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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A006054 a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

Original entry on oeis.org

0, 0, 1, 2, 5, 11, 25, 56, 126, 283, 636, 1429, 3211, 7215, 16212, 36428, 81853, 183922, 413269, 928607, 2086561, 4688460, 10534874, 23671647, 53189708, 119516189, 268550439, 603427359, 1355888968, 3046654856, 6845771321, 15382308530, 34563733525
Offset: 0

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Comments

Let u(k), v(k), w(k) be defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)+w(k), v(k+1)=u(k)+v(k), w(k+1)=u(k); then {u(n)} = 1,1,3,6,14,31,... (A006356 with an extra initial 1), {v(n)} = 0,1,2,5,11,25,... (this sequence with its initial 0 deleted) and {w(n)} = {u(n)} prefixed by an extra 0 = A077998 with an extra initial 0. - Benoit Cloitre, Apr 05 2002. Also u(k)^2+v(k)^2+w(k)^2 = u(2k). - Gary W. Adamson, Dec 23 2003
Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then A006054 counts walks of length n between the vertex of degree 1 and the vertex of degree 3. - Paul Barry, Oct 02 2004
Form the digraph with matrix [1,1,0; 1,0,1; 1,1,1]. A006054(n) counts walks of length n between the vertices with loops. - Paul Barry, Oct 15 2004
Nonzero terms = INVERT transform of (1, 1, 2, 2, 3, 3, ...). Example: 56 = (1, 1, 2, 5, 11, 25) dot (3, 3, 2, 2, 1, 1) = (3 + 3 + 4 + 10 + 11 + 25). - Gary W. Adamson, Apr 20 2009
-a(n+1) appears in the formula for the nonpositive powers of rho:= 2*cos(Pi/7), the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with sigma:=rho^2-1, the ratio of the larger heptagon diagonal to the side length, as follows. rho^(-n) = C(n)*1 + C(n-1)*rho - a(n+1)*sigma, n >= 0, with C(n)=A077998(n), C(-1):=0. See the Steinbach reference, and a comment under A052547.
If, with the above notations, the power basis of the field Q(rho) is taken one has for nonpositive powers of rho, rho^(-n) = a(n+2)*1 + A077998(n-1)*rho - a(n+1)*rho^2. For nonnegative powers see A006053. See also the Steinbach reference. - Wolfdieter Lang, May 06 2011
a(n) appears also in the nonnegative powers of sigma,(defined in the above comment, where also the basis is given). See a comment in A106803.
The sequence b(n):=(-1)^(n+1)*a(n) forms the negative part (i.e., with nonpositive indices) of the sequence (-1)^n*A006053(n+1). In this way we obtain what we shall call the Ramanujan-type sequence number 2a for the argument 2*Pi/7 (see the comment to Witula's formula in A006053). We have b(n) = -2*b(n-1) + b(n-2) + b(n-3) and b(n) * 49^(1/3) = (c(1)/c(4))^(1/3) * (c(1))^(-n) + (c(2)/c(1))^(1/3) * (c(2))^(-n) + (c(4)/c(2))^(1/3) * (c(4))^(-n) = (c(2)/c(1))^(1/3) * (c(1))^(-n+1) + (c(4)/c(2))^(1/3) * (c(2))^(-n+1) + (c(1)/c(4))^(1/3) * (c(4))^(-n+1), where c(j) := 2*cos(2*Pi*j/7) (for the proof, see the comments to A215112). - Roman Witula, Aug 06 2012
(1, 1, 2, 5, 11, 25, 56, ...) * (1, 0, 1, 0, 1, ...) = the variant of A006356: (1, 1, 3, 6, 14, 31, ...). - Gary W. Adamson, May 15 2013
The limit of a(n+1)/a(n) for n -> infinity is, for all generic sequences with this recurrence of signature (2,1,-1), sigma = rho^2-1, approximately 2.246979603, the length ratio (largest diagonal)/side in the regular heptagon (7-gon). For rho = 2*cos(Pi/7) and sigma see a comment above, and the P. Steinbach reference. Proof: a(n+1)/a(n) = 2 + 1/(a(n)/a(n-1)) - 1/((a(n)/a(n-1))*(a(n-1)/a(n-2))), leading in the limit to sigma^3 -2*sigma^2 - sigma + 1, which is solved by sigma = rho^2-1, due to C(7, rho) = 0 , with the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho (see A187360). - Wolfdieter Lang, Nov 07 2013
Numbers of straight-chain aliphatic amino acids involving single, double or triple bonds (allowing adjacent double bonds) when cis/trans isomerism is neglected. - Stefan Schuster, Apr 19 2018
Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0) = 1. Also, A(r,n)=0 for n<0. Let A_1(r,n) = Sum_{j=0..n} A(r,j). Then the expansion of 1/(1 - 2*x - x^2 + x^3) is A_1(0,n) + A_1(1,n-2) + A_1(n-4) + ... = a(n) without the initial two 0's. In general, the expansion of 1/(1 - 2*x -x^k + x^(k+1)) is equal to Sum_{j>=0} A_1(j, n-j*k). - Gregory L. Simay, May 25 2018
For n>1, a(n) is the number of ways to tile a strip of length n-1 with one color of squares and dominos, two colors of trominos and quadrominos, 3 colors of 5-minos and 6-minos, and so on. - Greg Dresden and Zhiyu Zhang, Jun 26 2025

