cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 31-40 of 84 results. Next

A080923 First differences of A003946.

Original entry on oeis.org

1, 3, 8, 24, 72, 216, 648, 1944, 5832, 17496, 52488, 157464, 472392, 1417176, 4251528, 12754584, 38263752, 114791256, 344373768, 1033121304, 3099363912, 9298091736, 27894275208, 83682825624, 251048476872, 753145430616
Offset: 0

Views

Author

Paul Barry, Feb 26 2003

Keywords

Comments

Sum of consecutive pairs of elements of A025192.
The alternating sign sequence with g.f. (1-x^2)/(1+3x) gives the diagonal sums of A110168. - Paul Barry, Jul 14 2005
Let M = an infinite lower triangular matrix with the odd integers (1,3,5,...) in every column, with the leftmost column shifted up one row. Then A080923 = lim_{n->inf} M^n. - Gary W. Adamson, Feb 18 2010
a(n+1), n >= 0, with o.g.f. ((1-x^2)/(1-3*x)-1)/x = (3-x)/(1-3*x) provides the coefficients in the formal power series for tan(3*x)/tan(x) = (3-z)/(1-3*z) = Sum_{n>=0} a(n+1)*z^n, with z = tan(x)^2. Convergence holds for 0 <= z < 1/3, i.e., |x| < Pi/6, approximately 0.5235987758. For the numerator and denominator of this o.g.f. see A034867 and A034839, respectively. - Wolfdieter Lang, Jan 18 2013

Crossrefs

Essentially the same as A005051, A026097 and A083583.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^2) / (1 - 3 x), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 05 2013 *)

Formula

G.f.: (1-x^2)/(1-3*x).
G.f.: 1/(1 - 3*x + x^2 - 3*x^3 + x^4 - 3*x^5 + ...). - Gary W. Adamson, Jan 06 2011
a(n) = 2^3*3^(n-2), n >= 2, a(0) = 1, a(1) = 3. - Wolfdieter Lang, Jan 18 2013

A110593 a(1) = 3, a(n+1) = 2*(3^n).

Original entry on oeis.org

3, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366, 118098, 354294, 1062882, 3188646, 9565938, 28697814, 86093442, 258280326, 774840978, 2324522934, 6973568802, 20920706406, 62762119218, 188286357654, 564859072962, 1694577218886
Offset: 1

Views

Author

Jonathan Vos Post, Jul 29 2005

Keywords

Comments

Same as A025192 for n > 1. - Georg Fischer, Oct 21 2018

Crossrefs

Programs

  • Mathematica
    Rest[CoefficientList[Series[3 x + 6 x^2/(1 - 3 x), {x, 0, 50}], x]] (* G. C. Greubel, Sep 01 2017 *)
    Join[{3},NestList[3#&,6,30]] (* Harvey P. Dale, Aug 14 2024 *)
  • PARI
    x='x+O('x^50); Vec(3*x + 6*x^2/(1-3*x)) \\ G. C. Greubel, Sep 01 2017

Formula

a(n) = A008776(n-1) for n>1. - R. J. Mathar, Apr 24 2007
G.f.: 3*x + 6*x^2/(1-3*x). - R. J. Mathar, Nov 18 2007

A180142 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + x - x^2)/(1 - 3*x - 3*x^2).

Original entry on oeis.org

1, 4, 14, 54, 204, 774, 2934, 11124, 42174, 159894, 606204, 2298294, 8713494, 33035364, 125246574, 474845814, 1800277164, 6825368934, 25876938294, 98106921684, 371951579934, 1410175504854, 5346381254364, 20269670277654, 76848154596054, 291353474621124
Offset: 0

Views

Author

Johannes W. Meijer, Aug 13 2010

Keywords

Comments

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a rook on the eight side and corner squares but on the central square the rook goes berserk and turns into a berserker, see A180140.
The sequence above corresponds to 16 A[5] vectors with decimal values between 3 and 384. These vectors lead for the corner squares to A123620 and for the central square to A155116.
This sequence appears among the members of a family of sequences with g.f. (1 + x - k*x^2)/(1 - 3*x + (k-4)*x^2). Berserker sequences that are members of this family are 4*A007482 (k=2; with leading 1 added), A180142 (k=1; this sequence), A000302 (k=0), A180140 (k=-1) and 4*A154964 (k=-2; n>=1 and a(0)=1). Some other members of this family are 2*A180148 (k=3; with leading 1 added), 4*A025192 (k=4; with leading 1 added), 2*A005248 (k=5; with leading 1 added) and A123932 (k=6).

