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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A200312 a(n) = A000108(n)*A006130(n), where A000108 is the Catalan numbers and A006130(n) = A006130(n-1) + 3*A006130(n-2).

Original entry on oeis.org

1, 1, 8, 35, 266, 1680, 12804, 93093, 726440, 5635058, 45063668, 362121760, 2955642508, 24284658100, 201428123040, 1680921310635, 14119413718770, 119205791509200, 1011387051005100, 8617021562542470, 73704123363739440, 632601537174078420
Offset: 0

Views

Author

Paul D. Hanna, Nov 16 2011

Keywords

Comments

More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), S(0)=1, |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

Examples

			G.f.: A(x) = 1 + x + 2*4*x^2 + 5*7*x^3 + 14*19*x^4 + 42*40*x^5 + 132*97*x^6 + 429*217*x^7 + ... + A000108(n)*A006130(n)*x^n + ...
where the g.f. of A006130, F(x) = 1/(1-x-3*x^2), begins:
F(x) = 1 + x + 4*x^2 + 7*x^3 + 19*x^4 + 40*x^5 + 97*x^6 + 217*x^7 + ...
		

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!(Sqrt((1-2*x - Sqrt(1-4*x-48*x^2))/26)/x)); // G. C. Greubel, Jul 27 2018
  • Mathematica
    CoefficientList[Series[Sqrt[(1 - 2*x - Sqrt[1 - 4*x - 48*x^2])/26]/x, {x, 0, 30}], x] (* G. C. Greubel, Jul 27 2018 *)
  • PARI
    {a(n)=binomial(2*n, n)/(n+1)*polcoeff(1/(1-x-3*x^2+x*O(x^n)),n)}
    
  • PARI
    {a(n)=polcoeff(sqrt((1-2*x - sqrt(1-4*x-48*x^2+x^3*O(x^n)))/26)/x,n)}
    
  • PARI
    {a(n)=polcoeff(serreverse(x*sqrt(1-12*x^2+x^2*O(x^n)) - x^2)/x,n)}
    
  • PARI
    {a(n)=polcoeff((1/x)*serreverse(x-x^2 - 6*x^3*sum(m=0,n\2,binomial(2*m,m)/(m+1)*3^m*x^(2*m))+x^3*O(x^n)),n)}
    

Formula

G.f.: sqrt( (1-2*x - sqrt(1-4*x-48*x^2))/26 )/x.
G.f.: (1/x)*Series_Reversion( x*sqrt(1-12*x^2) - x^2 ).
G.f.: (1/x)*Series_Reversion( x-x^2 - 6*x^3*Sum_{n>=0} A000108(n)*3^n*x^(2*n) ).
G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x)^2 + 13*x^2*A(x)^4).
Conjecture: n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -12*(2*n-1)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 17 2011
a(n) = ( ((1+sqrt(13))/2)^(n+1) - ((1-sqrt(13))/2)^(n+1) )/sqrt(13) * binomial(2*n+1,n)/(2*n+1). - Paul D. Hanna, Sep 25 2012
0 = +a(n)*(+110592*a(n+3) -9216*a(n+4) -7392*a(n+5) +858*a(n+6)) +a(n+1)*(+6912*a(n+3) -1968*a(n+4) -910*a(n+5) +154*a(n+6)) +a(n+2)*(-240*a(n+3) -2*a(n+4) +41*a(n+5) -4*a(n+6)) +a(n+3)*(+6*a(n+3) +5*a(n+4) +3*a(n+5) -a(n+6)) for all n in Z. - Michael Somos, Jul 28 2018

A175291 Pisano period of A006130 modulo n.

Original entry on oeis.org

1, 3, 1, 6, 24, 3, 24, 6, 3, 24, 120, 6, 156, 24, 24, 12, 16, 3, 90, 24, 24, 120, 22, 6, 120, 156, 9, 24, 28, 24, 240, 24, 120, 48, 24, 6, 171, 90, 156, 24, 336, 24, 42, 120, 24, 66, 736, 12, 168, 120, 16, 156, 52, 9, 120, 24, 90, 84, 3480, 24, 20, 240, 24, 48, 312, 120, 748, 48, 22, 24, 5040, 6, 888, 171, 120, 90, 120, 156, 39
Offset: 1

Views

Author

R. J. Mathar, Mar 24 2010

Keywords

Examples

			Reading 0, 1, 1, 4, 7, 19, 40, 97, 217, 508, 1159, 2683, 6160, 14209, ... modulo n=7 gives 0, 1, 1, 4, 0, 5, 5, 6, 0, 4, 4, 2, 0, 6, 6, 3, 0, 2, 2, 1, 0, 3, 3, 5, 0, 1, 1, 4, 0, 5, 5, 6, 0, 4, 4, 2, 0, 6, 6, 3, 0, ... with period a(n=7)=24.
		