Examples

			G.f. = x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 56*x^7 + 126*x^8 + 283*x^9 + ... - _Michael Somos_, Jun 25 2018
		

References

  • Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a006054 n = a006053_list !! n
    a006054_list = 0 : 0 : 1 : zipWith (+) (map (2 *) $ drop 2 a006054_list)
       (zipWith (-) (tail a006054_list) a006054_list)
    -- Reinhard Zumkeller, Oct 14 2011
  • Maple
    A006054:=z**2/(1-2*z-z**2+z**3); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    LinearRecurrence[{2, 1, -1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2012 *)
  • Maxima
    a(n):=if n<2 then 0 else if n=2 then 1 else b(n-2);
    b(n):=sum(sum(binomial(j,n-3*k+2*j)*(-1)^(j-k)*binomial(k,j)*2^(-n+3*k-j),j,0,k),k,1,n); /* Vladimir Kruchinin, May 05 2011 */
    
  • PARI
    x='x+O('x^66);
    concat([0, 0], Vec(x^2/(1-2*x-x^2+x^3))) \\ Joerg Arndt, May 05 2011
    

Formula

G.f.: x^2/(1-2*x-x^2+x^3).
Sum_{k=0..n+2} a(k) = A077850(n). - Philippe Deléham, Sep 07 2006
Let M = the 3 X 3 matrix [1,1,0; 1,2,1; 0,1,2], then M^n*[1,0,0] = [A080937(n-1), A094790(n), A006054(n-1)]. E.g., M^3*[1,0,0] = [5,9,5] = [A080937(2), A094790(3), A006054(2)]. - Gary W. Adamson, Feb 15 2006
a(n) = round(k*A006356(n-1)), for n>1, where k = 0.3568958678... = 1/(1+2*cos(Pi/7)). - Gary W. Adamson, Jun 06 2008
a(n+1) = A187070(2n+1) = A187068(2n+3). - L. Edson Jeffery, Mar 10 2011
a(n+3) = Sum_{k=1..n} Sum_{j=0..k} binomial(j,n-3*k+2*j)*(-1)^(j-k)*binomial(k,j)*2^(-n+3*k-j); a(0)=0, a(1)=0, a(2)=1. - Vladimir Kruchinin, May 05 2011
7*a(n) = (c(2)-c(4))*(1+c(1))^n + (c(4)-c(1))*(1+c(2))^n + (c(1)-c(2))*(1+c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula et al. papers. - Roman Witula, Aug 07 2012
a(n) = -A006053(1-n) for all n in Z. - Michael Somos, Jun 25 2018

A006053 a(n) = a(n-1) + 2*a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 9, 14, 28, 47, 89, 155, 286, 507, 924, 1652, 2993, 5373, 9707, 17460, 31501, 56714, 102256, 184183, 331981, 598091, 1077870, 1942071, 3499720, 6305992, 11363361, 20475625, 36896355, 66484244, 119801329, 215873462, 388991876, 700937471
Offset: 0

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a(n+1) = S(n) for n>=1, where S(n) is the number of 01-words of length n, having first letter 1, in which all runlengths of 1's are odd. Example: S(4) counts 1000, 1001, 1010, 1110. See A077865. - Clark Kimberling, Jun 26 2004
For n>=1, number of compositions of n into floor(j/2) kinds of j's (see g.f.). - Joerg Arndt, Jul 06 2011
Counts walks of length n between the first and second nodes of P_3, to which a loop has been added at the end. Let A be the adjacency matrix of the graph P_3 with a loop added at the end. A is a 'reverse Jordan matrix' [0,0,1; 0,1,1; 1,1,0]. a(n) is obtained by taking the (1,2) element of A^n. - Paul Barry, Jul 16 2004
Interleaves A094790 and A094789. - Paul Barry, Oct 30 2004
a(n) appears in the formula for the nonnegative powers of rho:= 2*cos(Pi/7), the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with sigma:=rho^2-1, the ratio of the larger heptagon diagonal to the side length, as follows. rho^n = C(n)*1 + C(n+1)*rho + a(n)*sigma, n>=0, with C(n) = A052547(n-2). See the Steinbach reference, and a comment under A052547. - Wolfdieter Lang, Nov 25 2010
If with the above notations the power basis <1,rho,rho^2> of Q(rho) is used, nonnegative powers of rho are given by rho^n = -a(n-1)*1 + A052547(n-1)*rho + a(n)*rho^2. For negative powers see A006054. - Wolfdieter Lang, May 06 2011
-a(n-1) also appears in the formula for the nonpositive powers of sigma (see the above comment for the definition, and the Steinbach basis <1,rho,sigma>) as follows: sigma^(-n) = A(n)*1 -a(n+1)*rho -A(n-1)*sigma, with A(n) = A052547(n), A(-1):=0. - Wolfdieter Lang, Nov 25 2010

Examples

			G.f. = x^2 + x^3 + 3*x^4 + 4*x^5 + 9*x^6 + 14*x^7 + 28*x^8 + 47*x^9 + ...
Regarding the description "number of compositions of n into floor(j/2) kinds of j's," the a(6)=9 compositions of 6 are (2a, 2a, 2a), (3a, 3a), (2a, 4a), (2a, 4b), (4a, 2a), (4b, 2a), (6a), (6b), (6c). - _Bridget Tenner_, Feb 25 2022
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. Witula, E. Hetmaniok and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications (2012).

