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.

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A052534 Expansion of (1-x)*(1+x)/(1-2*x-x^2+x^3).

Original entry on oeis.org

1, 2, 4, 9, 20, 45, 101, 227, 510, 1146, 2575, 5786, 13001, 29213, 65641, 147494, 331416, 744685, 1673292, 3759853, 8448313, 18983187, 42654834, 95844542, 215360731, 483911170, 1087338529, 2443227497, 5489882353, 12335653674
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Pairwise sums of A006356. Cf. A033303, A077850. - Ralf Stephan, Jul 06 2003
Number of (3412, P)-avoiding involutions in S_{n+1}, where P={1342, 1423, 2314, 3142, 2431, 4132, 3241, 4213, 21543, 32154, 43215, 15432, 53241, 52431, 42315, 15342, 54321}. - Ralf Stephan, Jul 06 2003
Number of 31- and 22-avoiding words of length n on alphabet {1,2,3} which do not end in 3 (e.g., at n=3, we have 111, 112, 121, 132, 211, 212, 232, 321 and 332). See A028859, A001519. - Jon Perry, Aug 04 2003
Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then the sequence 1,1,2,4,... with g.f. (1-x-x^2)/(1-2x-x^2+x^3) counts closed walks of length n at the degree 3 vertex. - Paul Barry, Oct 02 2004
a(n) is the number of Motzkin (n+1)-sequences whose flatsteps all occur at level <=1 and whose height is <=2. For example, a(5)=45 counts all 51 Motzkin 6-paths except FUUFDD, UFUFDD, UUFDDF, UUFDFD, UUFFDD, UUUDDD (the first five violate the flatstep restriction and the last violates the height restriction). - David Callan, Dec 09 2004
From Paul Barry, Nov 03 2010: (Start)
The g.f. of 1,1,2,4,9,... can be expressed as 1/(1-x/(1-x/(1-x^2))) and as 1/(1-x-x^2/(1-x-x^2)).
The second expression shows the link to the Motzkin numbers. (End)
From Emeric Deutsch, Oct 31 2010: (Start)
a(n) is the number of compositions of n into odd summands when we have two kinds of 1's. Proof: the g.f. of the set S={1,1',3,5,7,...} is g=2x+x^3/(1-x^2) and the g.f. of finite sequences of elements of S is 1/(1-g). Example: a(4)=20 because we have 1+3, 1'+3, 3+1, 3+1', and 2^4=16 of sums x+y+z+u, where x,y,z,u are taken from {1,1'}.
(End)
a(n-1) is the top left entry of the n-th power of any of the six 3 X 3 matrices [1, 1, 0; 1, 1, 1; 0, 1, 0] or [1, 1, 1; 0, 1, 1; 1, 1, 0] or [1, 0, 1; 1, 1, 1; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 0, 1, 1] or [1, 0, 1; 0, 0, 1; 1, 1, 1] or [1, 1, 0; 1, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

Examples

			G.f. = 1 + 2*x + 4*x^2 + 9*x^3 + 20*x^4 + 45*x^5 + 101*x^6 + 227*x^7 + 510*x^8 + ... - _Michael Somos_, Dec 12 2023
		

Crossrefs

Programs

  • GAP
    a:=[1,2,4];; for n in [4..40] do a[n]:=2*a[n-1]+a[n-2]-a[n-3]; od; a; # G. C. Greubel, May 09 2019
  • Magma
    [n le 3 select 2^(n-1) else 2*Self(n-1)+Self(n-2)-Self(n-3): n in [1..40]]; // Vincenzo Librandi, Mar 17 2015
    
  • Maple
    spec := [S,{S=Sequence(Union(Z,Prod(Z,Sequence(Prod(Z,Z)))))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    LinearRecurrence[{2,1,-1},{1,2,4},40] (* Roman Witula, Aug 07 2012 *)
    CoefficientList[Series[(1-x^2)/(1-2x-x^2+x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 17 2015 *)
    a[ n_] := {0, 1, 0} . MatrixPower[{{1, 1, 1}, {1, 1, 0}, {1, 0, 0}}, n+1] . {0, 1, 0}; (* Michael Somos, Dec 12 2023 *)
  • Maxima
    h(n):=if n=0 then 1 else sum(sum(binomial(k,j)*binomial(j,n-3*k+2*j)*2^(3*k-n-j)*(-1)^(k-j),j,0,k),k,1,n); a(n):=if n<2 then h(n) else h(n)-h(n-2); /* Vladimir Kruchinin, Sep 09 2010 */
    