Crossrefs

Cf. A180141 (corner squares), A180140 (side squares), A180147 (central square).

Programs

  • Maple
    with(LinearAlgebra): nmax:=23; m:=2; A[5]:=[0,0,0,0,0,0,0,1,1]: A:= Matrix([[0,1,1,1,0,0,1,0,0], [1,0,1,0,1,0,0,1,0], [1,1,0,0,0,1,0,0,1], [1,0,0,0,1,1,1,0,0], A[5], [0,0,1,1,1,0,0,0,1], [1,0,0,1,0,0,0,1,1], [0,1,0,0,1,0,1,0,1], [0,0,1,0,0,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);
    # second Maple program:
    a:= n-> ceil((<<0|1>, <3|3>>^n. <<2/3, 4>>)[1,1]):
    seq(a(n), n=0..25);  # Alois P. Heinz, Jul 14 2021
  • Mathematica
    LinearRecurrence[{3, 3}, {1, 4, 14}, 26] (* Jean-François Alcover, Jan 18 2025 *)

Formula

G.f.: (1 + x - x^2)/(1 - 3*x - 3*x^2).
a(n) = 3*a(n-1) + 3*a(n-2) for n >= 2 with a(0)=1, a(1)=4 and a(2)=14.
a(n) = (6-2*A)*A^(-n-1)/21 + (6-2*B)*B^(-n-1)/21 with A=(-3+sqrt(21))/6 and B=(-3-sqrt(21))/6.
Lim_{k->infinity} a(2*n+k)/a(k) = 2*A000244(n)/(A003501(n) - A004254(n)*sqrt(21)) for n >= 1.
Lim_{k->infinity} a(2*n-1+k)/a(k) = 2*A000244(n)/(A004253(n)*sqrt(21) - 3*A030221(n-1)) for n >= 1.

A193723 Mirror of the fusion triangle A193722.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 18, 21, 8, 1, 54, 81, 45, 11, 1, 162, 297, 216, 78, 14, 1, 486, 1053, 945, 450, 120, 17, 1, 1458, 3645, 3888, 2295, 810, 171, 20, 1, 4374, 12393, 15309, 10773, 4725, 1323, 231, 23, 1, 13122, 41553, 58320, 47628, 24948, 8694, 2016, 300, 26, 1
Offset: 0

Views

Author

Clark Kimberling, Aug 04 2011

Keywords

Comments

A193723 is obtained by reversing the rows of the triangle A193722.
Triangle T(n,k), read by rows, given by [2,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
From Philippe Deléham, Nov 14 2011: (Start)
Riordan array ((1-x)/(1-3x), x/(1-3x)).
Product A200139*A007318 as infinite lower triangular arrays. (End)

Examples

			First six rows:
    1;
    2,   1;
    6,   5,   1;
   18,  21,   8,   1;
   54,  81,  45,  11,   1;
  162, 297, 216,  78,  14,   1;
		

Crossrefs

Cf. A084938, A193722, A052924 (antidiagonal sums), Diagonals: A000012, A016789, A081266, Columns: A025192, A081038.

Programs

  • Mathematica
    z = 9; a = 1; b = 1; c = 1; d = 2;
    p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
    t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
    w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
    g[n_] := CoefficientList[w[n, x], {x}]
    TableForm[Table[Reverse[g[n]], {n, -1, z}]]
    Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193722 *)
    TableForm[Table[g[n], {n, -1, z}]]
    Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)

Formula

Write w(n,k) for the triangle at A193722. The triangle at A193723 is then given by w(n,n-k).
T(n,k) = T(n-1,k-1) + 3*T(n-1,k) with T(0,0)=T(1,1)=1 and T(1,0)=2. - Philippe Deléham, Oct 05 2011
From Philippe Deléham, Nov 14 2011: (Start)
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for x=-2,-1,0,1,2,3,4,5,6,7,8,9 respectively.
T(n,k) = Sum_{j>=0} T(n-1-j,k-1)*3^j.
G.f.: (1-x)/(1-(3+y)*x). (End)

A206901 Number of nonisomorphic graded posets with 0 of rank n with no 3-element antichain.

