12, 19, 21, 15, 6, 1, 9, 18, 31, 40, 36, 21, 7, 1, 13, 27, 49, 71, 76, 57, 28, 8, 1, 19, 40, 76, 120, 147, 133, 85, 36, 9, 1, 28, 59, 116, 196, 267, 280, 218, 121, 45, 10, 1, 41, 87, 175, 312, 463, 547, 498, 339, 166, 55, 11, 1

Offset: 0

Paul Barry, Oct 22 2004

Inverse matrix is A099569.

Subtriangle of the triangle in A144903. - Philippe Deléham, Dec 29 2013

			Rows begin:
   1;
   1,  1;
   1,  2,   1;
   2,  3,   3,   1;
   3,  5,   6,   4,   1;
   4,  8,  11,  10,   5,   1;
   6, 12,  19,  21,  15,   6,   1;
   9, 18,  31,  40,  36,  21,   7,   1;
  13, 27,  49,  71,  76,  57,  28,   8,   1;
  19, 40,  76, 120, 147, 133,  85,  36,   9,   1;
  28, 59, 116, 196, 267, 280, 218, 121,  45,  10,   1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000217, A050407, A099568 (row sums), A099569, A144903, A144904.

Columns: A000930, A077868, A050228, A226405, A144898, A144899, A144900, A144901, A144902.

T:= func< n,k | (&+[Binomial(n-2*j, k+j): j in [0..Floor(n/3)]]) >;
[[T(n,k): k in [0..n]]: n in [0..15]]; // G. C. Greubel, Jul 27 2022

T[n_, 0]:=T[n,0]=HypergeometricPFQ[{(1-n)/3,(2-n)/3,-n/3}, {(1-n)/2,-n/2}, -27/4];
T[n_, k_]:= T[n,k]= If[k==n, 1, T[n-1,k-1] +T[n-1,k]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 28 2017 *)

@CachedFunction
def A099567(n, k): return sum( binomial(n-2*j, k+j) for j in (0..(n//3)) )
flatten([[A099567(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 27 2022

Number triangle T(n, k) = Sum_{j=0..floor(n/3)} binomial(n-2*j, k+j).
Columns have g.f. (1/(1-x-x^3))*(x/(1-x))^k.
Sum_{k=0..n} T(n, k) = A099568(n).
T(n,0) = A000930(n), T(n,n) = 1, T(n,k) = T(n-1,k-1) + T(n-1,k) for 0Philippe Deléham, Dec 29 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 3*x + 3*x^2/2! + x^3/3!) = 2 + 5*x + 11*x^2/2! + 21*x^3/3! + 36*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
From G. C. Greubel, Jul 27 2022: (Start)
T(n, n-1) = n, for n >= 1.
T(n, n-2) = A000217(n-1), for n >= 2.
T(n, n-3) = A050407(n+1), for n >= 3.
T(2*n, n) = A144904(n+1), for n >= 1. (End)

A144903 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of x/((1-x-x^3)*(1-x)^(k-1)).

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 1, 0, 1, 3, 3, 2, 1, 0, 1, 4, 6, 5, 3, 1, 0, 1, 5, 10, 11, 8, 4, 2, 0, 1, 6, 15, 21, 19, 12, 6, 3, 0, 1, 7, 21, 36, 40, 31, 18, 9, 4, 0, 1, 8, 28, 57, 76, 71, 49, 27, 13, 6, 0, 1, 9, 36, 85, 133, 147, 120, 76, 40, 19, 9, 0, 1, 10, 45, 121, 218, 280, 267, 196, 116, 59, 28, 13

Offset: 0

Alois P. Heinz, Sep 24 2008

			Square array (A(n,k)) begins:
  0, 0,  0,  0,  0,   0,   0 ... A000004;
  1, 1,  1,  1,  1,   1,   1 ... A000012;
  0, 1,  2,  3,  4,   5,   6 ... A001477;
  0, 1,  3,  6, 10,  15,  21 ... A000217;
  1, 2,  5, 11, 21,  36,  57 ... A050407;
  1, 3,  8, 19, 40,  76, 133 ... ;
  1, 4, 12, 31, 71, 147, 200 ... A027658;
Antidiagonal triangle (T(n,k)) begins as:
  0;
  0,  1;
  0,  1,  0;
  0,  1,  1,  0;
  0,  1,  2,  1,  1;
  0,  1,  3,  3,  2,  1;
  0,  1,  4,  6,  5,  3,  1;
  0,  1,  5, 10, 11,  8,  4,  2;
  0,  1,  6, 15, 21, 19, 12,  6,  3;

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Rows 0-4, 6 give: A000004, A000012, A001477, A000217, A050407(n+3), A027658.
Columns 0-9 give: A078012 and A135851(n+2), A078012(n+2) and A135851(n+4), A077868(n-1) for n>0, A050228(n-1) for n>0, A226405, A144898, A144899, A144900, A144901, A144902.
Main diagonal gives: A144904.
Cf. A000930.