  • PARI
    my(x='x+O('x^40)); Vec((1-x^2)/(1-2*x-x^2+x^3)) \\ G. C. Greubel, May 09 2019
    
  • PARI
    {a(n) = [0, 1, 0] * [1, 1, 1; 1, 1, 0; 1, 0, 0]^(n+1) * [0, 1, 0]~}; /* Michael Somos, Dec 12 2023 */
    
  • SageMath
    ((1-x^2)/(1-2*x-x^2+x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 09 2019
    

Formula

G.f.: (1 - x^2)/(1 - 2*x - x^2 + x^3).
a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0)=1, a(1)=2, a(2)=4.
a(n) = Sum_{alpha = RootOf(1-2*x-x^2+x^3)} (1/7)*(2 + alpha)*alpha^(-1-n).
a(n) = central term in the (n+1)-th power of the 3 X 3 matrix (shown in the example of A066170): [1 1 1 / 1 1 0 / 1 0 0]. E.g. a(6) = 101 since the central term in M^7 = 101. - Gary W. Adamson, Feb 01 2004
a(n) = A006054(n+2) - A006054(n). - Vladimir Kruchinin, Sep 09 2010
a(n) = A077998(n+2) - 2*A006054(n+2), which implies 7*a(n-2) = (2 + c(4) - 2*c(2))*(1 + c(1))^n + (2 + c(1) - 2*c(4))*(1 + c(2))^n + (2 + c(2) - 2*c(1))*(1 + c(4))^n, where c(j)=2*Cos(2Pi*j/7), a(-2)=a(-1)=1 since A077998 and A006054 are equal to the respective quasi-Fibonacci numbers. [Witula, Slota and Warzynski] - Roman Witula, Aug 07 2012
a(n+1) = A033303(n+1) - A033303(n). - Roman Witula, Sep 14 2012
a(n) = A006054(n+2)-A006054(n). - R. J. Mathar, Nov 23 2020
a(n) = A028495(-1-n) for all n in Z. - Michael Somos, Dec 12 2023

A052975 Expansion of (1-2*x)*(1-x)/(1-5*x+6*x^2-x^3).

Original entry on oeis.org

1, 2, 6, 19, 61, 197, 638, 2069, 6714, 21794, 70755, 229725, 745889, 2421850, 7863641, 25532994, 82904974, 269190547, 874055885, 2838041117, 9215060822, 29921113293, 97153242650, 315454594314, 1024274628963, 3325798821581, 10798800928441, 35063486341682
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 7 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 3. - Herbert Kociemba, Jun 11 2004
Counts all paths of length (2*n), n>=0, starting at the initial node and ending on the nodes 1, 2, 3, 4 and 5 on the path graph P_6, see the second Maple program. - Johannes W. Meijer, May 29 2010

Crossrefs

Programs

  • Magma
    I:=[1,2,6]; [n le 3 select I[n] else 5*Self(n-1)-6*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015
    
  • Maple
    spec := [S,{S=Sequence(Prod(Union(Sequence(Prod(Sequence(Z),Z)),Sequence(Z)),Z))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
    with(GraphTheory):G:=PathGraph(6): A:= AdjacencyMatrix(G): nmax:=25; n2:=2*nmax+1: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[k,1],k=1..5); od: seq(a(2*n),n=0..nmax); # Johannes W. Meijer, May 29 2010
  • Mathematica
    LinearRecurrence[{5,-6,1}, {1,2,6}, 50] (* Roman Witula, Aug 09 2012 *)
    CoefficientList[Series[(1 - 2 x) (1 - x)/(1 - 5 x + 6 x^2 - x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 18 2015 *)
  • PARI
    x='x+O('x^30); Vec((1-2*x)*(1-x)/(1-5*x+6*x^2-x^3)) \\ G. C. Greubel, Apr 19 2018

