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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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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

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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

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

Original entry on oeis.org

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1
Offset: 0

Views

Author

Philippe Deléham, Jun 09 2007

Keywords

Comments

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011
Row sums: A057078. - Philippe Deléham, Jun 11 2007
Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012
If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

Examples

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020
		

Crossrefs

Programs

  • Maple
    # The function RiordanSquare is defined in A321620.
    RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
    LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019
  • Mathematica
    max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)
  • Sage
    @CachedFunction
    def A129818(n,k):
        if n< 0: return 0
        if n==0: return 1 if k == 0 else 0
        h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
        return A129818(n-1,k-1) - A129818(n-2,k) - h
    for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Formula

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).
Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009
O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
From Wolfdieter Lang, Dec 20 2010: (Start)
Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).
T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)
T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012
A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019
Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020

A085260 Ratio-determined insertion sequence I(0.0833344) (see the link below).

Original entry on oeis.org

1, 12, 155, 2003, 25884, 334489, 4322473, 55857660, 721827107, 9327894731, 120540804396, 1557702562417, 20129592507025, 260127000028908, 3361521407868779, 43439651302265219, 561353945521579068
Offset: 1

Views

Author

John W. Layman, Jun 23 2003

Keywords

Comments

This sequence is the ratio-determined insertion sequence (RDIS) "twin" to A078362 (see the link for an explanation of "twin"). See A082630 or A082981 for recent examples of RDIS sequences.
a(n) = L(n,13), where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(11)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Seems to be positive values of x (or y) satisfying x^2 - 13xy + y^2 + 11 = 0. - Colin Barker, Feb 10 2014
It appears that the b-file, formulas and programs are based on the conjectured, so far apparently unproved recurrence relation. - M. F. Hasler, Nov 05 2018
Nonnegative y values in solutions to the Diophantine equation 11*x^2 - 15*y^2 = -4. The corresponding x values are in A126866. Note that a(n+1)^2 - a(n)*a(n+2) = -11. - Klaus Purath, Mar 21 2025

Crossrefs

Row 13 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,12]; [n le 2 select I[n] else 13*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 18 2018
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 13 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
    LinearRecurrence[{13,-1}, {1,12}, 30] (* G. C. Greubel, Jan 18 2018 *)
  • PARI
    my(x='x+O('x^30)); Vec(x*(1-x)/(1-13*x+x^2)) \\ G. C. Greubel, Jan 18 2018
    

Formula

It appears that the sequence satisfies a(n+1) = 13*a(n) - a(n-1). [Corrected by M. F. Hasler, Nov 05 2018]
If the recurrence a(n+2) = 13*a(n+1) - a(n) holds then for n > 0, a(n)*a(n+3) = 143 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
G.f.: x*(1-x)/(1 - 13*x + x^2). - Philippe Deléham, Nov 17 2008
For n>1, a(n) is the numerator of the continued fraction [1,11,1,11,...,1,11] with (n-1) repetitions of 1,11. - Greg Dresden, Sep 10 2019
Showing 1-3 of 3 results.