Crossrefs

Programs

  • Haskell
    a006053 n = a006053_list !! n
    a006053_list = 0 : 0 : 1 : zipWith (+) (drop 2 a006053_list)
       (zipWith (-) (map (2 *) $ tail a006053_list) a006053_list)
    -- Reinhard Zumkeller, Oct 14 2011
    
  • Magma
    [ n eq 1 select 0 else n eq 2 select 0 else n eq 3 select 1 else Self(n-1) +2*Self(n-2) -Self(n-3): n in [1..40] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    a[0]:=0: a[1]:=0: a[2]:=1: for n from 3 to 40 do a[n]:=a[n-1]+2*a[n-2]-a[n-3] od:seq(a[n], n=0..40); # Emeric Deutsch
    A006053:=z**2/(1-z-2*z**2+z**3); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    LinearRecurrence[{1,2,-1}, {0,0,1}, 50]  (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
  • PARI
    {a(n) = if( n<0, n = -1-n; polcoeff( -1 / (1 - 2*x - x^2 + x^3) + x * O(x^n), n), polcoeff( x^2 / (1 - x - 2*x^2 + x^3) + x * O(x^n), n))}; /* Michael Somos, Nov 30 2014 */
    
  • SageMath
    @CachedFunction
    def a(n): # a = A006053
        if (n<3): return (n//2)
        else: return a(n-1) + 2*a(n-2) - a(n-3)
    [a(n) for n in range(41)] # G. C. Greubel, Feb 12 2023

Formula

G.f.: x^2/(1 - x - 2*x^2 + x^3). - Emeric Deutsch, Dec 14 2004
a(n) = c^(n-2) - a(n-1)*(c-1) + (1/c)*a(n-2) for n > 3 where c = 2*cos(Pi/7). Example: a(7) = 14 = c^5 - 9*(c-1) + 4/c = 18.997607... - 7.21743962... + 2.219832528... - Gary W. Adamson, Jan 24 2010
G.f.: -1 + 1/(1 - Sum_{j>=1} floor(j/2)*x^j). - Joerg Arndt, Jul 06 2011
a(n+2) = A094790(n/2+1)*(1+(-1)^n)/2 + A094789((n+1)/2)*(1-(-1)^n)/2. - Paul Barry, Oct 30 2004
First differences of A028495. - Floor van Lamoen, Nov 02 2005
a(n) = A187065(2*n+1); a(n+1) = A187066(2*n+1) = A187067(2*n). - L. Edson Jeffery, Mar 16 2011
a(n) = 2^n*(c(1)^(n-1)*(c(1)+c(2)) + c(3)^(n-1)*(c(3)+c(6)) + c(5)^(n-1)*(c(5)+c(4)) )/7, with c(j):=cos(Pi*j/7). - Herbert Kociemba, Dec 18 2011
a(n+1)*(-1)^n*49^(1/3) = (c(1)/c(4))^(1/3)*(2*c(1))^n + (c(2)/c(1))^(1/3)*(2*c(2))^n + (c(4)/c(2))^(1/3)*(2c(4))^n = (c(2)/c(1))^(1/3)*(2*c(1))^(n+1) + (c(4)/c(2))^(1/3)*(c(2))^(n+1) + (c(1)/c(4))^(1/3)*(2*c(4))^(n+1), where c(j) := cos(2Pi*j/7); for the proof, see Witula et al.'s papers. - Roman Witula, Jul 21 2012
The previous formula connects the sequence a(n) with A214683, A215076, A215100, A120757. We may call a(n) the Ramanujan-type sequence number 2 for the argument 2*Pi/7. - Roman Witula, Aug 02 2012
a(n) = -A006054(1-n) for all n in Z. - Michael Somos, Nov 30 2014
G.f.: x^2 / (1 - x / (1 - 2*x / (1 + 5*x / (2 - x / (5 - 2*x))))). - Michael Somos, Jan 20 2017
a(n) ~ r*c^n, where r=0.241717... is one of the roots of 49*x^3-7*x+1, and c=2*cos(Pi/7) (as in Gary W. Adamson's formula). - Daniel Checa, Nov 04 2022
a(2n-1) = 2*a(n+1)*a(n) - a(n)^2 - a(n-1)^2. - Richard Peterson, May 25 2023

Extensions

More terms from Emeric Deutsch, Dec 14 2004
Typo in definition fixed by Reinhard Zumkeller, Oct 14 2011

A028495 Expansion of g.f. (1-x^2)/(1-x-2*x^2+x^3).