Formula

G.f.: (1-2*x)*(1-x)/(1-5*x+6*x^2-x^3).
a(n) = A028495(2*n). - Floor van Lamoen, Nov 02 2005
a(n) = Sum (1/7*(2-3*_alpha+_alpha^2)*_alpha^(-1-n), _alpha=RootOf(-1+5*_Z-6*_Z^2+_Z^3))
From Herbert Kociemba, Jun 11 2004: (Start)
a(n) = (2/7)*Sum_{r=1..6} sin(r*3*Pi/7)^2*(2*cos(r*Pi/7))^(2*n).
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3). (End)
a(n) = 2^n*A(n;1/2) = (1/7)*(s(2)^2*c(4)^(2n) + s(4)^2*c(1)^(2n) + s(1)^2*c(2)^(2n)), where c(j):=2*cos(2Pi*j/7) and s(j):=2*sin(2*Pi*j/7). Here A(n;d), n in N, d in C denotes the respective quasi-Fibonacci number - see A121449 and Witula-Slota-Warzynski paper for details (see also A094789, A085810, A077998, A006054, A121442). I note that my and the respective Herbert Kociemba's formulas are "compatible". - Roman Witula, Aug 09 2012
a(n) = A005021(n)-3*A005021(n-1)+2*A005021(n-2). - R. J. Mathar, Feb 27 2019

A078038 Expansion of (1-x)/(1+x-2*x^2-x^3).

Original entry on oeis.org

1, -2, 4, -7, 13, -23, 42, -75, 136, -244, 441, -793, 1431, -2576, 4645, -8366, 15080, -27167, 48961, -88215, 158970, -286439, 516164, -930072, 1675961, -3019941, 5441791, -9805712, 17669353, -31838986, 57371980, -103380599, 186285573, -335674791, 604865338, -1089929347, 1963985232
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

From Johannes W. Meijer, May 29 2010: (Start)
The absolute values of the a(n) represent the number of ways White can force checkmate in exactly (n+1) moves, n>=0, ignoring the fifty-move and the triple repetition rules, in the following chess position: White Ka1, Ra8, Bc1, Nb8, pawns a6, a7, b2, c6, d2, f6 and h6; Black Ke8, pawns b3, c7, d3, f7 and h7. (After Noam D. Elkies, see link; diagram 5).
The absolute values of the a(n) represent all paths of length n starting at the third (or fourth) node on the path graph P_6, see the Maple program.
(End)
For n>=1, abs(a(n-1)) is the number of compositions where there is no rise between every second pair of parts, starting with the second and third part; see example. Also, abs(a(n-1)) is the number of compositions of n where there is no fall between every second pair of parts, starting with the second and third part; see example. [Joerg Arndt, May 21 2013]

Examples

			From _Joerg Arndt_, May 21 2013: (Start)
There are abs(a(6-1))=23 compositions of 6 where there is no rise between every second pair of parts:
          v   v    <--= no rise over these positions
01:  [ 1 1 1 1 1 1 ]
02:  [ 1 1 1 2 1 ]
03:  [ 1 1 1 3 ]
04:  [ 1 2 1 1 1 ]
05:  [ 1 2 1 2 ]
06:  [ 1 2 2 1 ]
07:  [ 1 3 1 1 ]
08:  [ 1 3 2 ]
09:  [ 1 4 1 ]
10:  [ 1 5 ]
11:  [ 2 1 1 1 1 ]
12:  [ 2 1 1 2 ]
13:  [ 2 2 1 1 ]
14:  [ 2 2 2 ]
15:  [ 2 3 1 ]
16:  [ 2 4 ]
17:  [ 3 1 1 1 ]
18:  [ 3 2 1 ]
19:  [ 3 3 ]
20:  [ 4 1 1 ]
21:  [ 4 2 ]
22:  [ 5 1 ]
23:  [ 6 ]
There are abs(a(6-1))=23 compositions of 6 where there is no fall between every second pair of parts, starting with the second and third part:
          v   v    <--= no fall over these positions
01:  [ 1 1 1 1 1 1 ]
02:  [ 1 1 1 1 2 ]
03:  [ 1 1 1 3 ]
04:  [ 1 1 2 1 1 ]
05:  [ 1 1 2 2 ]
06:  [ 1 1 3 1 ]
07:  [ 1 1 4 ]
08:  [ 1 2 2 1 ]
09:  [ 1 2 3 ]
10:  [ 1 5 ]
11:  [ 2 1 1 1 1 ]
12:  [ 2 1 1 2 ]
13:  [ 2 1 2 1 ]
14:  [ 2 1 3 ]
15:  [ 2 2 2 ]
16:  [ 2 4 ]
17:  [ 3 1 1 1 ]
18:  [ 3 1 2 ]
19:  [ 3 3 ]
20:  [ 4 1 1 ]
21:  [ 4 2 ]
22:  [ 5 1 ]
23:  [ 6 ]
(End)
		