A000930:= func< n | (&+[Binomial(n-2*j,j): j in [0..Floor(n/3)]]) >;
A144903:= func< n,k | k eq 0 select 0 else (&+[Binomial(n-k+j-2,j)*A000930(k-j-1) : j in [0..k-1]]) >;
[A144903(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Aug 01 2022

A:= proc(n,k) coeftayl (x/ (1-x-x^3)/ (1-x)^(k-1), x=0, n) end:
seq(seq(A(n, d-n), n=0..d), d=0..13);

(* First program *)
a[n_, k_] := SeriesCoefficient[x/((1-x-x^3)*(1-x)^(k-1)), {x, 0, n}];
Table[a[n-k, k], {n,0,12}, {k,n,0,-1}]//Flatten (* Jean-François Alcover, Jan 15 2014 *)
(* Second Program *)
A000930[n_]:= A000930[n]= Sum[Binomial[n-2*j,j], {j,0,Floor[n/3]}];
T[n_, k_]:= T[n, k]= If[k==0, 0, Sum[Binomial[n-k+j-2,j]*A000930[k-j-1], {j,0,k- 1}]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2022 *)

def A000930(n): return sum(binomial(n-2*j,j) for j in (0..(n//3)))
def A144903(n,k):
    if (k==0): return 0
    else: return sum(binomial(n-k+j-2,j)*A000930(k-j-1) for j in (0..k-1))
flatten([[A144903(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Aug 01 2022

G.f. of column k: x/((1-x-x^3)*(1-x)^(k-1)).
A(n, n) = A144904(n).
From G. C. Greubel, Aug 01 2022: (Start)
A(n, k) = Sum_{j=0..n-1} binomial(k+j-2, j)*A000930(n-j-1), with A(0, k) = 0.
T(n, k) = Sum_{j=0..k-1} binomial(n-k-j-2, j)*A000930(k-j-1), with T(n, 0) = 0.
T(2*n, n) = A144904(n). (End)

A144898 Expansion of x/((1-x-x^3)*(1-x)^4).

Original entry on oeis.org

0, 1, 5, 15, 36, 76, 147, 267, 463, 775, 1262, 2011, 3150, 4867, 7438, 11268, 16951, 25358, 37766, 56047, 82945, 122482, 180553, 265798, 390880, 574358, 843432, 1237966, 1816384, 2664311, 3907237, 5729077, 8399372, 12313154, 18049371, 26456513, 38778103

Offset: 0

Alois P. Heinz, Sep 24 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,11,-9,7,-4,1).

5th column of A144903.
Cf. A000930, A050228, A077868, A144899, A144900, A144901, A144902, A144903, A144904, A226405.
Cf. A078012, A099567, A135851.

A144898:= func< n | n eq 0 select 0 else (&+[Binomial(n-2*j+3, j+4): j in [0..Floor((n+3)/3)]]) >;
[A144898(n): n in [0..40]]; // G. C. Greubel, Jul 27 2022

a:= n-> (Matrix(7, (i, j)-> if i=j-1 then 1 elif j=1 then [5, -10, 11, -9, 7, -4, 1][i] else 0 fi)^n)[1, 2]: seq(a(n), n=0..40);

CoefficientList[Series[ x/((1-x-x^3)(1-x)^4), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2013 *)

def A144898(n): return sum(binomial(n-2*j+3, j+4) for j in (0..((n+3)//3)))
[A144898(n) for n in (0..40)] # G. C. Greubel, Jul 27 2022

G.f.: x/((1-x-x^3)*(1-x)^4).
From G. C. Greubel, Jul 27 2022: (Start)
a(n) = Sum_{j=0..floor((n+3)/3)} binomial(n-2*j+3, j+4).
a(n) = A099567(n+3, 4). (End)

A144899 Expansion of x/((1-x-x^3)*(1-x)^5).

Original entry on oeis.org

0, 1, 6, 21, 57, 133, 280, 547, 1010, 1785, 3047, 5058, 8208, 13075, 20513, 31781, 48732, 74090, 111856, 167903, 250848, 373330, 553883, 819681, 1210561, 1784919, 2628351, 3866317, 5682701, 8347012, 12254249, 17983326, 26382698, 38695852, 56745223, 83201736

Offset: 0

Alois P. Heinz, Sep 24 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,21,-20,16,-11,5,-1).