Crossrefs

Extensions

a(9) corrected by R. J. Mathar, Apr 18 2010

A176261 Triangle T(n,k) = A006130(k) - A006130(n) + A006130(n-k) read by rows.

Original entry on oeis.org

1, 1, 1, 1, -2, 1, 1, -2, -2, 1, 1, -11, -11, -11, 1, 1, -20, -29, -29, -20, 1, 1, -56, -74, -83, -74, -56, 1, 1, -119, -173, -191, -191, -173, -119, 1, 1, -290, -407, -461, -470, -461, -407, -290, 1, 1, -650, -938, -1055, -1100, -1100, -1055, -938, -650, 1
Offset: 0

Views

Author

Roger L. Bagula, Apr 13 2010

Keywords

Comments

Row sums are s(n) = {1, 2, 0, -2, -31, -96, -341, -964, -2784, -7484, -20041, ...}, obey s(n) = 3*s(n-1) + 3*s(n-2) - 11*s(n-3) - 3*s(n-4) + 9*s(n-5) and have g.f. (1-x+3*x^3-9*x^2)/((1-x)*(1-x-3*x^2)^2).

Examples

			Triangle begins as:
  1;
  1,     1;
  1,    -2,     1;
  1,    -2,    -2,     1;
  1,   -11,   -11,   -11,     1;
  1,   -20,   -29,   -29,   -20,     1;
  1,   -56,   -74,   -83,   -74,   -56,     1;
  1,  -119,  -173,  -191,  -191,  -173,  -119,     1;
  1,  -290,  -407,  -461,  -470,  -461,  -407,  -290,     1;
  1,  -650,  -938, -1055, -1100, -1100, -1055,  -938,  -650,     1;
  1, -1523, -2171, -2459, -2567, -2603, -2567, -2459, -2171, -1523, 1;
		

Crossrefs

Cf. A006130.

Programs

Formula

T(n,k) = T(n,n-k).
T(n,k) = A006130(k) - A006130(n) + A006130(n-k), where A006130(n) = Sum_{j=0..n} binomial(n-j, j)*3^j. - G. C. Greubel, Nov 24 2019

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

Views

Author

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Crossrefs

Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
    
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
    T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
    Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
    Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
    trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle
    def A039599_triangle(n) :
        D = [0]*(n+2); D[1] = 1
        b = True ; h = 1
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            if b : print([D[z] for z in (1..h-1)])
            b = not b
    A039599_triangle(10)  # Peter Luschny, May 01 2012
    

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A015518 a(n) = 2*a(n-1) + 3*a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082
Offset: 0

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Author

Keywords

Comments

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths: 1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.

Programs

  • Magma
    [Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011
    
  • Mathematica
    Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)
  • Maxima
    a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */
  • PARI
    a(n)=round(3^n/4)
    
  • Python
    for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018
    
  • Sage
    [round(3^n/4) for n in range(0,27)]
    

Formula

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + 1/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction). (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) = x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

Extensions

More terms from Emeric Deutsch, Apr 01 2004
Edited by Ralf Stephan, Aug 30 2004

A109466 Riordan array (1, x(1-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0
Offset: 0

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Author

Philippe Deléham, Aug 28 2005

Keywords

Comments

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

Examples

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)
		

Crossrefs

Cf. A026729 (unsigned version), A000108, A030528, A124644.