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 19, 33, 61, 108, 197, 352, 638, 1145, 2069, 3721, 6714, 12087, 21794, 39254, 70755, 127469, 229725, 413908, 745889, 1343980, 2421850, 4363921, 7863641, 14169633, 25532994, 46008619, 82904974, 149389218, 269190547, 485064009, 874055885
Offset: 0

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Comments

Form the graph with matrix A = [0,1,1; 1,0,0; 1,0,1] (P_3 with a loop at an extremity). Then A028495 counts closed walks of length n at the degree 3 vertex. - Paul Barry, Oct 02 2004
Equals INVERT transform of (1, 1, 0, 1, 0, 1, 0, 1, ...). - Gary W. Adamson, Apr 28 2009
From Johannes W. Meijer, May 29 2010: (Start)
a(n) is 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 Kc8, pawns b3, c7, d3, f7 and h7. (After Noam D. Elkies, see link; diagram 5).
Counts all paths of length n, n>=0, starting at the initial node on the path graph P_6, see the second Maple program. (End)
a(n) is the number of length n-1 binary words such that each maximal block of 1's has odd length. a(4) = 6 because we have: 000, 001, 010, 100, 101, 111. - Geoffrey Critzer, Nov 17 2012
a(n) is the number of compositions of n where increments can only appear at every second position, starting with the second and third part, see example. Also, a(n) is the number of compositions of n where there is no fall between every second pair of parts, starting with the first and second part; see example. - Joerg Arndt, May 21 2013
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 1, 0; 1, 0, 1; 0, 1, 0] or of the 3 X 3 matrix [1, 0, 1; 0, 0, 1; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
Range of row n of the circular Pascal array of order 7. - Shaun V. Ault, Jun 05 2014
a(n) is the number of compositions of n into parts from {1,2,4,6,8,10,...}. Example: a(4)= 6 because we have 4, 22, 211, 121, 112, and 1111. - Emeric Deutsch, Aug 17 2016
In general, a(n,m) = (2^n/(m+1))*Sum_{r=1..m} (1-(-1)^r)*cos(Pi*r/(m+1))^n*(1+cos(Pi*r/(m+1))) gives the number of paths of length n starting at the initial node on the path graph P_m. Here we have m=6. - Herbert Kociemba, Sep 15 2020
a(n-1) is the number of triangular dcc-polyominoes having area n (see Baril et al. at page 11). - Stefano Spezia, Oct 14 2023
a(n) is the number of permutations p of [n] with p(j)Alois P. Heinz, Mar 29 2024

Examples

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 19*x^6 + 33*x^7 + 61*x^8 + ...
From _Joerg Arndt_, May 21 2013: (Start)
There are a(6)=19 compositions of 6 where increments can only appear at every second position:
  01:  [ 1 1 1 1 1 1 ]
  02:  [ 1 1 1 1 2 ]
  03:  [ 1 1 2 1 1 ]
  04:  [ 1 1 2 2 ]
  05:  [ 1 1 3 1 ]
  06:  [ 1 1 4 ]
  07:  [ 2 1 1 1 1 ]
  08:  [ 2 1 2 1 ]
  09:  [ 2 1 3 ]
  10:  [ 2 2 1 1 ]
  11:  [ 2 2 2 ]
  12:  [ 3 1 1 1 ]
  13:  [ 3 1 2 ]
  14:  [ 3 2 1 ]
  15:  [ 3 3 ]
  16:  [ 4 1 1 ]
  17:  [ 4 2 ]
  18:  [ 5 1 ]
  19:  [ 6 ]
There are a(6)=19 compositions of 6 where there is no fall between every second pair of parts, starting with the first and second part:
  01:  [ 1 1 1 1 1 1 ]
  02:  [ 1 1 1 1 2 ]
  03:  [ 1 1 1 2 1 ]
  04:  [ 1 1 1 3 ]
  05:  [ 1 1 2 2 ]
  06:  [ 1 1 4 ]
  07:  [ 1 2 1 1 1 ]
  08:  [ 1 2 1 2 ]
  09:  [ 1 2 3 ]
  10:  [ 1 3 1 1 ]
  11:  [ 1 3 2 ]
  12:  [ 1 4 1 ]
  13:  [ 1 5 ]
  14:  [ 2 2 1 1 ]
  15:  [ 2 2 2 ]
  16:  [ 2 3 1 ]
  17:  [ 2 4 ]
  18:  [ 3 3 ]
  19:  [ 6 ]
(End)
19 = (1, 0, 1, 0, 1, 1) dot (1, 1, 2, 3, 6, 10) = (1 + 0 + 2 + 0 + 6 + 10). Cf. comment of Apr 28 2009. - _Gary W. Adamson_, Aug 10 2016
		

Crossrefs

Programs

  • Maple
    spec := [S,{S=Sequence(Union(Prod(Sequence(Prod(Z,Z)),Z,Z),Z))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
    with(GraphTheory): P:=6: G:= PathGraph(P): A:=AdjacencyMatrix(G): nmax:=34; for n from 0 to nmax do B(n):=A^n; a(n):=add(B(n)[1,k], k=1..P) od: seq(a(n), n=0..nmax); # Johannes W. Meijer, May 29 2010
    a := (-1)^(3/7) - (-1)^(4/7):
    b := (-1)^(5/7) - (-1)^(2/7):
    c := (-1)^(1/7) - (-1)^(6/7):
    f := n -> (a^n * (2 + a) + b^n * (2 + b) + c^n * (2 + c))/7:
    seq(simplify(f(n)), n=0..36); # Peter Luschny, Sep 16 2020
  • Mathematica
    LinearRecurrence[{1, 2, -1}, {1, 1, 2}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 11 2012 *)
    CoefficientList[Series[(1-x^2)/(1-x-2x^2+x^3),{x,0,40}],x] (* Harvey P. Dale, Dec 23 2018 *)
    a[n_,m_]:= 2^(n+1)/(m+1) Module[{x=(Pi r)/(m+1)},Sum[Cos[x]^n (1+Cos[x]),{r,1,m,2}]]
    Table[a[n,6],{n,0,40}]//Round (* Herbert Kociemba, Sep 15 2020 *) (* Herbert Kociemba, Sep 14 2020 *)
  • PARI
    {a(n) = if( n<0, n = -1-n; polcoeff( (1 - x^2) / (1 - 2*x - x^2 + x^3) + x * O(x^n), n), polcoeff( (1 - x^2) / (1 - x - 2*x^2 + x^3) + x * O(x^n), n))} /* Michael Somos, Apr 05 2012 */
    
