A008574 -id:A008574 - cp's OEIS frontend

A257088 a(2n) = 4n if n>0, a(2n + 1) = 2n + 1, a(0) = 1.

1, 1, 4, 3, 8, 5, 12, 7, 16, 9, 20, 11, 24, 13, 28, 15, 32, 17, 36, 19, 40, 21, 44, 23, 48, 25, 52, 27, 56, 29, 60, 31, 64, 33, 68, 35, 72, 37, 76, 39, 80, 41, 84, 43, 88, 45, 92, 47, 96, 49, 100, 51, 104, 53, 108, 55, 112, 57, 116, 59, 120, 61, 124, 63, 128

Offset: 0

Michael Somos, Apr 16 2015

			G.f. = 1 + x + 4*x^2 + 3*x^3 + 8*x^4 + 5*x^5 + 12*x^6 + 7*x^7 + 16*x^8 + ...

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

Cf. A001082, A005408, A007310, A008574, A040001, A215947.

CF. A257083 (partial sums), A246695.

import Data.List (transpose)
a257088 n = a257088_list !! n
a257088_list = concat $ transpose [a008574_list, a005408_list]
-- Reinhard Zumkeller, Apr 17 2015

a[ n_] := Which[ n < 1, Boole[n == 0], OddQ[n], n, True, 2 n];
a[ n_] := SeriesCoefficient[ (1 + x + 2*x^2 + x^3 + x^4) / (1 - 2*x^2 + x^4), {x, 0, n}];

PARI
```
{a(n) = if( n<1, n==0, n%2, n, 2*n)};
```

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

Euler transform of length 4 sequence [ 1, 3, -1, -1].
a(n) is multiplicative with a(2^e) = 2^(e+1) if e>0, otherwise a(p^e) = p^e.
G.f.: (1 + x + 2*x^2 + x^3 + x^4) / (1 - 2*x^2 + x^4).
G.f.: (1 - x^3) * (1 - x^4) / ((1 - x) * (1 - x^2)^3).
MOBIUS transform of A215947 is [1, 4, 3, 8, 5, ...].
a(n) = n * A040001(n) if n>0.
a(n) + a(n-1) = A007310(n) if n>0.
a(n) = A001082(n+1) - A001082(n) if n>0.
Binomial transform with a(0)=0 is A128543 if n>0.
a(2*n) = A008574(n). a(2*n + 1) = A005408(n).
a(n) = A022998(n) if n>0. - R. J. Mathar, Apr 19 2015
From Amiram Eldar, Jan 28 2025: (Start)
Dirichlet g.f.: (1+2^(1-s)) * zeta(s-1).
Sum_{k=1..n} a(k) ~ (3/4) * n^2. (End)
a(n) = gcd(n^n, 2*n). - Mia Boudreau, Jun 27 2025

A265045 Coordination sequence for a 6.6.6 point in the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6} of the plane by squares and dominoes (hexagons).

Original entry on oeis.org

1, 3, 7, 11, 14, 18, 23, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232

Offset: 0

N. J. A. Sloane and Susanna Cuyler, Dec 27 2015

This tiling is 3-transitive but not 3-uniform since the polygons are not regular. It is a common floor-tiling.

The coordination sequences with respect to the points of types 4.6.6 (labeled "C" in the illustration), 6.6.6 ("B"), 6.6.6.6 ("A") are A265046, A265045, and A008574, respectively. The present sequence is for a "B" point.

Colin Barker, Table of n, a(n) for n = 0..1000
N. J. A. Sloane, A portion of the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6}
N. J. A. Sloane, A portion of the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6} showing the three types of point
N. J. A. Sloane, Hand-drawn illustration showing a(0) to a(10)
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008574, A265046.

LinearRecurrence[{2,-1},{1,3,7,11,14,18,23,28,32},60] (* Harvey P. Dale, Sep 23 2017 *)

Vec((1+x)*(1+2*x^2-2*x^3+x^4+x^6-x^7)/(1-x)^2 + O(x^100)) \\ Colin Barker, Jan 01 2016

For n >= 7 all three sequences equal 4n. (For n >= 7 the n-th shell contains n-1 points in the interior of each quadrant plus 4 points on the axes.)
From Colin Barker, Jan 01 2016: (Start)
a(n) = 2*a(n-1)-a(n-2) for n>8.
a(n) = 4*n for n>6.
G.f.: (1+x)*(1+2*x^2-2*x^3+x^4+x^6-x^7) / (1-x)^2.
(End)

A265046 Coordination sequence for a 4.6.6 point in the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6} of the plane by squares and dominoes (hexagons).