Crossrefs

Cf. A028495, A068911, A094790, A287381 (absolute values).

Programs

  • Maple
    with(GraphTheory): G:= PathGraph(6): A:=AdjacencyMatrix(G): nmax:=36; for n from 0 to nmax do B(n):=A^n; a(n):=add(B(n)[3,k], k=1..6) od: seq(a(n), n=0..nmax); # Johannes W. Meijer, May 29 2010
  • Mathematica
    LinearRecurrence[{-1, 2, 1}, {1, -2, 4}, 40] (* Jean-François Alcover, Jan 08 2019 *)
    a[n_]:=Sum[(-(-2)^(n+1)Cos[(Pi r)/7]^n Cot[(Pi r)/14]Sin[(3Pi r)/7])/7,{r,1,5,2}]
    Table[a[n],{n,0,40}]//Round (* Herbert Kociemba, Sep 17 2020 *)
  • PARI
    Vec((1-x)/(1+x-2*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

Formula

a(n+3) = -a(n+2)+2*a(n+1)+a(n), a(0)=1, a(1)=-2, a(2)=4. - Wouter Meeussen, Jan 02 2005
a(n) = (-1)^n * (A006053(n+1) + A006053(n+2)). G.f. of |a(n)|: (1+x)/(x^3 - 2*x^2 - x + 1). - Ralf Stephan, Aug 19 2013
a(n) = Sum_{r=1..6} ((-2)^n*(1-(-1)^r)*cos(Pi*r/7)^n*cot(Pi*r/14)*sin(3*Pi*r/7))/7. - Herbert Kociemba, Sep 17 2020

A060557 Row sums of triangle A060556.

Original entry on oeis.org

1, 3, 10, 33, 108, 352, 1145, 3721, 12087, 39254, 127469, 413908, 1343980, 4363921, 14169633, 46008619, 149389218, 485064009, 1574993356, 5113971944, 16604963593, 53915979657, 175064088671
Offset: 0

Views

Author

Wolfdieter Lang, Apr 06 2001

Keywords

Comments

Equals the INVERT transform of A045623: (1, 2, 5, 12, 28, ...). - Gary W. Adamson, Oct 26 2010

Crossrefs

a(n)=A028495(2n+1).
Cf. A053975.
Cf. A052975 (row sums of triangle A060102).
Cf. A045623. - Gary W. Adamson, Oct 26 2010

Programs

  • Mathematica
    a[0] = 1; a[1] = 3; a[2] = 10; a[n_] := a[n] = 5*a[n-1] - 6*a[n-2] + a[n-3]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Jul 05 2013, after Floor van Lamoen *)
    LinearRecurrence[{5,-6,1},{1,3,10},30] (* Harvey P. Dale, Nov 29 2013 *)
  • PARI
    { f="b060557.txt"; a0=1; a1=3; a2=10; write(f, "0 1"); write(f, "1 3"); write(f, "2 10"); for (n=3, 500, write(f, n, " ", a=5*a2 - 6*a1 + a0); a0=a1; a1=a2; a2=a; ) } \\ Harry J. Smith, Jul 07 2009