  • PARI
    a(n)=([0,1,0;0,0,1;-1,2,1]^n*[1;1;2])[1,1] \\ Charles R Greathouse IV, Aug 25 2016

Formula

Recurrence: {a(0)=1, a(1)=1, a(2)=2, a(n)-2*a(n+1)-a(n+2)+a(n+3)=0}.
a(n) = Sum_(1/7*(1+2*_alpha)*_alpha^(-1-n), _alpha=RootOf(_Z^3-2*_Z^2-_Z+1)).
a(n) = A094718(6, n). - N. J. A. Sloane, Jun 12 2004
a(n) = a(n-1) + Sum_{k=1..floor(n/2)} a(n-2*k). - Floor van Lamoen, Oct 29 2005
a(n) = 5*a(n-2) - 6*a(n-4) + a(n-6). - Floor van Lamoen, Nov 02 2005
a(n) = A006053(n+2) - A006053(n). - R. J. Mathar, Nov 16 2007
a(2*n) = A052975(n), a(2*n+1) = A060557(n). - Johannes W. Meijer, May 29 2010
G.f.: 1 / (1 - x / (1 - x / (1 + x / (1 + x / (1 - x))))). - Michael Somos, Apr 05 2012
a(-1 - n) = A052534(n). - Michael Somos, Apr 05 2012
a(n) = (2^n/7)*Sum_{r=1..6} (1-(-1)^r)*cos(Pi*r/7)^n*(1+cos(Pi*r/7)). - Herbert Kociemba, Sep 15 2020

Extensions

More terms from James Sellers, Jun 05 2000

A055588 a(n) = 3*a(n-1) - a(n-2) - 1 with a(0) = 1 and a(1) = 2.

Original entry on oeis.org

1, 2, 4, 9, 22, 56, 145, 378, 988, 2585, 6766, 17712, 46369, 121394, 317812, 832041, 2178310, 5702888, 14930353, 39088170, 102334156, 267914297, 701408734, 1836311904, 4807526977, 12586269026, 32951280100, 86267571273
Offset: 0

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Author

Wolfdieter Lang, May 30 2000; Barry E. Williams, Jun 04 2000

Keywords

Comments

Number of directed column-convex polyominoes with area n+2 and having two cells in the bottom row. - Emeric Deutsch, Jun 14 2001
a(n) is the length of the list generated by the substitution: 3->3, 4->(3,4,6), 6->(3,4,6,6): {3, 4}, {3, 3, 4, 6}, {3, 3, 3, 4, 6, 3, 4, 6, 6}, {3, 3, 3, 3, 4, 6, 3, 4, 6, 6, 3, 3, 4, 6, 3, 4, 6, 6, 3, 4, 6, 6}, etc. - Wouter Meeussen, Nov 23 2003
Equals row sums of triangle A144955. - Gary W. Adamson, Sep 27 2008
Equals the INVERT transform of A034943 and the INVERTi transform of A094790. - Gary W. Adamson, Apr 01 2011

Crossrefs

Partial sums of A001519.
Apart from the first term, same as A052925.

Programs

  • GAP
    List([0..40], n-> Fibonacci(2*n)+1 ); # G. C. Greubel, Jun 06 2019
  • Magma
    [Fibonacci(2*n)+1: n in [0..40]]; // Vincenzo Librandi, Sep 30 2017
    
  • Maple
    g:=z/(1-3*z+z^2): gser:=series(g, z=0, 43): seq(abs(coeff(gser, z, n)+1), n=0..27); # Zerinvary Lajos, Mar 22 2009
  • Mathematica
    Table[Fibonacci[2n] +1, {n, 0, 40}] (* or *) LinearRecurrence[{4, -4, 1}, {1, 2, 4}, 40] (* Vincenzo Librandi, Sep 30 2017 *)
  • PARI
    vector(40, n, n--; fibonacci(2*n)+1) \\ G. C. Greubel, Jun 06 2019
    
  • Sage
    [lucas_number1(n,3,1)+1 for n in range(40)] # Zerinvary Lajos, Jul 06 2008
    

Formula

a(n) = (((3 + sqrt(5))/2)^n - ((3 - sqrt(5))/2)^n)/sqrt(5) + 1.
a(n) = Sum_{m=0..n} A055587(n, m) = 1 + A001906(n).
G.f.: (1 - 2*x)/((1 - 3*x + x^2)*(1-x)).
From Paul Barry, Oct 07 2004: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3);
a(n) = Sum_{k=0..floor(n/3)} binomial(n-k, 2*k)2^(n-3*k). (End)
From Paul Barry, Oct 26 2004: (Start)
a(n) = A001906(n) + 1.
a(n) = Sum_{k=0..n} Fibonacci(2*k+2)*(2*0^(n-k) - 1).
a(n) = A008346(2*n). (End)
a(n) = Sum_{k=0..2*n+1} ((-1)^(k+1))*Fibonacci(k). - Michel Lagneau, Feb 03 2014
E.g.f.: cosh(x) + sinh(x) + 2*exp(3*x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, May 14 2024
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/10) (A019827). - Amiram Eldar, Nov 28 2024

A005021 Random walks (binomial transform of A006054).