Programs

  • Magma
    /* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Formula

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A006131 a(n) = a(n-1) + 4*a(n-2), a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 5, 9, 29, 65, 181, 441, 1165, 2929, 7589, 19305, 49661, 126881, 325525, 833049, 2135149, 5467345, 14007941, 35877321, 91909085, 235418369, 603054709, 1544728185, 3956947021, 10135859761, 25963647845, 66507086889, 170361678269
Offset: 0

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Author

Keywords

Comments

Length-n words with letters {0,1,2,3,4} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Equals INVERTi transform of A063727: (1, 2, 8, 24, 80, 256, 832, ...). - Gary W. Adamson, Aug 12 2010
a(n) is equal to the permanent of the n X n Hessenberg matrix with 1's along the main diagonal, 2's along the superdiagonal and the subdiagonal, and 0's everywhere else. - John M. Campbell, Jun 09 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 5*a(n-2) equals the number of 5-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 8, 1, 6, 8, 48, 2, 24, 6,120, 8, 12, 48, 24, 4,136, 24, 18, 6, ... - R. J. Mathar, Aug 10 2012
This is one of only two Lucas-type sequences whose 8th term is a square. The other one is A097705. - Michel Marcus, Dec 07 2012
Numerators of stationary probabilities for the M2/M/1 queue. In this queue, customers arrives in groups of 2. Intensity of arrival = 1. Service rate = 4. There is only one server and an infinite queue. - Igor Kleiner, Nov 02 2018
Number of 4-compositions of n+2 with 1 not allowed as a part; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From M. Eren Kesim, May 13 2021: (Start)
a(n) is equal to the number of n-step walks from a universal vertex to another (itself or the other) on the diamond graph. It is also equal to the number of (n+1)-step walks from vertex A to vertex B on the graph below.
B--C
| /|
|/ |
A--D
(End)
From Wolfdieter Lang, Jan 03 2024: (Start)
This sequence {a(n-1)}, with a(-1) = 0, appears in the formula for powers of phi17 := (1 + sqrt(17))/2 = A222132, the fundamental (integer) algebraic number of Q(sqrt(17)): phi17^n = A052923(n) + a(n-1)*phi17, for n >= 0.
Limit_{n->oo} a(n+1)/a(n) = phi17. (End)

Examples

			G.f. = 1 + x + 5*x^2 + 9*x^3 + 29*x^4 + 65*x^5 + 181*x^6 + 441*x^7 + 1165*x^8 + ...
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    a:=[1,1];; for n in [3..30] do a[n]:=a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Dec 26 2019
    
  • Magma
    [ n eq 1 select 1 else n eq 2 select 1 else Self(n-1)+4*Self(n-2): n in [1..40] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    A006131:=-1/(-1+z+4*z**2); # conjectured by Simon Plouffe in his 1992 dissertation
    seq( simplify((2/I)^n*ChebyshevU(n, I/4)), n=0..30); # G. C. Greubel, Dec 26 2019
  • Mathematica
    m = 16; f[n_] = Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; Table[FullSimplify[ExpandAll[f[n]]], {n, 0, 15}]; N[%] (* Roger L. Bagula, Nov 21 2008 *)
    a[n_]:=(MatrixPower[{{1,4},{1,0}},n].{{1},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{1, 4}, {1, 1}, 29] (* Jean-François Alcover, Sep 25 2017 *)
    Table[2^n*Fibonacci[n+1, 1/2], {n,0,30}] (* G. C. Greubel, Dec 26 2019 *)
  • PARI
    a(n)=([0,1; 4,1]^n*[1;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016
    
  • PARI
    vector(31, n, (2/I)^(n-1)*polchebyshev(n-1, 2, I/4) ) \\ G. C. Greubel, Dec 26 2019
    