Original entry on oeis.org

1, 3, 5, 8, 13, 18, 23, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232

Offset: 0

N. J. A. Sloane and Susanna Cuyler, Dec 27 2015

This tiling is 3-transitive but not 3-uniform since the polygons are not regular. It is a common floor-tiling.

The coordination sequences with respect to the points of types 4.6.6 (labeled "C" in the illustration), 6.6.6 ("B"), 6.6.6.6 ("A") are A265046, A265045, and A008574, respectively. The present sequence is for a "C" point.

Colin Barker, Table of n, a(n) for n = 0..1000
N. J. A. Sloane, A portion of the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6}
N. J. A. Sloane, A portion of the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6} showing the three types of point
N. J. A. Sloane, Hand-drawn illustration showing a(0) to a(8)
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008574, A265045.

Vec((1+x)*(1+x^3+x^4-x^5+x^6-x^7)/(1-x)^2+ O(x^100)) \\ Colin Barker, Jan 01 2016

For n >= 7 all three sequences equal 4n. (For n >= 7 the n-th shell contains n-1 points in the interior of each quadrant plus 4 points on the axes.)
From Colin Barker, Jan 01 2016: (Start)
a(n) = 2*a(n-1)-a(n-2) for n>8.
a(n) = 4*n for n>6.
G.f.: (1+x)*(1+x^3+x^4-x^5+x^6-x^7) / (1-x)^2.
(End)

A085690 Number of intersections between a sphere inscribed in a cube and the n X n X n cubes resulting from a cubic lattice subdivision of the enclosing cube.

Original entry on oeis.org

8, 26, 56, 98, 152, 194, 272, 362, 440, 530, 656, 746, 872, 1034, 1160, 1298, 1496, 1658, 1856, 1994, 2240, 2450, 2624, 2906, 3128, 3362, 3656, 3890, 4208, 4442, 4760, 5090, 5360, 5714, 6032, 6362, 6752, 7106, 7496, 7826, 8216, 8618, 9080, 9458, 9896

Offset: 2

Hugo Pfoertner, Jul 17 2003

nonn

A concise description of the problem is given by Clive Tooth in the Seaman, Tooth link. Sequence terms up to n=10 were first given by Dave Seaman. Cubes having at least one vertex on the sphere and all other vertices either all inside or all outside the sphere are counted as 1/2. a(n) is asymptotic to (3/2)*Pi*n^2. (Clive Tooth) The terms a(2),...,a(6) are identical with A005897(n-1) (points on surface of cube with square grid on its faces).

			a(2)=8 because all 8 cubes resulting from a 2*2*2 subdivision of a cube are intersected by a sphere inscribed in the large cube.
a(4)=56: 8 central cubes of 4*4*4=64 not intersected.

Hugo Pfoertner, Table of n, a(n) for n = 2..1000
Hugo Pfoertner, FORTRAN program to count intersections.
Dave Seaman, Clive Tooth, Sphere/Cube Intersections. Discussion in Newsgroup sci.math.

Cf. A005897, A008574.

Fortran
```
! See Links.
(C#) // See Links.
```

Corrected overflow in program and b-file by Hugo Pfoertner, Apr 09 2016

A086460 Square array read by antidiagonals: T(n,k)=nk+0^n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 4, 3, 0, 1, 4, 6, 6, 4, 0, 1, 5, 8, 9, 8, 5, 0, 1, 6, 10, 12, 12, 10, 6, 0, 1, 7, 12, 15, 16, 15, 12, 7, 0, 1, 8, 14, 18, 20, 20, 18, 14, 8, 0, 1, 9, 16, 21, 24, 25, 24, 21, 16, 9, 0, 1, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 0, 1, 11, 20, 27, 32, 35, 36

Offset: 0

Paul Barry, Jul 21 2003

Rows include A028310, A004277, A008486, A008574, A008706, A008458. Main diagonal is n^2+0^n (A000290, preceded by extra 1). Inverse binomial transform of array A049513.