Original entry on oeis.org

1, 5, 19, 66, 221, 728, 2380, 7753, 25213, 81927, 266110, 864201, 2806272, 9112264, 29587889, 96072133, 311945595, 1012883066, 3288813893, 10678716664, 34673583028, 112584429049, 365559363741, 1186963827439, 3854047383798, 12514013318097, 40632746115136
Offset: 0

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Author

Keywords

Comments

Number of walks of length 2n+5 in the path graph P_6 from one end to the other one. Example: a(1)=5 because in the path ABCDEF we have ABABCDEF, ABCBCDEF, ABCDCDEF, ABCDEDEF and ABCDEFEF. - Emeric Deutsch, Apr 02 2004
Since a(n) is the binomial transform of A006054 from formula (3.63) in the Witula-Slota-Warzynski paper, it follows that a(n)=A(n;1)*(B(n;-1)-C(n;-1))-B(n;1)*B(n;-1)+C(n;1)*(A(n;-1)-B(n;-1)+C(n;-1)), where A(n;1)=A077998(n), B(n;1)=A006054(n+1), C(n;1)=A006054(n), A(n;-1)=A121449(n), B(n+1;-1)=-A085810(n+1), C(n;-1)=A215404(n) and A(n;d), B(n;d), C(n;d), n in N, d in C, denote the quasi-Fibonacci numbers defined and discussed in comments in A121449 and in the cited paper. - Roman Witula, Aug 09 2012
From Wolfdieter Lang, Mar 30 2020: (Start)
With offset -4 this sequence 6, 1, 0, 0, 1, 5, ... appears in the formula for the n-th power of the 3 X 3 tridiagonal Matrix M_3 = Matrix([1,1,0], [1,2,1], [0,1,2]) from A332602: (M_3)^n = a(n-2)*(M_3)^2 - (6*a(n-3) - a(n-4))*M_3 + a(n-3)*1_3, with the 3 X 3 unit matrix 1_3, for n >= 0. Proof from Cayley-Hamilton: (M_3)^n = 5*(M_3)^3 - 6*M_3 + 1_3 (see A332602 for the characteristic polynomial Phi(3, x)), and the recurrence (M_3)^n = M_3*(M_3)^(n-1). For (M_3)^n[1,1] = 2*a(n-2) - 5*a(n-3) + a(n-4), for n >= 0, see A080937(n).
The formula for a(n) in terms of r = rho(7) = A160389 given below shows that a(n)/a(n-1) converges to rho(7)^2 = A116425 = 3.2469796... for n -> infinity. This is because r - 2/r = 0.692..., and r - 1 - 1/r = 0.137... .
(End)

References

  • W. Feller, An Introduction to Probability Theory and its Applications, 3rd ed, Wiley, New York, 1968, p. 96.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Double partial sums of A060557. Bisection of A052547.

Programs

  • Magma
    I:=[1,5,19]; [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
    a:=k->sum(binomial(5+2*k,7*j+k-2),j=ceil((2-k)/7)..floor((7+k)/7))-sum(binomial(5+2*k,7*j+k-1),j=ceil((1-k)/7)..floor((6+k)/7)): seq(a(k),k=0..25);
    A005021:=-(z-1)*(z-5)/(-1+5*z-6*z**2+z**3); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence apart from the initial 1
  • Mathematica
    LinearRecurrence[{5,-6,1}, {1,5,19}, 50] (* Roman Witula, Aug 09 2012 *)
    CoefficientList[Series[1/(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/(1-5*x+6*x^2-x^3)) \\ G. C. Greubel, Apr 19 2018

Formula

G.f.: 1/(1-5x+6x^2-x^3). - Emeric Deutsch, Apr 02 2004
a(n) = 5*a(n-1) -6*a(n-2) +a(n-3). - Emeric Deutsch, Apr 02 2004
a(n) = Sum_{j=-infinity..infinity} (binomial(5+2*k, 7*j+k-2) - binomial(5+2*k, 7*j+k-1)) (a finite sum).
a(n-2) = 2^n*C(n;1/2)=(1/7)*((c(2)-c(4))*(c(4))^(2n) + (c(4)-c(1))*(c(1))^(2n) + (c(1)-c(2))*(c(2))^(2n)), where a(-2)=a(-1):=0, c(j):=2*cos(2Pi*j/7). This formula follows from the Binet formula for C(n;d)--one of the quasi-Fibonacci numbers (see comments in A121449 and the formula (3.17) in the Witula-Slota-Warzynski paper). - Roman Witula, Aug 09 2012
In terms of the algebraic number r = rho(7) = 2*cos(Pi/7) = A160389 of degree 3 the preceding formula gives a(n) = r^(2*(n+2))*(A1(r) + A2(r)*(r - 2/r)^(2*(n+1)) = A3(r)*(r - 1 - 1/r)^(2*(n+1)))/7, for n >= -4 (see a comment above for this offset), with A1(r) = -r^2 + 2*r + 1, A2(r) = -r^2 - r + 2, and A3(r) = 2*r^2 - r - 3. - Wolfdieter Lang, Mar 30 2020

Extensions

a(25)-a(26) from Vincenzo Librandi, Sep 18 2015

A080937 Number of Catalan paths (nonnegative, starting and ending at 0, step +/-1) of 2*n steps with all values <= 5.