6th column of A144903.
Cf. A000930, A050228, A077868, A144898, A144900, A144901, A144902, A144903, A144904, A226405.
Cf. A078012, A099567, A135851.

A144899:= func< n | n eq 0 select 0 else (&+[Binomial(n-2*j+4, j+5): j in [0..Floor((n+4)/3)]]) >;
[A144899(n): n in [0..40]]; // G. C. Greubel, Jul 27 2022

a:= n-> (Matrix(8, (i, j)-> if i=j-1 then 1 elif j=1 then [6, -15, 21, -20, 16, -11, 5, -1][i] else 0 fi)^n)[1, 2]: seq(a(n), n=0..40);

CoefficientList[Series[x/((1-x-x^3)(1-x)^5), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2013 *)

def A144899(n): return sum(binomial(n-2*j+4, j+5) for j in (0..((n+4)//3)))
[A144899(n) for n in (0..40)] # G. C. Greubel, Jul 27 2022

G.f.: x/((1-x-x^3)*(1-x)^5).
From G. C. Greubel, Jul 27 2022: (Start)
a(n) = Sum_{j=0..floor((n+4)/3)} binomial(n-2*j+4, j+5).
a(n) = A099567(n+4, 5). (End)

A226405 Expansion of x/((1-x-x^3)*(1-x)^3).

Original entry on oeis.org

0, 1, 4, 10, 21, 40, 71, 120, 196, 312, 487, 749, 1139, 1717, 2571, 3830, 5683, 8407, 12408, 18281, 26898, 39537, 58071, 85245, 125082, 183478, 269074, 394534, 578418, 847927, 1242926, 1821840, 2670295, 3913782, 5736217, 8407142, 12321590, 18058510, 26466393

Offset: 0

Alois P. Heinz, Jun 06 2013

From Bruno Berselli, Jun 07 2013: (Start)
A050228(n)  = a(n)   -a(n-1), n>0.
A077868(n-1)= a(n) -2*a(n-1)   +a(n-2), n>1.
A000217(n)  = a(n)   -a(n-1)             -a(n-3), n>2.
A000930(n-1)= a(n) -3*a(n-1) +3*a(n-2)   -a(n-3), n>2.
n           = a(n) -2*a(n-1)   +a(n-2)   -a(n-3)   +a(n-4), n>3.
1           = a(n) -3*a(n-1) +3*a(n-2) -2*a(n-3) +2*a(n-4)   -a(n-5), n>4.
0           = a(n) -4*a(n-1) +6*a(n-2) -5*a(n-3) +4*a(n-4) -3*a(n-5) +a(n-6), n>5.
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-4,3,-1).

4th column of A144903.
Cf. A000217, A078012, A099567, A135851.
Cf. A000930, A050228, A077868, A144898, A144899, A144900, A144901, A144902, A144903, A144904, A226405.

A226405:= func< n | n eq 0 select 0 else (&+[Binomial(n-2*j+2, j+3): j in [0..Floor((n+2)/3)]]) >;
[A226405(n): n in [0..40]]; // G. C. Greubel, Jul 27 2022

a:= n-> (Matrix(6, (i, j)-> if i=j-1 then 1 elif j=1 then [4, -6, 5, -4, 3, -1][i] else 0 fi)^n)[1, 2]: seq(a(n), n=0..40);

LinearRecurrence[{4,-6,5,-4,3,-1}, {0,1,4,10,21,40}, 40]  (* Bruno Berselli, Jun 07 2013 *)
CoefficientList[Series[x/((1-x-x^3)*(1-x)^3), {x, 0, 50}], x] (* G. C. Greubel, Apr 28 2017 *)

my(x='x+O('x^50)); Vec(x/((1-x-x^3)*(1-x)^3)) \\ G. C. Greubel, Apr 28 2017

def A226405(n): return sum(binomial(n-2*j+2, j+3) for j in (0..((n+2)//3)))
[A226405(n) for n in (0..40)] # G. C. Greubel, Jul 27 2022

G.f.: x/((1-x-x^3)*(1-x)^3).
From G. C. Greubel, Jul 27 2022: (Start)
a(n) = Sum_{j=0..floor((n+2)/3)} binomial(n-2*j+2, j+3).
a(n) = A099567(n+2, 3). (End)

A048516 Array T read by diagonals: T(m,n)=number of subsets S of {1,2,3,...,m+n-1} such that |S|>1 and |a-b|>=m for all distinct a and b in S, m=1,2,3,...; n=1,2,3,...