  • Python
    def A006131_list(n):
        list = [1, 1] + [0] * (n - 2)
        for i in range(2, n):
            list[i] = list[i - 1] + 4 * list[i - 2]
        return list
    print(A006131_list(29)) # M. Eren Kesim, Jul 19 2021
  • Sage
    [lucas_number1(n,1,-4) for n in range(1, 30)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: 1/(1 - x - 4*x^2).
a(n) = (((1+sqrt(17))/2)^(n+1) - ((1-sqrt(17))/2)^(n+1))/sqrt(17).
a(n+1) = Sum_{k=0..ceiling(n/2)} 4^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^(n-k)/2. - Paul Barry, Aug 28 2005
a(n) = A102446(n)/2. - Zerinvary Lajos, Jul 09 2008
a(n) = Sum_{k=0..n} A109466(n,k)*(-4)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = Product_{k=1..floor((n - 1)/2)} (1 + 16*cos(k*Pi/n)^2). - Roger L. Bagula, Nov 21 2008
Limiting ratio a(n+1)/a(n) is (1 + sqrt(17))/2 = 2.561552812... - Roger L. Bagula, Nov 21 2008
The fraction b(n) = a(n)/2^n satisfies b(n) = 1/2 b(n-1) + b(n-2); g.f. 1/(1-x/2-x^2); b(n) = (( (1+sqrt(17))/4 )^(n+1) - ( (1-sqrt(17))/4 )^(n+1))*2/sqrt(17). - Franklin T. Adams-Watters, Nov 30 2009
G.f.: G(0)/(2-x), where G(k) = 1 + 1/(1 - x*(17*k-1)/(x*(17*k+16) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013
G.f.: Q(0)/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + 4*x)/( x*(4*k+3 + 4*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
G.f.: Q(0)/2, where Q(k) = 1 + 1/(1 - x*(k+1 + 4*x)/( x*(k+3/2 + 4*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 14 2013
G.f.: 1 / (1 - x / (1 - 4*x / (1 + 4*x))). - Michael Somos, Sep 15 2013
a(n) = (Sum_{1<=k<=n+1, k odd} C(n+1,k)*17^((k-1)/2))/2^n. - Vladimir Shevelev, Feb 05 2014
a(n) = 2^n*Fibonacci(n+1, 1/2) = (2/i)^n*ChebyshevU(n, i/4). - G. C. Greubel, Dec 26 2019
E.g.f.: exp(x/2)*(sqrt(17)*cosh(sqrt(17)*x/2) + sinh(sqrt(17)*x/2))/sqrt(17). - Stefano Spezia, Dec 27 2019
a(n) = A344236(n) + A344261(n). - M. Eren Kesim, May 13 2021
With an initial 0 prepended, the sequence [0, 1, 1, 5, 9, 29, 65, ...] satisfies the congruences a(n*p^k) == e*a(n*p^(k-1)) (mod p^k) for positive integers k and n and all primes p, where e = +1 for the primes p listed in A296938, e = 0 when p = 17, otherwise e = -1. - Peter Bala, Dec 28 2022
a(n) = A052923(n+2)/4. - Wolfdieter Lang, Jan 03 2024
From Peter Bala, Jun 27 2025: (Start)
The following products telescope:
Product_{k >= 0} (1 + 4^k/a(2*k+1)) = 1 + sqrt(17).
Product_{k >= 1} (1 - 4^k/a(2*k+1)) = 1/18 * (1 + sqrt(17)).
Product_{k >= 0} (1 + (-4)^k/a(2*k+1)) = (1/17) * (17 + sqrt(17)).
Product_{k >= 1} (1 - (-4)^k/a(2*k+1)) = (1/18) * (17 + sqrt(17)). (End)

Extensions

More terms from Roger L. Bagula, Sep 26 2006

A105476 Number of compositions of n when each even part can be of two kinds.

Original entry on oeis.org

1, 1, 3, 6, 15, 33, 78, 177, 411, 942, 2175, 5001, 11526, 26529, 61107, 140694, 324015, 746097, 1718142, 3956433, 9110859, 20980158, 48312735, 111253209, 256191414, 589951041, 1358525283, 3128378406, 7203954255, 16589089473, 38200952238, 87968220657
Offset: 0

Views

Author

Emeric Deutsch, Apr 09 2005

Keywords

Comments

Row sums of A105475.
Starting (1, 3, 6, 15, ...) = sum of (n-1)-th row terms of triangle A140168. - Gary W. Adamson, May 10 2008
a(n) is also the number of compositions of n using 1's and 2's such that each run of like numbers can be grouped arbitrarily. For example, a(4) = 15 because 4 = (1)+(1)+(1)+(1) = (1+1)+(1)+(1) = (1)+(1+1)+(1) = (1)+(1)+(1+1) = (1+1)+(1+1) = (1+1+1)+(1) = (1)+(1+1+1) = (1+1+1+1) = (2)+(1)+(1) = (1)+(2)+(1) = (1)+(1)+(2) = (2)+(1+1) = (1+1)+(2) = (2)+(2) = (2+2). - Martin J. Erickson (erickson(AT)truman.edu), Dec 09 2008
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 69, 261, 321 and 324, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A006138. - Johannes W. Meijer, Aug 15 2010
Inverse INVERT transform of the left shifted sequence gives A000034.
Eigensequence of the triangle
1,
2, 1,
1, 2, 1,
2, 1, 2, 1,
1, 2, 1, 2, 1,
2, 1, 2, 1, 2, 1,
1, 2, 1, 2, 1, 2, 1,
2, 1, 2, 1, 2, 1, 2, 1 ... - Paul Barry, Feb 10 2011
Pisano period lengths: 1, 3, 1, 6, 24, 3, 24, 6, 1, 24, 120, 6, 156, 24, 24, 12, 16, 3, 90, 24, ... - R. J. Mathar, Aug 10 2012