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 131, 417, 1341, 4334, 14041, 45542, 147798, 479779, 1557649, 5057369, 16420730, 53317085, 173118414, 562110290, 1825158051, 5926246929, 19242396629, 62479659622, 202870165265, 658715265222, 2138834994142, 6944753544643, 22549473023585
Offset: 0

Views

Author

Henry Bottomley, Feb 25 2003

Keywords

Comments

With interpolated zeros (1,0,1,0,2,...), counts closed walks of length n at start or end node of P_6. The sequence (0,1,0,2,...) counts walks of length n between the start and second node. - Paul Barry, Jan 26 2005
HANKEL transform of sequence and the sequence omitting a(0) is the sequence A130716. This is the unique sequence with that property. - Michael Somos, May 04 2012
From Wolfdieter Lang, Mar 30 2020: (Start)
a(n) is also the upper left entry of the n-th power of the 3 X 3 tridiagonal matrix M_3 = Matrix([1,1,0], [1,2,1], [0,1,2]) from A332602: a(n) = ((M_3)^n)[1,1].
Proof: (M_3)^n = b(n-2)*(M_3)^2 - (6*b(n-3) - b(n-4))*M_3 + b(n-3)*1_3, for n >= 0, with b(n) = A005021(n), for n >= -4. For the proof of this see a comment in A005021. Hence (M_3)^n[1,1] = 2*b(n-2) - 5*b(n-3) + b(n-4), for n >= 0. This proves the 3 X 3 part of the conjecture in A332602 by Gary W. Adamson.
The formula for a(n) given below in terms of r = rho(7) = A160389 proves that a(n)/a(n-1) converges to rho(7)^2 = A116425 = 3.2469796..., because r - 2/r = 0.6920... < 1, and r^2 - 3 = 0.2469... < 1. This limit was conjectured in A332602 by Gary W. Adamson.
(End)

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 131*x^6 + 417*x^7 + 1341*x^8 + ...
		

Crossrefs

Cf. A033191 which essentially provide the same sequence for different limits and tend to A000108.

Programs

  • Magma
    I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-6*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 09 2016
  • Maple
    a:= n-> (<<0|1|0>, <0|0|1>, <1|-6|5>>^n. <<1, 1, 2>>)[1, 1]:
    seq(a(n), n=0..35);  # Alois P. Heinz, Nov 09 2012
  • Mathematica
    nn=56;Select[CoefficientList[Series[(1-4x^2+3x^4)/(1-5x^2+6x^4-x^6), {x,0,nn}], x],#>0 &] (* Geoffrey Critzer, Jan 26 2014 *)
    LinearRecurrence[{5,-6,1},{1,1,2},30] (* Jean-François Alcover, Jan 09 2016 *)
  • PARI
    a=vector(99); a[1]=1; a[2]=2;a[3]=5; for(n=4,#a,a[n]=5*a[n-1]-6*a[n-2] +a[n-3]); a \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    {a(n) = if( n<0, n = -n; polcoeff( (1 - 3*x + x^2) / (1 - 6*x + 5*x^2 - x^3) + x * O(x^n), n), polcoeff( (1 - 4*x + 3*x^2) / (1 - 5*x + 6*x^2 - x^3) + x * O(x^n), n))} /* Michael Somos, May 04 2012 */
    

Formula

a(n) = A080934(n,5).
G.f.: (1-4*x+3*x^2)/(1-5*x+6*x^2-x^3). - Ralf Stephan, May 13 2003
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3). - Herbert Kociemba, Jun 11 2004
a(n) = A096976(2*n). - Floor van Lamoen, Nov 02 2005
a(n) = (4/7-4/7*cos(1/7*Pi)^2)*(4*(cos(Pi/7))^2)^n + (1/7-2/7*cos(1/7*Pi) + 4/7*cos(1/7*Pi)^2)*(4*(cos(2*Pi/7))^2)^n + (2/7+2/7*cos(1/7*Pi))*(4*(cos(3*Pi/7))^2)^n for n>=0. - Richard Choulet, Apr 19 2010
G.f.: 1 / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x))))). - Michael Somos, May 04 2012
a(-n) = A038213(n). a(n + 2) * a(n) - a(n + 1)^2 = a(1 - n). Convolution inverse is A123183 with A123183(0)=1. - Michael Somos, May 04 2012
From Wolfdieter Lang, Mar 30 2020: (Start)
In terms of the algebraic number r = rho(7) = A160389 of degree 3 the formula given by Richard Choulet becomes a(n) = (1/7)*(r)^(2*n)*(C1(r) + C2(r)*(r - 2/r)^(2*n) + C3(r)*(r^2 - 3)^(2*n)), with C1(r) = 4 - r^2, C2(r) = 1 - r + r^2, and C3 = 2 + r.
a(n) = ((M_3)^n)[1,1] = 2*b(n-2) - 5*b(n-3) + b(n-4), for n >= 0, with the 3 X 3 tridiagonal matrix M_3 = Matrix([1,1,0], [1,2,1], [0,1,2]) from A332602, and b(n) = A005021(n) (with offset n >= -4). (End)

A094789 Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 7 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = 1, s(2n+1) = 4.