Formula

a(n) = Sum_{m=0..n} A060556(n, m).
G.f.: (1-x)^2/(1 - 5*x + 6*x^2 - x^3).
a(n) = 5a(n-1) - 6a(n-2) + a(n-3). - Floor van Lamoen, Nov 02 2005

A068914 Square array read by antidiagonals of number of k step walks (each step +-1 starting from 0) which are never more than n or less than 0.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 4, 3, 2, 1, 1, 0, 1, 4, 5, 3, 2, 1, 1, 0, 1, 8, 8, 6, 3, 2, 1, 1, 0, 1, 8, 13, 9, 6, 3, 2, 1, 1, 0, 1, 16, 21, 18, 10, 6, 3, 2, 1, 1, 0, 1, 16, 34, 27, 19, 10, 6, 3, 2, 1, 1, 0, 1, 32, 55, 54, 33, 20, 10, 6, 3, 2, 1, 1, 0, 1, 32, 89
Offset: 0

Views

Author

Henry Bottomley, Mar 06 2002

Keywords

Comments

The (n,k)-entry of the square array is p(n,k) in the R. Kemp reference (see Table 1 on p. 160 and Theorem 2 on p. 159). - Emeric Deutsch, Jun 16 2011

Examples

			Rows start:
1,0,0,0,0,...;
1,1,1,1,1,...;
1,1,2,2,4,...;
1,1,2,3,5,...;
etc.
		

Crossrefs

Rows include effectively A000007, A000012, A016116, A000045, A038754, A028495, A030436, A061551. Central and lower diagonals are A001405. Cf. A068913 for starting in the middle rather than an edge.
Reflected version of A094718.

Programs

  • Maple
    v := ((1-sqrt(1-4*z^2))*1/2)/z: G := proc (k) options operator, arrow: (1+v^2)*(1-v^(k+1))/((1-v)*(1+v^(k+2))) end proc: a := proc (n, k) options operator, arrow: coeff(series(G(k), z = 0, 80), z, n) end proc: for n from 0 to 15 do seq(a(n, k), k = 0 .. 15) end do; # yields the first 16 entries of the first 16 rows of the square array
    v := ((1-sqrt(1-4*z^2))*1/2)/z: G := proc (k) options operator, arrow: (1+v^2)*(1-v^(k+1))/((1-v)*(1+v^(k+2))) end proc: a := proc (n, k) options operator, arrow: coeff(series(G(k), z = 0, 80), z, n) end proc: for n from 0 to 13 do seq(a(n-i, i), i = 0 .. n) end do; # yields the first 14 antidiagonals of the square array in triangular form
  • Mathematica
    v = (1-Sqrt[1-4z^2])/(2z); f[k_] = (1+v^2)*(1-v^(k+1))/((1-v)*(1+v^(k+2))) ; m = 14; a = Table[ PadRight[ CoefficientList[ Series[f[k], {z, 0, m}], z], m], {k, 0, m}]; Flatten[Table[a[[n+1-k, k]], {n, m}, {k, n, 1, -1}]][[;; 95]] (* Jean-François Alcover, Jul 13 2011, after Emeric Deutsch *)
  • PARI
    T(n,k) = sum(j=floor(-n/(k+2)), ceil(n/(k+2)), (-1)^j*binomial(n,floor((n+(k+2)*j)/2))); \\ Stefano Spezia, May 08 2020

Formula

An explicit expression for the (n,k)-entry of the square array can be found in the R. Kemp reference (Theorem 2 on p. 159). - Emeric Deutsch, Jun 16 2011
The g.f. of column k is (1 + v^2)*(1 - v^(k+1))/((1 - v)*(1 + v^(k+2))), where v = (1 - sqrt(1-4*z^2))/(2*z) (see p. 159 of the R. Kemp reference). - Emeric Deutsch, Jun 16 2011

A113435 a(n) = a(n-1) + Sum_{k=0..n/3} a(n-3k) with a(0)=1.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 7, 11, 16, 26, 41, 62, 98, 154, 237, 371, 581, 901, 1406, 2197, 3418, 5329, 8317, 12956, 20196, 31501, 49096, 76532, 119338, 186029, 289997, 452141, 704861, 1098826, 1713111, 2670692, 4163483, 6490879, 10119152, 15775426
Offset: 0

Views

Author

Floor van Lamoen, Nov 04 2005

Keywords

Comments

If presented in three rows a(3n), a(3n+1) and a(3n+2) each term is the sum of the previous term in the sequence and the partial sum of its row.