Examples

			a(3)=6 because we have (3),(1,2),(1,2'),(2,1),(2',1) and (1,1,1).
		

Crossrefs

Programs

  • GAP
    a:=[1,3];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; Concatenation([1], a); # G. C. Greubel, Jan 15 2020
  • Magma
    I:=[1,1,3]; [n le 3 select I[n] else Self(n-1)+3*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Jul 21 2013
    
  • Magma
    R:=PowerSeriesRing(Integers(), 33); Coefficients(R!( 1/(1-(x/(1-x))-x^2/(1-x^2)))); // Marius A. Burtea, Jan 15 2020
    
  • Maple
    G:=(1-z^2)/(1-z-3*z^2): Gser:=series(G,z=0,35): 1,seq(coeff(Gser,z^n),n=1..33);
  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-x-3x^2), {x,0,35}], x] (* or *) Join[{1}, LinearRecurrence[{1, 3}, {1, 3}, 50]] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011; typo fixed by Vincenzo Librandi, Jul 21 2013 *)
    Table[Round[Sqrt[3]^(n-3)*(2*Sqrt[3]*Fibonacci[n+1, 1/Sqrt[3]] +Fibonacci[n, 1/Sqrt[3]])], {n, 0, 40}] (* G. C. Greubel, Jan 15 2020 *)
  • PARI
    Vec((1-x^2)/(1-x-3*x^2)+O(x^40)) \\ Charles R Greathouse IV, Jun 13 2013
    
  • Sage
    def A105476_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1-x^2)/(1-x-3*x^2) ).list()
    A105476_list(40) # G. C. Greubel, Jan 15 2020
    

Formula

G.f.: (1-x^2) / (1-x-3*x^2).
a(n) = a(n-1) + 3*a(n-2) for n>=3.
a(n) = 3*A006138(n-2), n>=2.
a(n) = ((2+sqrt(13))*(1+sqrt(13))^n - (2-sqrt(13))*(1-sqrt(13))^n)/(3*2^n*sqrt(13)) for n>0. - Bruno Berselli, May 24 2011
G.f.: 1/(1 - Sum_{k>=1} x^k*(1+x^k) ). - Joerg Arndt, Mar 09 2014
G.f.: 1/(1 - (x/(1-x)) - x^2/(1-x^2)) = 1/(1 - (x+2*x^2+x^3+2*x^4+x^5+2*x^6+...) ); in general 1/(1 - Sum_{j>=1} m(j)*x^j ) is the g.f. for compositions with m(k) sorts of part k. - Joerg Arndt, Feb 16 2015
a(n) = 3^((n-1)/2)*( 2*sqrt(3)*Fibonacci(n, 1/sqrt(3)) + Fibonacci(n, 1/sqrt(3)) ). - G. C. Greubel, Jan 15 2020
E.g.f.: 1/3 + (2/39)*exp(x/2)*(13*cosh((sqrt(13)*x)/2) + 2*sqrt(13)*sinh((sqrt(13)*x)/2)). - Stefano Spezia, Jan 15 2020

A175654 Eight bishops and one elephant on a 3 X 3 chessboard. G.f.: (1 - x - x^2)/(1 - 3*x - x^2 + 6*x^3).