Original entry on oeis.org

1, 4, 14, 47, 155, 507, 1652, 5373, 17460, 56714, 184183, 598091, 1942071, 6305992, 20475625, 66484244, 215873462, 700937471, 2275930827, 7389902771, 23994866364, 77910846021, 252974934692, 821404463698, 2667083556359
Offset: 1

Views

Author

Herbert Kociemba, Jun 11 2004

Keywords

Comments

In general, a(n) = (2/m)*Sum_{r=1..m-1} sin(r*j*Pi/m)*sin(r*k*Pi/m)*(2*cos(r*Pi/m))^(2n+1) counts (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < m and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = j, s(2n+1) = k.
With interpolated zeros (0,0,0,1,0,4,0,14,...) counts walks of length n between the start and fourth nodes on P_6. - Paul Barry, Jan 26 2005
The Hankel transforms of this sequence or of this sequence with the first term omitted give 1, -2, 1, 1, -2, 1, ... . - Wathek Chammam, Nov 16 2011
Diagonal of the square array A216201. - Philippe Deléham, Mar 28 2013

Crossrefs

Programs

  • Magma
    I:=[1,4,14]; [n le 3 select I[n] else 5*Self(n-1)-6*Self(n-2)+Self(n-3): n in [1..45]]; // Vincenzo Librandi, Nov 10 2014
    
  • Mathematica
    f[n_] := FullSimplify[ TrigToExp[(2/7)Sum[ Sin[Pi*k/7]Sin[4Pi*k/7](2Cos[Pi*k/7])^(2n + 1), {k, 1, 6}]]]; Table[ f[n], {n, 25}] (* Robert G. Wilson v, Jun 18 2004 *)
    LinearRecurrence[{5,-6,1}, {1,4,14}, 50] (* Roman Witula, Aug 09 2012 *)
    CoefficientList[Series[(x - 1) / (- 1 + 5 x - 6 x^2 + x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
  • PARI
    Vec(x*(x-1)/(-1 + 5*x - 6*x^2 + x^3) + O(x^40)) \\ Michel Marcus, Nov 10 2014

Formula

a(n) = (2/7)*Sum_{k = 1..6} sin(Pi*k/7)*sin(4*Pi*k/7)*(2*cos(Pi*k/7))^(2n + 1).
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3).
G.f.: x*(x-1)/(-1 + 5*x - 6*x^2 + x^3). - Corrected by Vincenzo Librandi, Nov 10 2014
a(n) = 2^n*B(n; 1/2) = (1/7)*((c(1) - c(4))*(c(4))^(2n) + (c(2) - c(1))*(c(1))^(2n) + (c(4) - c(2))*(c(2))^(2n)), where c(j) := 2*cos(2*Pi*j/7). Here B(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 A052975, A085810, A077998, A006054, A121442). - Roman Witula, Aug 09 2012
a(n+1) = A216201(n,n+2) = A216201(n,n+3). - Philippe Deléham, Mar 28 2013

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

A216201 Square array T, read by antidiagonals : T(n,k) = 0 if n-k>=3 or if k-n>=4, T(2,0) = T(1,0) = T(0,0) = T(0,1) = T(0,2) = T(0,3) = 1, T(n,k) = T(n-1,k) + T(n,k-1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 0, 0, 4, 6, 3, 0, 0, 4, 10, 9, 0, 0, 0, 0, 14, 19, 9, 0, 0, 0, 0, 14, 33, 28, 0, 0, 0, 0, 0, 0, 47, 61, 28, 0, 0, 0, 0, 0, 0, 47, 108, 89, 0, 0, 0, 0, 0, 0, 0, 0, 155, 197, 89, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Mar 12 2013

Keywords

Examples

			Square array begins:
1, 1, 1,  1,  0,   0,   0,   0,   0,   0, 0, 0, 0, ... row n = 0
1, 2, 3,  4,  4,   0,   0,   0,   0,   0, 0, 0, 0, ... row n = 1
1, 3, 6, 10, 14,  14,   0,   0,   0,   0, 0, 0, 0, ... row n = 2
0, 3, 9, 19, 33,  47,  47,   0,   0,   0, 0, 0, 0, ... row n = 3
0, 0, 9, 28, 61, 108, 155, 155,   0,   0, 0, 0, 0, ... row n = 4
0, 0, 0, 28, 89, 197, 352, 507, 507,   0, 0, 0, 0, ... row n = 5
0, 0, 0,  0, 89, 286, 638,1147,1652,1652, 0, 0, 0, ... row n = 6
...
		

References

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

Crossrefs

Formula

T(n,n) = A052975(n).
T(n,n+1) = A060557(n).
T(n+1,n) = T(n+2,n) = A094790(n+1).
T(n,n+2) = T(n,n+3) = A094789(n+1).
Sum_{k, 0<=k<=n} T(n-k,k) = (-1)^n*A078038(n).

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 ]
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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
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