Crossrefs

Partial sums of A176848.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^3)/(1 - x - 2*x^3 + x^4), {x,0,50}], x] (* G. C. Greubel, Mar 10 2017 *)
    LinearRecurrence[{1,0,2,-1},{1,1,1,2},40] (* Harvey P. Dale, Dec 17 2023 *)
  • PARI
    x='x+O(x^50); Vec((1 - x^3)/(1 - x - 2*x^3 + x^4)) \\ G. C. Greubel, Mar 10 2017

Formula

a(n) = a(n-1) + 2*a(n-3) - a(n-4) = 7*a(n-3) - 5*a(n-6) + 11*a(n-9) - a(n-12).
G.f.: (1-x^3)/(1-x-2*x^3+x^4).
G.f.: 1/(1-x) + x^3*Q(0)/(2-2*x) , where Q(k) = 1 + 1/(1 - x*(4*k+1 + 2*x^2 - x^3)/( x*(4*k+3 + 2*x^2 - x^3 ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 11 2013

A113439 a(n) = a(n-1) + Sum_{k=1..floor(n/4)} a(n-4k), with a(0)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 4, 5, 8, 12, 17, 23, 34, 50, 72, 101, 146, 212, 306, 436, 627, 905, 1305, 1871, 2689, 3872, 5577, 8014, 11521, 16576, 23858, 34309, 49337, 70968, 102108, 146868, 211233, 303832, 437080, 628708, 904306, 1300737, 1871065, 2691401
Offset: 0

Views

Author

Floor van Lamoen, Nov 04 2005

Keywords

Comments

If presented in four rows a(4k), a(4k+1), a(4k+2) and a(4k+3), each term is the sum of the previous term in the sequence and the partial sum of its row; see Example section.

Examples

			From _Jon E. Schoenfield_, Mar 11 2017: (Start)
Table of values T(j,k) = a(4k+j) in 4 rows:
.
  j | k=0     1     2     3     4     5     6     7
----+--------------------------------------------------
  0 |   1     2     8    34   146   627  2689 11521 ...
  1 |   1     3    12    50   212   905  3872 16576 ...
  2 |   1     4    17    72   306  1305  5577 23858 ...
  3 |   1     5    23   101   436  1871  8014 34309 ...
.
    T(2,4) = T(1,4) + T(2,0) + T(2,1) + T(2,2) + T(2,3)
      306  =   212  +    1   +    4   +   17   +   72
(End)
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^4)/(1 - x - 2*x^4 + x^5), {x,0,50}], x] (* G. C. Greubel, Mar 11 2017 *)
    LinearRecurrence[{1,0,0,2,-1},{1,1,1,1,2},50] (* Harvey P. Dale, Nov 10 2019 *)
  • PARI
    x='x+O('x^50); Vec((1-x^4)/(1-x-2*x^4+x^5)) \\ G. C. Greubel, Mar 11 2017

Formula

a(n) = a(n-1) + 2*a(n-4) - a(n-5).
a(n) = 9*a(n-4) - 28*a(n-8) + 38*a(n-12) - 20*a(n-16) +a(n-20).
G.f.: (1-x^4)/(1-x-2*x^4+x^5).