Original entry on oeis.org

1, 2, 6, 14, 36, 86, 210, 500, 1194, 2822, 6660, 15638, 36642, 85604, 199626, 464630, 1079892, 2506550, 5811762, 13462484, 31159914, 72071654, 166599972, 384912086, 888906306, 2052031172, 4735527306, 10925175254, 25198866036, 58108609526, 133973643090
Offset: 0

Views

Author

Johannes W. Meijer, Aug 06 2010; edited Jun 21 2013

Keywords

Comments

a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 or 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a bishop on the eight side and corner squares but on the center square the bishop flies into a rage and turns into a raging elephant.
In chaturanga, the old Indian version of chess, one of the pieces was called gaja, elephant in Sanskrit. The Arabs called the game shatranj and the elephant became el fil in Arabic. In Spain chess became chess as we know it today but surprisingly in Spanish a bishop isn't a Christian bishop but a Moorish elephant and it still goes by its original name of el alfil.
On a 3 X 3 chessboard there are 2^9 = 512 ways for an elephant to fly into a rage on the central square (off the center the piece behaves like a normal bishop). The elephant is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program and A180140. For the corner squares the 512 elephants lead to 46 different elephant sequences, see the overview of elephant sequences and the crossreferences.
The sequence above corresponds to 16 A[5] vectors with decimal values 71, 77, 101, 197, 263, 269, 293, 323, 326, 329, 332, 353, 356, 389, 449 and 452. These vectors lead for the side squares to A000079 and for the central square to A175655.

References

  • Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.
  • David Hooper and Kenneth Whyld, The Oxford Companion to Chess, pp. 74, 366, 1992.

Crossrefs

Cf. Elephant sequences corner squares [decimal value A[5]]: A040000 [0], A000027 [16], A000045 [1], A094373 [2], A000079 [3], A083329 [42], A027934 [11], A172481 [7], A006138 [69], A000325 [26], A045623 [19], A000129 [21], A095121 [170], A074878 [43], A059570 [15], A175654 [71, this sequence], A026597 [325], A097813 [58], A057711 [27], 2*A094723 [23; n>=-1], A002605 [85], A175660 [171], A123203 [186], A066373 [59], A015518 [341], A134401 [187], A093833 [343].

Programs

  • Magma
    [n le 3 select Factorial(n) else 3*Self(n-1) +Self(n-2) -6*Self(n-3): n in [1..41]]; // G. C. Greubel, Dec 08 2021
    
  • Maple
    nmax:=28; m:=1; A[1]:=[0,0,0,0,1,0,0,0,1]: A[2]:=[0,0,0,1,0,1,0,0,0]: A[3]:=[0,0,0,0,1,0,1,0,0]: A[4]:=[0,1,0,0,0,0,0,1,0]: A[5]:=[0,0,1,0,0,0,1,1,1]: A[6]:=[0,1,0,0,0,0,0,1,0]: A[7]:=[0,0,1,0,1,0,0,0,0]: A[8]:=[0,0,0,1,0,1,0,0,0]: A[9]:=[1,0,0,0,1,0,0,0,0]: A:=Matrix([A[1], A[2], A[3], A[4], A[5], A[6], A[7], A[8], A[9]]): 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);
  • Mathematica
    LinearRecurrence[{3,1,-6}, {1,2,6}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2012 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -6,1,3]^n*[1;2;6])[1,1] \\ Charles R Greathouse IV, Oct 03 2016
    
  • Sage
    [( (1-x-x^2)/((1-2*x)*(1-x-3*x^2)) ).series(x,n+1).list()[n] for n in (0..40)] # G. C. Greubel, Dec 08 2021

Formula

G.f.: (1 - x - x^2)/(1 - 3*x - x^2 + 6*x^3).
a(n) = 3*a(n-1) + a(n-2) - 6*a(n-3) with a(0)=1, a(1)=2 and a(2)=6.
a(n) = ((6+10*A)*A^(-n-1) + (6+10*B)*B^(-n-1))/13 - 2^n with A = (-1+sqrt(13))/6 and B = (-1-sqrt(13))/6.
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n)*2*A000244(n)/(A075118(n) - A006130(n-1)*sqrt(13)).
a(n) = b(n) - b(n-1) - b(n-2), where b(n) = Sum_{k=1..n} Sum_{j=0..k} binomial(j,n-3*k+2*j)*(-6)^(k-j)*binomial(k,j)*3^(3*k-n-j), n>0, b(0)=1, with a(0) = b(0), a(1) = b(1) - b(0). - Vladimir Kruchinin, Aug 20 2010
a(n) = 2*A006138(n) - 2^n = 2*(A006130(n) + A006130(n-1)) - 2^n. - G. C. Greubel, Dec 08 2021
E.g.f.: 2*exp(x/2)*(13*cosh(sqrt(13)*x/2) + 3*sqrt(13)*sinh(sqrt(13)*x/2))/13 - cosh(2*x) - sinh(2*x). - Stefano Spezia, Feb 12 2023