Original entry on oeis.org

1, 2, 8, 39, 199, 1027, 5316, 27539, 142694, 739416, 3831589, 19855045, 102887673, 533158028, 2762794601, 14316644946, 74188042696, 384438233215, 1992137140383, 10323141778619, 53493935746148, 277202543857995, 1436447874880342, 7443591492820888
Offset: 0

Views

Author

David Nacin, Feb 13 2012

Keywords

Comments

We do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length.

Crossrefs

Cf. A124292 (counts with unique maximal element).
Cf. A025192, A206902 (adding a uniformity condition in the sense of the Retakh et al. paper with and without maximal elements).

Programs

  • Mathematica
    m = {{3, 3, 1, 0}, {1, 3, 0, 0}, {2, 3, 1, 0}, {6, 9, 2, 0}}; Table[MatrixPower[m, n][[4,3]], {n, 1, 40}]
  • Python
    def a(n,adict={0:1,1:2,2:8}):
        if n in adict:
            return adict[n]
        adict[n]=7*a(n-1)-10*a(n-2)+3*a(n-3)
        return adict[n]

Formula

a(n+3) = 7a(n+2) - 10a(n+1) + 3a(n), a(0)=1, a(1)=2, a(2)=8.
G.f.: (1-5x+4x^2)/(1-7x+10x^2-3x^3).

A206902 Number of nonisomorphic graded posets with 0 and uniform Hasse diagram of rank n with no 3-element antichain.

Original entry on oeis.org

1, 2, 8, 36, 166, 768, 3554, 16446, 76102, 352152, 1629536, 7540458, 34892452, 161460114, 747134894, 3457265922, 15998031616, 74028732924, 342557973998, 1585140808368, 7335025230994, 33941839649382, 157061283704438, 726779900373936, 3363075935260696
Offset: 0

Views

Author

David Nacin, Feb 13 2012

Keywords

Comments

We do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length.
Uniform (in the definition) used in the sense of Retakh, Serconek and Wilson (see paper in Links lines). - David Nacin, Mar 01 2012

Crossrefs

Cf. A025192 (adding a unique maximal element).
Cf. A124292, A206901 (dropping uniformity with and without maximal element).

Programs

  • GAP
    a:=[2,8,36];; for n in [4..30] do a[n]:=6*a[n-1]-7*a[n-2]+3*a[n-3]; od; Concatenation([1], a); # G. C. Greubel, May 21 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-4*x +3*x^2-x^3)/(1-6*x+7*x^2-3*x^3) )); // G. C. Greubel, May 21 2019
    
  • Mathematica
    LinearRecurrence[{6,-7,3}, {1,2,8,36}, 30] (* Vincenzo Librandi, Feb 27 2012 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-4*x+3*x^2-x^3)/(1-6*x+7*x^2-3*x^3)) \\ G. C. Greubel, May 21 2019
    
  • Python
    def a(n, adict={1:2,2:8,3:36}):
        if n in adict:
            return adict[n]
        adict[n]=6*a(n-1)-7*a(n-2)+3*a(n-3)
        return adict[n]
    
  • Sage
    ((1-4*x+3*x^2-x^3)/(1-6*x+7*x^2-3*x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 21 2019
    

Formula

a(n) = 6*a(n-1) - 7*a(n-2) + 3*a(n-3), a(1)=2, a(2)=8, a(3)=36.
G.f.: (1 -4*x +3*x^2 -x^3)/(1 -6*x +7*x^2 -3*x^3).

A265185 Non-vanishing traces of the powers of the adjacency matrix for the simple Lie algebra B_4: 2 * ((2 + sqrt(2))^n + (2 - sqrt(2))^n).

Original entry on oeis.org

4, 8, 24, 80, 272, 928, 3168, 10816, 36928, 126080, 430464, 1469696, 5017856, 17132032, 58492416, 199705600, 681837568, 2327939072, 7948081152, 27136446464, 92649623552, 316325601280, 1080003158016, 3687361429504, 12589439401984, 42983034748928
Offset: 0

