A077868 -id:A077868 - cp's OEIS frontend

A344002 Erroneous version of A077868 (if initial 0 is ignored).

0, 1, 2, 3, 5, 8, 12, 18, 27, 30, 49, 77, 118

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), 150-157. [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.]

A171861 Expansion of x(1+x+x^2) / ( (x-1)(x^3+x^2-1) ).

Original entry on oeis.org

1, 2, 4, 6, 9, 13, 18, 25, 34, 46, 62, 83, 111, 148, 197, 262, 348, 462, 613, 813, 1078, 1429, 1894, 2510, 3326, 4407, 5839, 7736, 10249, 13578, 17988, 23830, 31569, 41821, 55402, 73393, 97226, 128798, 170622, 226027, 299423, 396652, 525453, 696078, 922108

Offset: 1

Ed Pegg Jr, Oct 16 2010

Number of wins in Penney's game if the two players start HHT and TTT and HHT beats TTT.

HHT beats TTT 70% of the time. - Geoffrey Critzer, Mar 01 2014

			a(n) enumerates length n+2 sequences on {H,T} that end in HHT but do not contain the contiguous subsequence TTT.
a(3)=4 because we have: TTHHT, THHHT, HTHHT, HHHHT.
a(4)=6 because we have: TTHHHT, THTHHT, THHHHT, HTTHHT, HTHHHT, HHHHHT. - _Geoffrey Critzer_, Mar 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..1000
Wikipedia, Penney's game
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1).

Related sequences are A000045 (HHH beats HHT, HTT beats TTH), A006498 (HHH beats HTH), A023434 (HHH beats HTT), A000930 (HHH beats THT, HTH beats HHT), A000931 (HHH beats TTH), A077868 (HHT beats HTH), A002620 (HHT beats HTT), A000012 (HHT beats THH), A004277 (HHT beats THT), A070550 (HTH beats HHH), A000027 (HTH beats HTT), A097333 (HTH beats THH), A040000 (HTH beats TTH), A068921 (HTH beats TTT), A054405 (HTT beats HHH), A008619 (HTT beats HHT), A038718 (HTT beats THT), A128588 (HTT beats TTT).

Cf. A164315 (essentially the same sequence).

A171861 := proc(n) option remember; if n <=4 then op(n,[1,2,4,6]); else procname(n-1)+procname(n-2)-procname(n-4) ; end if; end proc:

nn=44;CoefficientList[Series[x(1+x+x^2)/(1-x-x^2+x^4),{x,0,nn}],x] (* Geoffrey Critzer, Mar 01 2014 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,1,1]^(n-1)*[1;2;4;6])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

a(n) = a(n-1) +a(n-2) -a(n-4) = A000931(n+10)-3 = A134816(n+6)-3 = A078027(n+12)-3.

a(n) = A164315(n-1). - Alois P. Heinz, Oct 12 2017

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)

A050228 a(n) is the number of subsequences {s(k)} of {1,2,3,...n} such that s(k+1)-s(k) is 1 or 3.

Original entry on oeis.org

1, 3, 6, 11, 19, 31, 49, 76, 116, 175, 262, 390, 578, 854, 1259, 1853, 2724, 4001, 5873, 8617, 12639, 18534, 27174, 39837, 58396, 85596, 125460, 183884, 269509, 394999, 578914, 848455, 1243487, 1822435, 2670925, 3914448, 5736920, 8407883

Offset: 1

John W. Layman, Dec 20 1999

The second differences c(n) of {a(n)} satisfy c(n)=c(n-1)+c(n-3) and give A000930 with the first 5 terms deleted.

Partial sums of A077868. - Paul Barry, Sep 16 2004

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 = 1..1000
Hung Viet Chu, Various sequences from counting subsets, arXiv:2005.10081 [math.CO], 2020-2021.
Z. Kasa, On scattered subword complexity, arXiv preprint arXiv:1104.4425 [cs.DM], 2011.
Index entries for linear recurrences with constant coefficients, signature (3,-3,2,-2,1).

Cf. A000930, A077868, A144898, A144899, A144900, A144901, A144902, A144903, A144904, A226405.

Cf. A078012, A099567, A135851.