A187067 Let i be in {1,2,3} and let r >= 0 be an integer. Let p = {p_1, p_2, p_3} = {-2,0,1}, n = 2*r + p_i and define a(-2)=0. Then, a(n) = a(2*r + p_i) gives the quantity of H_(7,3,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x = sqrt(2*cos(Pi/7)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 3, 3, 4, 6, 9, 10, 14, 19, 28, 33, 47, 61, 89, 108, 155, 197, 286, 352, 507, 638, 924, 1145, 1652, 2069, 2993, 3721, 5373, 6714, 9707, 12087, 17460, 21794, 31501, 39254, 56714, 70755, 102256
Offset: 0

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Author

L. Edson Jeffery, Mar 09 2011

Keywords

Comments

Theory. (Start)
1. Definitions. Let T_(7,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/7,(7-j)*Pi/7) and Area(T_(7,j,0)) = sin(j*Pi/7), j in {1,2,3}. Associated with T_(7,j,0) are its angle coefficients (j, 7-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(7,j,0) along a line extending between its two corners with even angle coefficient; let H_(7,j,0) denote this half-tile. Similarly, a T_(7,j,r) tile is a linearly scaled version of T_(7,j,0) with sides of length x^r and Area(T_(7,j,r)) = x^(2*r)*sin(j*Pi/7), r>=0 an integer, where x is the positive, constant square root x = sqrt(2*cos(j*Pi/7)); likewise let H_(7,j,r) denote the corresponding half-tile. Often H_(7,i,r) (i in {1,2,3}) can be subdivided into an integral number of each equivalence class H_(7,j,0). But regardless of whether or not H_(7,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3, in which the entry m_(i,j) gives the quantity of H_(7,j,0) tiles that should be present in a subdivided H_(7,i,r) tile. The number x^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_1)^r, where
U_1= (0 1 0)
(1 0 1)
(0 1 1).
2. The sequence. Let r>=0, and let C_r be the r-th "block" defined by C_r = {a(2*r-2), a(2*r), a(2*r+1)}. Note that C_r - C_(r-1) - 2*C_(r-2) + C_(r-3) = {0,0,0}, with C_0 = {a(-2),a(0),a(1)} = {0,0,1}. Let n = 2*r + p_i. Then a(n) = a(2*r + p_i) = m_(i,3), where M = (m_(i,j)) = (U_1)^r was defined above. Hence the block C_r corresponds component-wise to the third column of M, and a(n) = m_(i,3) gives the quantity of H_(7,3,0) tiles that should appear in a subdivided H_(7,i,r) tile. (End)
Combining blocks A_r, B_r and C_r, from A187065, A187066 and this sequence, respectively, as matrix columns [A_r, B_r, C_r] generates the matrix (U_1)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r), B_(-r), C_(-r)] = (U_1)^(-r) of (U_1)^r. Therefore, the three sequences need not be causal.
Since a(2*r-2) = a(2*(r-1)) for all r, this sequence arises by concatenation of third-column entries m_(2,3) and m_(3,3) from successive matrices M = (U_1)^r.
a(2*n) = A006053(n+1), a(2*n+1) = A028495(n).

Examples

			Suppose r=3. Then
C_r = C_3 = {a(2*r-2), a(2*r), a(2*r+1)} = {a(4), a(6), a(7)} = {1,3,3},
corresponding to the entries in the third column of
M = (U_2)^3 = (0 2 1)
              (2 1 3)
              (1 3 3).
Choose i=2 and set n = 2*r + p_i. Then a(n) = a(2*r + p_i) = a(6+0) = a(6) = 3, which equals the entry in row 2 and column 3 of M. Hence a subdivided H_(7,2,3) tile should contain a(6) = m_(2,3) = 3 H_(7,3,0) tiles.
		

Crossrefs

Formula

Recurrence: a(n) = a(n-2) + 2*a(n-4) - a(n-6).
G.f.: x(1 + x - x^4)/(1 - x^2 - 2*x^4 + x^6).
Closed-form: a(n) = -(1/14)*((X_1 + Y_1*(-1)^(n-1))*((w_2)^2 - (w_3)^2)*(w_1)^(n-1) + (X_2 + Y_2*(-1)^(n-1))*((w_3)^2 - (w_1)^2)*(w_2)^(n-1) + (X_3 + Y_3*(-1)^(n-1))*((w_1)^2 - (w_2)^2)*(w_3)^(n-1)), where w_k = sqrt(2*(-1)^(k-1)*cos(k*Pi/7)), X_k = (w_k)^4 + (w_k)^3 - 1 and Y_k = (w_k)^4 - (w_k)^3 - 1, k=1,2,3.
For n>1, a(2n) = a(2n-1) + a(2n-4), a(2n+1) = a(2n-1) + a(2n-2). - Franklin T. Adams-Watters, Jan 06 2014