Views

Author

Tom Copeland, Dec 04 2015

Keywords

Comments

a(n) is the trace of the 2*n-th power of the adjacency matrix M for the simple Lie algebra B_4, given in the Damianou link. M = Matrix[row 1; row 2; row 3; row 4] = Matrix[0,1,0,0; 1,0,1,0; 0,1,0,2; 0,0,1,0]. Equivalently, the trace tr(M^(2*k)) is the sum of the 2*n-th powers of the eigenvalues of M. The eigenvalues are the zeros of the characteristic polynomial of M, which is det(x*I - M) = x^4 - 4*x^2 + 2 = A127672(4,x), and are (+-) sqrt(2 + sqrt(2)) and (+-) sqrt(2 - sqrt(2)), or the four unique values generated by 2*cos((2*n+1)*Pi/8). Compare with A025192 for B_3. The odd power traces vanish.
-log(1 - 4*x^2 + 2*x^4) = 8*x^2/2 + 24*x^4/4 + 80*x^6/6 + ... = Sum_{n>0} tr(M^k) x^k / k = Sum_{n>0} a(n) x^(2k) / 2k gives an aerated version of the sequence a(n), excluding a(0), and exp(-log(1 - 4*x + 2*x^2)) = 1 / (1 - 4*x + 2*x^2) is the e.g.f. for A007070.
As in A025192, the cycle index partition polynomials P_k(x[1],...,x[k]) of A036039 evaluated with the negated power sums, the aerated a(n), are P_2(0,-a(1)) = P_2(0,-8) = -8, P_4(0,-a(1),0,-a(2)) = P_4(0,-8,0,-24) = 48, and all other P_k(0,-a(1),0,-a(2),0,...) = 0 since 1 - 4*x^2 + 2*x^4 = 1 - 8*x^2/2! + 48*x^4/4! = det(I - x M) = exp(-Sum_{k>0} tr(M^k) x^k / k) = exp[P.(-tr(M),-tr(M^2),...)x] = exp[P.(0,-a(1),0,-a(2),...)x].
Because of the inverse relation between the Faber polynomials F_n(b1,b2,...,bn) of A263916 and the cycle index polynomials, F_n(0,-4,0,2,0,0,0,...) = tr(M^n) gives aerated a(n), excluding a(0). E.g., F_2(0,-4) = -2 * -4 = 8, F_4(0,-4,0,2) = -4 * 2 + 2 * (-4)^2 = 24, and F_6(0,-4,0,2,0,0) = -2*(-4)^3 + 6*(-4)*2 = 80.

Crossrefs

Programs

  • Magma
    [Floor(2 * ((2 + Sqrt(2))^n + (2 - Sqrt(2))^n)): n in [0..30]]; // Vincenzo Librandi, Dec 06 2015
    
  • Mathematica
    4 LinearRecurrence[{4, -2}, {1, 2}, 30] (* Vincenzo Librandi, Dec 06 2015 and slightly modified by Robert G. Wilson v, Feb 13 2018 *)
  • PARI
    my(x='x+O('x^30)); Vec((4-8*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Feb 12 2018

Formula

a(n) = 2 * ((2 + sqrt(2))^n + (2 - sqrt(2))^n) = Sum_{k=0..3} 2^(2n) (cos((2k+1)*Pi/8))^(2n) = 2*2^(2n) (cos(Pi/8)^(2n) + cos(3*Pi/8)^(2n)) = 2 Sum_{k=0..1} (exp(i(2k+1)*Pi/8) + exp(-i*(2k+1)*Pi/8))^(2n).
E.g.f.: 2 * e^(2*x) * (e^(sqrt(2)*x) + e^(-sqrt(2)*x)) = 4*e^(2*x)*cosh(sqrt(2)*x) = 2*(exp(4*x*cos(Pi/8)^2) + exp(4*x cos(3*Pi/8)^2) ).
a(n) = 4*A006012(n) = 8*A007052(n-1) = 2*A056236(n).
G.f.: (4-8*x)/(1-4*x+2*x^2). - Robert Israel, Dec 07 2015
Note the preceding o.g.f. is four times that of A006012 and the denominator is y^4 * A127672(4,1/y) with y = sqrt(x). Compare this with those of A025192 and A189315. - Tom Copeland, Dec 08 2015

Extensions

More terms from Vincenzo Librandi, Dec 06 2015

A085388 First differences of n^k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 6, 4, 0, 1, 4, 12, 18, 8, 0, 1, 5, 20, 48, 54, 16, 0, 1, 6, 30, 100, 192, 162, 32, 0, 1, 7, 42, 180, 500, 768, 486, 64, 0, 1, 8, 56, 294, 1080, 2500, 3072, 1458, 128, 0, 1, 9, 72, 448, 2058, 6480, 12500, 12288, 4374, 256, 0, 1, 10, 90, 648
Offset: 1

Views

Author

Paul Barry, Jun 30 2003

Keywords

Comments

T(n,k) is the number of k-digit numbers in base n; n,k >= 2. - Mohammed Yaseen, Nov 11 2022

Examples

			Rows begin
  1,   0,   0,   0,   0, ...
  1,   1,   2,   4,   8, ...
  1,   2,   6,  18,  54, ...
  1,   3,  12,  48, 192, ...
  1,   4,  20, 100, 500, ...
		