Original entry on oeis.org

0, 0, 1, 0, 4, 1, 0, 11, 3, 1, 0, 26, 7, 3, 1, 0, 57, 14, 6, 3, 1, 0, 120, 26, 11, 6, 3, 1, 0, 247, 46, 19, 10, 6, 3, 1, 0, 502, 79, 31, 16, 10, 6, 3, 1, 0, 1013, 133, 49, 25, 15, 10, 6, 3, 1, 0, 2036, 221, 76, 38, 22, 15, 10, 6, 3, 1, 0, 4083, 364

Offset: 1

Clark Kimberling

			Diagonals: {0}; {1,0}; {4,1,0}; ...

Sean A. Irvine, Java program (github)

A000295 (row 1), A001924 (row 2), A050228 (row 3).

T(m,n) = [x^m] 1/((1-x^2)*(1-x-x^n)). - Sean A. Irvine, Jun 19 2021

A335184 a(n) is the number of subsets of {1,2,...,n} with at least two elements and the difference between successive elements at least 6.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 3, 6, 10, 15, 21, 29, 40, 55, 75, 101, 134, 176, 230, 300, 391, 509, 661, 856, 1106, 1427, 1840, 2372, 3057, 3938, 5070, 6524, 8392, 10793, 13880, 17849, 22951, 29508, 37934, 48762, 62678, 80564, 103553, 133100, 171074, 219877, 282597, 363204, 466801, 599946, 771066, 990990

Offset: 0

Enrique Navarrete, May 25 2020

nonn

For n >= 6 the sequence contains the triangular numbers; for n >= 12 we have to add the tetrahedral numbers; for n >= 18 we have to add the numbers binomial(n,4) (starting with 0,1,5,...); for n >= 24 we have to add the numbers binomial(n,5) (starting with 0,1,6,..); in general, for n >= 6*k we have to add to the sequence the numbers binomial(n, k+1), k >= 1.
For example, a(26) = 1106 = 210+560+330+6, where 210 is a triangular number, 560 is a tetrahedral number, 330 is a number binomial(n,4) and 6 is a number binomial(m,5) (with the proper n, m due to shifts in names of the sequences).
The sequence counts sets with more than 2 elements such as {1,7,14}, {1,8,14}, {2,8,14,20}, etc. The first 3-element set is {1,7,13}, the first 4-element set is {1,7,13,19}, etc. Every time a larger set needs to be counted is when we have to add a term binomial(n, k+1).

			a(11) = 15 and the 15 subsets of {1,2,...11} with at least two elements and whose difference between successive elements is at least 6 are: {1,7}, {1,8}, {1,9}, {1,10}, {1,11}, {2,8}, {2,9}, {2,10}, {2,11}, {3,9}, {3,10}, {3,11}, {4,10}, {4,11}, {5,11}.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,1,-2,1).

Similar sequences with minimum difference 1..5 are A000295, A001924, A050228, A145131, A330910.

With[{k = 6}, Array[Count[Subsets[Range[# + k], {2, # + k}], ?(AllTrue[Differences@ #, # >= k &] &)] &, 16]] (* _Michael De Vlieger, Jun 26 2020 *)
LinearRecurrence[{3,-3,1,0,0,1,-2,1},{0,0,0,0,0,0,0,1},60] (* Harvey P. Dale, Nov 22 2022 *)

a(n) = {my(d=6); sum(k=0, (n-1)\d, binomial(n-d*k+k+1, k+2))} \\ Andrew Howroyd, Aug 11 2020

a(n) = Sum_{k=0..floor((n-1)/6)} binomial(n-6*k+k+1, k+2). - Andrew Howroyd, Aug 11 2020
From Colin Barker, May 26 2020: (Start)
G.f.: x^7 / ((1 - x)^2*(1 - x - x^6)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-6) - 2*a(n-7) + a(n-8) for n>=8.
(End)

cp's OEIS Frontend

A344003 Erroneous version of A050228 (if initial 0 is ignored).

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A077868 Expansion of 1/((1-x)*(1-x-x^3)).

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A099567 Riordan array (1/(1-x-x^3), 1/(1-x)).

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A144903 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of x/((1-x-x^3)*(1-x)^(k-1)).

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A144898 Expansion of x/((1-x-x^3)*(1-x)^4).

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A144899 Expansion of x/((1-x-x^3)*(1-x)^5).

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A226405 Expansion of x/((1-x-x^3)*(1-x)^3).

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