A050228:= func< n | n eq 0 select 0 else (&+[Binomial(n-2*j+1, j+2): j in [0..Floor((n+1)/3)]]) >;
[A050228(n): n in [1..40]]; // G. C. Greubel, Jul 27 2022

with(combstruct): SubSetSeqU := [T, {T=Subst(U,U), S=Set(U, card>=3), U=Sequence(Z, card>=3)}, unlabeled]: seq(count(SubSetSeqU, size=n), n=9..46); # Zerinvary Lajos, Mar 18 2008

Rest[CoefficientList[Series[1/((1-x)^2*(1-x-x^3)), {x, 0, 50}], x]] (* G. C. Greubel, Apr 27 2017 *)
LinearRecurrence[{3,-3,2,-2,1},{1,3,6,11,19},50] (* Harvey P. Dale, Apr 21 2020 *)

my(x='x+O('x^50)); Vec(x/((1-x)^3-x^3*(1-x)^2)) \\ G. C. Greubel, Apr 27 2017

def A050228(n): return sum(binomial(n-2*j+1, j+2) for j in (0..((n+1)//3)))
[A050228(n) for n in (1..40)] # G. C. Greubel, Jul 27 2022

From Paul Barry, Sep 16 2004: (Start)
G.f.: x/((1-x)^3 - x^3(1-x)^2).
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - 2*a(n-4) + a(n-5).
a(n-1) = Sum_{k=0..floor(n/3)} binomial(n-2*k, k+2). (End)
G.f. = 1/((1-x)^2*(1-x-x^3)). - N. J. A. Sloane, Jun 02 2021
a(n) = A000930(n+5) - n - 4. - Greg Dresden, Jun 20 2021
From G. C. Greubel, Jul 27 2022: (Start)
a(n) = Sum_{j=0..floor((n+1)/3)} binomial(n-2*j+1, j+2).
a(n) = A099567(n+1, 2). (End)

A098578 a(n) = Sum_{k=0..floor(n/4)} C(n-3*k,k+1).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 9, 13, 18, 25, 35, 49, 68, 94, 130, 180, 249, 344, 475, 656, 906, 1251, 1727, 2384, 3291, 4543, 6271, 8656, 11948, 16492, 22764, 31421, 43370, 59863, 82628, 114050, 157421, 217285, 299914, 413965, 571387, 788673, 1088588, 1502554

Offset: 0

Paul Barry, Sep 16 2004

Partial sums of A003269 (with leading zero).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-1)

Cf. A077868.

I:=[0,1,2,3,4]; [n le 5 select I[n] else 2*Self(n-1) - Self(n-2) + Self(n-4) - Self(n-5): n in [1..30]]; // G. C. Greubel, Feb 03 2018

CoefficientList[Series[x/((1-x)^2-x^4*(1-x)), {x,0,50}], x] (* or *) LinearRecurrence[{2,-1,0,1,-1}, {0,1,2,3,4}, 50] (* G. C. Greubel, Feb 03 2018 *)

x='x+O('x^30); concat([0], Vec(x/((1-x)^2-x^4*(1-x)))) \\ G. C. Greubel, Feb 03 2018

G.f.: x/((1-x)^2-x^4(1-x)) = x / ((x-1)*(x^4+x-1)).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - a(n-5).
a(n) = a(n-1) + a(n-4) + 1.

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)

A077941 Expansion of 1/(1-2*x+x^2+x^3).

Original entry on oeis.org

1, 2, 3, 3, 1, -4, -12, -21, -26, -19, 9, 63, 136, 200, 201, 66, -269, -805, -1407, -1740, -1268, 611, 4230, 9117, 13393, 13439, 4368, -18096, -53999, -94270, -116445, -84621, 41473, 284012, 611172, 896859, 898534, 289037, -1217319, -3622209, -6316136, -7792744, -5647143, 2814594

Offset: 0

N. J. A. Sloane, Nov 17 2002

With three leading zeros, is the inverse binomial transform of A077868, with three leading zeros. - Paul Barry, Oct 22 2004

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

Cf. A077990.

LinearRecurrence[{2, -1, -1}, {1, 2, 3}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)

Vec(1/(1-2*x+x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = sum{k=0..n+3, C(n+3, k)(-1)^(n+3-k)*sum{j=0..floor((k-2)/2), C(k-2-2j, j+1)}}. - Paul Barry, Oct 22 2004

a(n) = sum{k=0..floor(n/3), C(n+1-k,n-3k)*(-1)^k}. - Tani Akinari, Oct 10 2014

cp's OEIS Frontend

A344002 Erroneous version of A077868 (if initial 0 is ignored).

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A171861 Expansion of x*(1+x+x^2) / ( (x-1)*(x^3+x^2-1) ).

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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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A050228 a(n) is the number of subsequences {s(k)} of {1,2,3,...n} such that s(k+1)-s(k) is 1 or 3.

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A098578 a(n) = Sum_{k=0..floor(n/4)} C(n-3*k,k+1).

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

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A171861 Expansion of x(1+x+x^2) / ( (x-1)(x^3+x^2-1) ).