			Rows begin
1 0 0 0 0 ...
1 1 2 3 4 ...
1 2 4 6 8 ...
1 3 6 9 12 ...
1 4 8 12 16 ...

T(n, k)=nk+0^n

A129717 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 101's (n >= 0, 0 <= k <= floor((n-1)/2)). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 2, 3, 4, 1, 4, 4, 4, 8, 1, 4, 12, 5, 4, 16, 13, 1, 4, 20, 25, 6, 4, 24, 41, 19, 1, 4, 28, 61, 44, 7, 4, 32, 85, 85, 26, 1, 4, 36, 113, 146, 70, 8, 4, 40, 145, 231, 155, 34, 1, 4, 44, 181, 344, 301, 104, 9, 4, 48, 221, 489, 532, 259, 43, 1, 4, 52, 265, 670, 876, 560, 147, 10, 4, 56

Offset: 0

Emeric Deutsch, May 12 2007

Row n has 1+floor((n-1)/2) terms for n >= 1.
Row sums are the Fibonacci numbers (A000045).
T(n,1) = A008574(n-3).
T(n,2) = A001844(n-5).
T(n,3) = A005900(n-6).
T(n,4) = A006325(n-7).
T(n,5) = A033455(n-10).
T(n,k) = A129718(n,k+1) (since in each word: 1 + the number of 101's = number of runs of 1's).
Sum_{k>=0} k*T(n,k) = A004798(n-2).

			T(6,2)=5 because we have 110101, 101101, 101010, 101011 and 010101.
Triangle starts:
  1;
  2;
  3;
  4,  1;
  4,  4;
  4,  8,  1;
  4, 12,  5;

Michael De Vlieger, Table of n, a(n) for n = 0..10100 (rows 0 <= n <= 200, flattened.)

Cf. A000045, A008574, A001844, A005900, A006325, A033455, A129718, A004798.

T:=proc(n,k) if n=0 and k=0 then 1 elif n=1 and k=0 then 2 elif n=2 and k=0 then 3 elif n=3 and k=1 then 1 elif k

MapAt[{0, 1} + # &, #, 4] /. {} -> {1} &@ Table[If[n < 3, n + 1, Binomial[n - k - 1, k] + 2 Binomial[n - k - 2, k] + Binomial[n - k - 3, k]], {n, 0, 17}, {k, 0, Floor[(n - 1)/2]}] // Flatten (* Michael De Vlieger, Nov 15 2019 *)

G.f.: G(t,z) = (1+z)*(1 + z^2 - t*z^2)/(1 - z - t*z^2).
G.f. of col 0: (1+z)(1+z^2)/(1-z), leading to the partial sums of 1,1,1,1,0,0,0,...
G.f. of col k: z^(2k+1)*(1+z)^2/(1-z)^(k+1) (k >= 1).
T(n,k) = binomial(n-k-1, k) + 2*binomial(n-k-2, k) + binomial(n-k-3, k) for n >= 4 and 0 <= k < n/2.

A130323 A007318^(-1) * A130322.

Original entry on oeis.org

1, 3, 1, 5, 2, 1, 7, 3, 1, 1, 9, 4, 2, 0, 1, 11, 5, 2, 2, -1, 1, 13, 6, 3, 0, 3, -2, 1, 15, 7, 3, 3, -3, 5, -3, 1, 17, 8, 4, 0, 6, -8, 8, -4, 1, 19, 9, 4, 4, -6, 14, -16, 12, -5, 1

Offset: 0

Gary W. Adamson, May 24 2007

Row sums = A008574: (1, 4, 8, 12, 16, 20, 24, ...).

			First few rows of the triangle:
   1;
   3, 1;
   5, 2, 1;
   7, 3, 1, 1;
   9, 4, 2, 0,  1;
  11, 5, 2, 2, -1,  1;
  13, 6, 3, 0,  3, -2,   1;
  15, 7, 3, 3, -3,  5,  -3,  1;
  17, 8, 4, 0,  6, -8,   8, -4,  1;
  19, 9, 4, 4, -6, 14, -16, 12, -5, 1;
  ...