A113444 a(n) = a(n-1) + Sum_{0

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 9, 13, 18, 24, 31, 43, 60, 83, 113, 151, 206, 283, 389, 532, 721, 982, 1342, 1837, 2512, 3422, 4665, 6367, 8699, 11886, 16218, 22126, 30195, 41226, 56299, 76849, 104883, 143147, 195404, 266776, 364175, 497092, 678503, 926164
Offset: 0

Views

Author

Floor van Lamoen, Nov 04 2005

Keywords

Comments

If presented in five rows a(5n) a(5n+1).. a(5n+4) each term is the sum of the previous term in the sequence and the partial sum of its row.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^5)/(1 - x - 2*x^5 + x^6), {x,0,50}], x] (* G. C. Greubel, Mar 11 2017 *)
  • PARI
    x='x+O('x^50); Vec((1-x^5)/(1-x-2*x^5+x^6)) \\ G. C. Greubel, Mar 11 2017

Formula

G.f.: (1-x^5)/(1-x-2*x^5+x^6).
a(n) = a(n-1) + 2*a(n-5) - a(n-6).
a(n) = 11*a(n-5) -45*a(n-10) +90*a(n-15) -90*a(n-20) +37*a(n-25)-a(n-30).

A216054 Square array T, read by antidiagonals: T(n,k) = 0 if n-k >= 1 or if k-n >= 6, T(0,0) = T(0,1) = T(0,2) = T(0,3) = T(0,4) = T(0,5) = 1, T(n,k) = T(n-1,k) + T(n,k-1).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 2, 0, 0, 1, 4, 5, 0, 0, 0, 0, 5, 9, 5, 0, 0, 0, 0, 5, 14, 14, 0, 0, 0, 0, 0, 0, 19, 28, 14, 0, 0, 0, 0, 0, 0, 19, 47, 42, 0, 0, 0, 0, 0, 0, 0, 0, 66, 89, 42, 0, 0, 0, 0, 0, 0, 0, 0, 66, 155, 131, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 221, 286, 131, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Mar 16 2013

Keywords

Comments

A hexagon arithmetic of E. Lucas.

Examples

			Square array begins:
1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... row n=0
0, 1, 2, 3, 4, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... row n=1
0, 0, 2, 5, 9, 14, 19, 19, 0, 0, 0, 0, 0, 0, 0, ... row n=2
0, 0, 0, 5, 14, 28, 47, 66, 66, 0, 0, 0, 0, 0, 0, ... row n=3
0, 0, 0, 0, 14, 42, 89, 155, 221, 221, 0, 0, 0, 0, ... row n=4
0, 0, 0, 0, 0, 0, 42, 131, 286, 507, 728, 728, 0, 0, ... row n=5
0, 0, 0, 0, 0, 0, 131, 417, 924, 1652, 2380, 2380, 0, ... row n=6
...
		

References

  • E. Lucas, Théorie des nombres, A.Blanchard, Paris, 1958, Tome 1, p.89

Crossrefs

Cf. Similar sequences A216230, A216228, A216226, A216238

Programs

  • Mathematica
    Clear[t]; t[0, k_ /; k <= 5] = 1; t[n_, k_] /; k < n || k > n+5 = 0; t[n_, k_] := t[n, k] = t[n-1, k] + t[n, k-1]; Table[t[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Mar 18 2013 *)

Formula

T(n,n) = A080937(n).
T(n,n+1) = A080937(n+1).
T(n,n+2) = A094790(n+1).
T(n,n+3) = A094789(n+1).
T(n,n4) = T(n,n+5) = A005021(n).
Sum_{k, 0<=k<=n} T(n-k,k) = A028495(n).
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