Crossrefs

Diagonals include A053506, A085389, A085390.
Row-wise binomial transform is A083064.

Formula

T(n,k) = (n-1)*n^(k-1) + 0^k/n. - Corrected by Mohammed Yaseen, Nov 11 2022
T(n,0) = 1; T(n,k) = n^k - n^(k-1) for k >= 1. - Mohammed Yaseen, Nov 11 2022

Extensions

Offset corrected by Mohammed Yaseen, Nov 11 2022

A202390 Triangle T(n,k), read by rows, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 8, 3, 1, 10, 21, 17, 5, 1, 15, 45, 58, 35, 8, 1, 21, 85, 154, 144, 68, 13, 1, 28, 147, 350, 452, 330, 129, 21, 1, 36, 238, 714, 1195, 1198, 719, 239, 34, 1, 45, 366, 1344, 2799, 3611, 2959, 1506, 436, 55
Offset: 0

Views

Author

Philippe Deléham, Dec 18 2011

Keywords

Comments

T(n,n) = Fibonacci(n+1) = A000045(n+1).
A202390 is jointly generated with A208340 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=(x+1)*u(n-1,x)+(x+1)v(n-1,x). The alternating row sums of A202390, and also A208340, are 0 except for the first one. See the Mathematica section. - Clark Kimberling, Feb 27 2012

Examples

			Triangle begins:
  1
  1, 1
  1, 3, 2
  1, 6, 8, 3
  1, 10, 21, 17, 5
  1, 15, 45, 58, 35, 8
  1, 21, 85, 154, 144, 68, 13
  1, 28, 147, 350, 452, 330, 129, 21
		

Crossrefs

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 13;
    u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
    v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*v[n - 1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]  (* A202390 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]  (* A208340 *)
    Table[u[n, x] /. x -> 1, {n, 1, z}]  (*row sums*)
    Table[u[n, x] /. x -> -1, {n, 1, z}] (*alt. row sums*)

Formula

T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-2) - T(n-2,k) with T(0,0) = T(1,0) = T(1,1) = 1 and T(n,k) = 0 if k<0 or if n
G.f.: (1-x)/(1-(2+y)*x+(1-y^2)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = (-1)^n*A108411(n), A000007(n), A000012(n), A025192(n), A122558(n) for x = -2, -1, 0, 1, 2 respectively.

A206949 Number of nonisomorphic graded posets with 0 and non-uniform Hasse graph of rank n, with no 3-element antichain.

Original entry on oeis.org

0, 0, 0, 3, 24, 135, 657, 2961, 12744, 53244, 218025, 880308, 3518721, 13961727, 55097091, 216546048, 848476296, 3316800555, 12942852624, 50437433079, 196347606849, 763752142233, 2969021213928, 11536374392820, 44809232564673, 173997851613660, 675501426136017
Offset: 0

Author

David Nacin, Feb 13 2012

Keywords

Comments

Here, the term uniform is used in the sense of Retakh, Serconek and Wilson. Graded is used in terms of Stanley's definition that all maximal chains have the same length n.

References

  • Richard P. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Crossrefs

Cf. A206950 (maximal element removed).
Cf. A206947, A206948 (requiring exactly two elements in each rank level above 0 with and without maximal element).

Programs

  • Mathematica
    Join[{0}, LinearRecurrence[{9, -27, 30, -9}, {0, 0, 3, 24}, 40]]
  • Python
    def a(n,adict={0:0,1:0,2:0,3:3,4:24}):
        if n in adict:
            return adict[n]
        adict[n]=9*a(n-1)-27*a(n-2)+30*a(n-3)-9*a(n-4)
        return adict[n]

Formula

a(n) = 9*a(n-1) - 27*a(n-2) + 30*a(n-3) - 9*a(n-4), a(1)=0, a(2)=0, a(3)=3, a(4)=24.
G.f.: (3*(1-x)*x^3)/((1-3*x)*(1-6*x+9*x^2-3*x^3)).
a(n) = A124292(n+1) - A025192(n).
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