Cf. A130323, A007318, A008574.

Inverse binomial transform of triangle A130322, as infinite lower triangular matrices.

A131032 A097806 * A130296.

Original entry on oeis.org

1, 3, 1, 5, 2, 1, 7, 2, 2, 1, 9, 2, 2, 2, 1, 11, 2, 2, 2, 2, 1, 13, 2, 2, 2, 2, 2, 1, 15, 2, 2, 2, 2, 2, 2, 1, 17, 2, 2, 2, 2, 2, 2, 2, 1, 19, 2, 2, 2, 2, 2, 2, 2, 2, 1, 21, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 23, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1

Offset: 1

Gary W. Adamson, Jun 10 2007

Row sums give A008574.

A131033 = A130296 * A097806.

			First few rows of the triangle are:
1;
3, 1;
5, 2, 1;
7, 2, 2, 1;
9, 2, 2, 2, 1;
11, 2, 2, 2, 2, 1;
13, 2, 2, 2, 2, 2, 1;
...

Cf. A008574, A097806, A130296, A131033.

A097806 * A130296 as infinite lower triangular matrices. A097806 = the pairwise operator, A130296 = (1; 2,1; 3,1,1; ...).

Definition corrected and more terms from Georg Fischer, Oct 10 2021

A202241 Array F(n,m) read by antidiagonals: F(0,m)=1, F(n,0) = A130713(n), and column m+1 is recursively defined as the partial sums of column m.

Original entry on oeis.org

1, 2, 1, 1, 3, 1, 0, 4, 4, 1, 0, 4, 8, 5, 1, 0, 4, 12, 13, 6, 1, 0, 4, 16, 25, 19, 7, 1, 0, 4, 20, 41, 44, 26, 8, 1, 0, 4, 24, 61, 85, 70, 34, 9, 1, 0, 4, 28, 85, 146, 155, 104, 43, 10, 1, 0, 4, 32, 113, 231, 301, 259, 147, 53, 11, 1, 0, 4, 36, 145, 344, 532, 560, 406, 200, 64, 12, 1

Offset: 0

Paul Curtz, Dec 16 2011

The array F(n,m), beginning with row n=0, is:
  1, 1,  1,  1,   1,   1,    1,
  2, 3,  4,  5,   6,   7,    8,
  1, 4,  8, 13,  19,  26,   34,
  0, 4, 12, 25,  44,  70,  104,
  0, 4, 16, 41,  85, 155,  259,
  0, 4, 20, 61, 146, 301,  560,
  0, 4, 24, 85, 231, 532, 1092.
Columns after A130713, A113311, A008574 have signatures (3,-3,1), (4,-6,4,-1), (5,-10,10,-5,1), (6,-15,20,-15,6,-1) (from A135278(n+3)).
Inserting columns of zeros and pushing the columns down, plus alternating sign switches defines the following triangle T(n,2m) = (-1)^(m/2)*F(n-2m,m):
  1,
  2 0,
  1 0 -1,
  0 0 -3 0,
  0 0 -4 0 1,
  0 0 -4 0 4 0,
  0 0 -4 0 8 0 -1
The row sums in the triangle are (-1)^n*A099838(n).
The companion to A201863 is
  1
  1 0
  1 0   0
  1 0  -2 0
  1 0  -4 0  1
  1 0  -6 0  5 0
  1 0  -8 0 13 0   -2
  1 0 -10 0 25 0  -12 0
  1 0 -12 0 41 0  -38 0   4
  1 0 -14 0 61 0  -88 0  28 0
  1 0 -16 0 85 0 -170 0 104 0 -8
5th column: A001844; 7th column: -A035597=-2*A005900(n+1); 9th column: 4*A006325(n+2); 11th column: -8*(1,8,34,104) (from columns 4,5,6,7 of F(n,m)).
As a triangular array, this is the Riordan array ((1+x)^2, x/(1-x)). - Philippe Deléham, Feb 21 2012

			Triangle T(n,k) begins:
  1
  2, 1
  1, 3,  1
  0, 4,  4,  1
  0, 4,  8,  5,   1
  0, 4, 12, 13,   6,   1
  0, 4, 16, 25,  19,   7,   1
  0, 4, 20, 41,  44,  26,   8,  1
  0, 4, 24, 61,  85,  70,  34,  9,  1
  0, 4, 28, 85, 146, 155, 104, 43, 10, 1
- _Philippe Deléham_, Feb 21 2012

Muniru A Asiru, Table of n, a(n) for n = 0..5151

Cf. A130713 (column 0), A113311 (column 1), A008574 (column 2), A001844 (column 3), A005900 (column 4), A006325 (column 5), A033455 (column 6).