A015440 a(n) = a(n-1) + 5*a(n-2), with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 6, 11, 41, 96, 301, 781, 2286, 6191, 17621, 48576, 136681, 379561, 1062966, 2960771, 8275601, 23079456, 64457461, 179854741, 502142046, 1401415751, 3912125981, 10919204736, 30479834641, 85075858321, 237475031526, 662854323131, 1850229480761, 5164501096416
Offset: 0

Views

Author

Keywords

Comments

Original name: Generalized Fibonacci numbers.
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 6*a(n-2) equals the number of 6-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 3, 6, 6, 1, 6, 21, 12, 18, 3, 40, 6, 56, 21, 6, 24, 16, 18, 360, 6, .... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Jan 02 2024: (Start)
This sequence {a(n-1)}, with a(-1) = 0, appears in the formula for powers of phi21 := (1 + sqrt(21))/2 = A222134 = 2.791287..., together with A(n) = A365824(n), as phi21^n = A(n) + a(n-1)*phi21(n), for n >= 0.
Limit_{n->oo} a(n)/a(n-1) = phi21. (End)

Crossrefs

Programs

Formula

a(n) = a(n-1) + 5*a(n-2).
a(n) = (( (1+sqrt(21))/2 )^(n+1) - ( (1-sqrt(21))/2 )^(n+1))/sqrt(21).
a(n) = Sum_{k=0..ceiling(n/2)} 5^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
G.f.: 1/(1 - x - 5x^2). - R. J. Mathar, Sep 03 2008
a(n) = Sum_{k=0..n} A109466(n,k)*(-5)^(n-k). - Philippe Deléham, Oct 26 2008
From Jeffrey R. Goodwin, May 28 2011: (Start)
A special case of a more general class of Lucas sequences given by
U(n) = U(n-1) + (4^(m-1)-1)/3 U(n-2).
U(n) = (( (1+sqrt((4^(m)-1)/3))/2 )^(n+1) - ( (1-sqrt((4^(m)-1)/3))/2 )^(n+1))/sqrt((4^(m)-1)/3). Fix m = 2 to get the formula for the Fibonacci sequence, fix m = 3 to get the formula for a(n). (End)
G.f.: G(0)/(2-x), where G(k)= 1 + 1/(1 - x*(21*k-1)/(x*(21*k+20) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013
G.f.: Q(0)/x -1/x, where Q(k) = 1 + 5*x^2 + (k+2)*x - x*(k+1 + 5*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{k=1..n+1, k odd} binomial(n+1,k)*21^((k-1)/2))/2^n. - Vladimir Shevelev, Feb 05 2014
With an initial 0 prepended, the sequence [0, 1, 1, 6, 11, 41, 96, ...] satisfies the congruences a(n*p^k) == (3|p)*(7|p)*a(n*p^(k-1)) (mod p^k) for positive integers k and n and all primes p, where (n|p) denotes the Legendre symbol. See Young, Theorem 1, Corollary 1(i). - Peter Bala, Dec 28 2022
a(n) = sqrt(-5)^(n-1)*S(n-1,1/sqrt(-5)), for n >= 0, with the Chebyshev polynomial S(n, x) (see A049310). - Wolfdieter Lang, Nov 17 2023
From Peter Bala, Jun 27 2025: (Start)
The following products telescope:
Product_{k >= 0} (1 + 5^k/a(2*k+1)) = 1 + sqrt(21).
Product_{k >= 1} (1 - 5^k/a(2*k+1)) = 1/22 * (1 + sqrt(21)).
Product_{k >= 0} (1 + (-5)^k/a(2*k+1)) = (1/21) * (21 + sqrt(21)).
Product_{k >= 1} (1 - (-5)^k/a(2*k+1)) = (1/22) * (21 + sqrt(21)). (End)
E.g.f.: exp(x/2)*(sqrt(21)*cosh(sqrt(21)*x/2) + sinh(sqrt(21)*x/2))/sqrt(21). - Stefano Spezia, Jul 04 2025
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