Cf. A267633.

Flat(List([0..12],n->List([0..n],k->Binomial(n,n-k)+Binomial(n-1,n-k-1)-Binomial(n-2,n-k-2)-Binomial(n-3,n-k-3)))); # Muniru A Asiru, Mar 22 2018

A130713 := proc(n)
    if n <= 2 and n >=0 then
        op(n+1,[1,2,1]) ;
    else
        0;
    end if;
end proc:
A202241 := proc(n,m)
    option remember;
    if n < 0 then
        0 ;
    elif m = 0 then
        A130713(n);
    else
        procname(n,m-1)+procname(n-1,m) ;
    end if;
end proc:
for d from 0 to 12 do
    for m from 0 to d do
        printf("%d,",A202241(d-m,m)) ;
    end do:
end do: # R. J. Mathar, Dec 22 2011
C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if end proc:
for n from 0 to 10 do
     seq(C(n, n-k) + C(n-1, n-k-1) - C(n-2, n-k-2) - C(n-3, n-k-3), k = 0..n);
end do; # Peter Bala, Mar 20 2018

rows = 12;
T[0] = PadRight[{1, 2, 1}, rows];
T[n_ /; nJean-François Alcover, Jun 29 2019 *)

def Trow(n): return [binomial(n, n-k) + binomial(n-1, n-k-1) - binomial(n-2, n-k-2) - binomial(n-3, n-k-3) for k in (0..n)]
for n in (0..9): print(Trow(n)) # Peter Luschny, Mar 21 2018

F(1,m) = m+2.
F(2,m) = A034856(m+1).
F(3,m) = A000297(m-1).
Sum_{m=0..d} F(d-m,m) = A116453(d-3), d >= 3 (antidiagonal sums).
As a triangular array T(n,k), 0 <= k <= n, satisfies: T(n,k) = T(n-1,k) + T(n-1,k-1) with T(0,0) = 1, T(1,0) = 2, T(2,0) = 1, T(3,0) = 0. - Philippe Deléham, Feb 21 2012
Unsigned diagonals of A267633 (beginning with its main diagonal) appear to be the reverse rows of this entry's triangle beginning with the fourth row. - Tom Copeland, Jan 26 2016
T(n,k) = C(n, n-k) + C(n-1, n-k-1) - C(n-2, n-k-2) - C(n-3, n-k-3), where C(n, k) = n!/(k!*(n-k)!) if 0 <= k <= n, otherwise 0. - Peter Bala, Mar 20 2018

A257174 a(n) = 4n/3 if n = 3k and n!=0, otherwise a(n) = n except a(0) = 1.

Original entry on oeis.org

1, 1, 2, 4, 4, 5, 8, 7, 8, 12, 10, 11, 16, 13, 14, 20, 16, 17, 24, 19, 20, 28, 22, 23, 32, 25, 26, 36, 28, 29, 40, 31, 32, 44, 34, 35, 48, 37, 38, 52, 40, 41, 56, 43, 44, 60, 46, 47, 64, 49, 50, 68, 52, 53, 72, 55, 56, 76, 58, 59, 80, 61, 62, 84, 64, 65, 88

Offset: 0

Michael Somos, Apr 17 2015

			G.f. = 1 + x + 2*x^2 + 4*x^3 + 4*x^4 + 5*x^5 + 8*x^6 + 7*x^7 + 8*x^8 + ...

This is a divisibility sequence.

G. C. Greubel, Table of n, a(n) for n = 0..2500
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A008574, A280057.

I:=[1,1,2,4,4,5,8]; [n le 7 select I[n] else 2*Self(n-3)-Self(n-6): n in [1..80]]; // Vincenzo Librandi, Apr 28 2015

m:=60; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!((1+x^2)*(1+x+x^2+x^3+x^4)/(1-2*x^3+x^6))); // G. C. Greubel, Aug 02 2018

A257174:=n->`if`(n=0,1,(n/3)*(4+floor(n/3)-ceil(n/3))): seq(A257174(n), n=0..100); # Wesley Ivan Hurt, Apr 27 2015

a[ n_] := If[ n==0, 1, n + If[ Mod[n, 3] == 0, n/3, 0]];
a[ n_] := n + Which[ n==0, 1, Mod[n, 3] == 0, n/3, True, 0];
Join[{1}, LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 2, 4, 4, 5, 8}, 100]] (* Vincenzo Librandi, Apr 28 2015 *)
a[ n_] := If[ n==0, 1, Sign[n] SeriesCoefficient[ x / (1 - x)^2 + x^3 / (1 - x^3)^2, {x, 0, Abs@n}]]; (* Michael Somos, Dec 30 2016 *)
CoefficientList[Series[(1+x^2)*(1+x+x^2+x^3+x^4)/(1-2*x^3+x^6), {x, 0, 60}], x] (* G. C. Greubel, Aug 02 2018 *)

{a(n) = n + if( n==0, 1, n%3==0, n/3, 0)};

{a(n) = if( n==0, 1, sign(n) * polcoeff( (1 - x^4) * (1 - x^5) / ((1 - x) * (1 - x^2) * (1 - x^3)^2) + x * O(x^abs(n)), abs(n)))};

my(x='x+O('x^60)); Vec((1+x^2)*(1+x+x^2+x^3+x^4)/(1-2*x^3+x^6)) \\ G. C. Greubel, Aug 02 2018

Euler transform of length 5 sequence [1, 1, 2, -1, -1].
a(n) is multiplicative with a(0) = 1, a(3^e) = 4*3^(e-1) if e>0, a(p^e) = p^e otherwise.
G.f.: (1 - x^4) * (1 - x^5) / ((1 - x) * (1 - x^2) * (1 - x^3)^2).
G.f.: (1 + x^2) * (1 + x + x^2 + x^3 + x^4) / (1 - 2*x^3 + x^6).
a(3*n) = A008574(n).
a(n) = -a(-n) for all n in Z except n=0.
From Wesley Ivan Hurt, Apr 27 2015: (Start)
a(n) = 2*a(n-3)-a(n-6).
a(n) = n*(4+floor(n/3)+floor(-n/3))/3 for n>0. (End)
a(n) = (-1)^n * A280057(n). - Michael Somos, Dec 30 2016
G.f.: 1 + x / (1 - x)^2 + x^3 / (1 - x^3)^2. - Michael Somos, Dec 30 2016
0 = +22 + a(n)*(+21 + 3*a(n) + 7*a(n+1) -14*a(n+2)) + a(n+1)*(-6*a(n+1) + 7*a(n+2)) + a(n+2)*(-21 + 3*a(n+2)) if n>0. - Michael Somos, Dec 30 2016
Dirichlet g.f.: zeta(s-1)*(1+1/3^s). - Amiram Eldar, Dec 29 2022

cp's OEIS Frontend

A257088 a(2*n) = 4*n if n>0, a(2*n + 1) = 2*n + 1, a(0) = 1.

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A265045 Coordination sequence for a 6.6.6 point in the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6} of the plane by squares and dominoes (hexagons).

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A265046 Coordination sequence for a 4.6.6 point in the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6} of the plane by squares and dominoes (hexagons).

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A085690 Number of intersections between a sphere inscribed in a cube and the n X n X n cubes resulting from a cubic lattice subdivision of the enclosing cube.

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Fortran

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A086460 Square array read by antidiagonals: T(n,k)=nk+0^n.

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A129717 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 101's (n >= 0, 0 <= k <= floor((n-1)/2)). A Fibonacci binary word is a binary word having no 00 subword.

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Maple

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A130323 A007318^(-1) * A130322.

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A131032 A097806 * A130296.

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A257088 a(2n) = 4n if n>0, a(2n + 1) = 2n + 1, a(0) = 1.

A257174 a(n) = 4n/3 if n = 3k and n!=0, otherwise a(n) = n except a(0) = 1.