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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

A026009 Triangular array T read by rows: T(n,0) = 1 for n >= 0; T(1,1) = 1; and for n >= 2, T(n,k) = T(n-1,k-1) + T(n-1,k) for k = 1,2,...,[(n+1)/2]; T(n,n/2 + 1) = T(n-1,n/2) if n is even.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 4, 6, 3, 1, 5, 10, 9, 1, 6, 15, 19, 9, 1, 7, 21, 34, 28, 1, 8, 28, 55, 62, 28, 1, 9, 36, 83, 117, 90, 1, 10, 45, 119, 200, 207, 90, 1, 11, 55, 164, 319, 407, 297, 1, 12, 66, 219, 483, 726, 704, 297, 1, 13, 78, 285, 702, 1209, 1430, 1001, 1, 14, 91, 363, 987, 1911, 2639, 2431, 1001
Offset: 0

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Author

Keywords

Examples

			From _Jonathon Kirkpatrick_, Jul 01 2016: (Start)
Triangle begins:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3;
  1,  4,  6,   3;
  1,  5, 10,   9;
  1,  6, 15,  19,   9;
  1,  7, 21,  34,  28;
  1,  8, 28,  55,  62,   28;
  1,  9, 36,  83, 117,   90;
  1, 10, 45, 119, 200,  207,   90;
  1, 11, 55, 164, 319,  407,  297;
  1, 12, 66, 219, 483,  726,  704,  297;
  1, 13, 78, 285, 702, 1209, 1430, 1001;
  ... (End)
		

Crossrefs

Sums involving this sequence: A026010, A027287, A027288, A027289, A027290, A027291, A027292.

Programs

  • Magma
    [1] cat [Binomial(n,k) - Binomial(n,k-3): k in [0..Floor((n+2)/2)], n in [1..15]]; // G. C. Greubel, Mar 18 2021
  • Mathematica
    T[n_, k_]:= Binomial[n, k] - Binomial[n, k-3];
    Join[{1}, Table[T[n, k], {n,14}, {k,0,Floor[(n+2)/2]}]//Flatten] (* G. C. Greubel, Mar 18 2021 *)
  • Sage
    [1]+flatten([[binomial(n,k) - binomial(n,k-3) for k in (0..(n+2)//2)] for n in (1..15)]) # G. C. Greubel, Mar 18 2021
    

Formula

T(n, k) = binomial(n, k) - binomial(n, k-3). - Darko Marinov (marinov(AT)lcs.mit.edu), May 17 2001
Sum_{k=0..floor((n+2)/2)} T(n, k) = A026010(n). - G. C. Greubel, Mar 18 2021

A000782 a(n) = 2*Catalan(n) - Catalan(n-1).

Original entry on oeis.org

1, 3, 8, 23, 70, 222, 726, 2431, 8294, 28730, 100776, 357238, 1277788, 4605980, 16715250, 61020495, 223931910, 825632610, 3056887680, 11360977650, 42368413620, 158498860260, 594636663660, 2236748680998, 8433988655580, 31872759742852, 120699748759856
Offset: 1

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Author

Keywords

Comments

Number of Dyck (n+1)-paths that have a leading or trailing hill. - David Scambler, Aug 22 2012
a(n) is the number of parking functions of size n avoiding the patterns 132, 213, 312, and 321. - Lara Pudwell, Apr 10 2023
Number of Dyck (n+1)-paths that have exactly one return to the x-axis and/or a peak in the center of the path. - Roger Ford, May 15 2024

Crossrefs

Partial sums of A071735.
Essentially the same as A061557.

Programs

  • Magma
    [2*Catalan(n)-Catalan(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 10 2012
  • Mathematica
    CoefficientList[Series[(1+x*(1-(1-4*x)^(1/2))/(2*x)^1)*((1-(1-4*x)^(1/2))/(2*x))^2,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)

Formula

Expansion of x*(1 + x*C)*C^2, where C = (1 - (1 - 4*x)^(1/2))/(2*x) is the g.f. for the Catalan numbers, A000108.
Also, expansion of (1 + x^2*C^2)*C - 1, where C = (1 - (1 - 4*x)^(1/2))/(2*x) is the g.f. for Catalan numbers, A000108.
a(n) = (7*n - 5)/(n + 1) * C(n-1), where C(n) = A000108(n). - Ralf Stephan, Jan 13 2004
a(n) = leftmost column term of M^(n-1)*V, where M is a tridiagonal matrix with 1's in the super- and subdiagonals, (1, 2, 2, 2, ...) in the main diagonal, and the rest zeros; and V is the vector [1, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 16 2011
a(n) = A000108(n+1) - A026012(n-1). - David Scambler, Aug 22 2012

A059346 Difference array of Catalan numbers A000108 read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 2, 3, 5, 3, 4, 6, 9, 14, 6, 9, 13, 19, 28, 42, 15, 21, 30, 43, 62, 90, 132, 36, 51, 72, 102, 145, 207, 297, 429, 91, 127, 178, 250, 352, 497, 704, 1001, 1430, 232, 323, 450, 628, 878, 1230, 1727, 2431, 3432, 4862, 603, 835, 1158, 1608, 2236, 3114
Offset: 0

Views

Author

N. J. A. Sloane, Jan 27 2001

Keywords

Examples

			Array starts:
      1       1       2       5      14      42     132     429
      0       1       3       9      28      90     297    1001
      1       2       6      19      62     207     704    2431
      1       4      13      43     145     497    1727    6071
      3       9      30     102     352    1230    4344   15483
      6      21      72     250     878    3114   11139   40143
     15      51     178     628    2236    8025   29004  105477
     36     127     450    1608    5789   20979   76473  280221
     91     323    1158    4181   15190   55494  203748  751422
    232     835    3023   11009   40304  148254  547674 2031054
    603    2188    7986   29295  107950  399420 1483380 5527750
Triangle starts:
  1;
  0,  1;
  1,  1,  2;
  1,  2,  3,  5;
  3,  4,  6,  9, 14;
		

Crossrefs

Top row is A000108, leading diagonals give A005043, A001006, A005554.
Row sums are A106640.

Programs

  • Maple
    T := (n,k) -> (-1)^(n-k)*binomial(2*k,k)*hypergeom([k-n,k+1/2], [k+2], 4)/(k+1): seq(seq(simplify(T(n,k)), k=0..n), n=0..10);
    # Peter Luschny, Aug 16 2012, updated May 25 2021
  • Mathematica
    max = 11; t = Table[ Differences[ Table[ CatalanNumber[k], {k, 0, max}], n], {n, 0, max}]; Flatten[ Table[t[[n-k+1, k]], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Nov 15 2011 *)
  • Sage
    def T(n, k) :
        if k > n : return 0
        if n == k : return binomial(2*n, n)/(n+1)
        return T(n-1, k) - T(n, k+1)
    A059346 = lambda n,k: (-1)^(n-k)*T(n, k)
    for n in (0..5): [A059346(n,k) for k in (0..n)] # Peter Luschny, Aug 16 2012

Formula

T(n, k) = (-1)^(n-k)*binomial(2*k,k)/(k+1)*hypergeometric([k-n, k+1/2],[k+2], 4). - Peter Luschny, Aug 16 2012

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Feb 16 2001

A026029 Number of (s(0), s(1), ..., s(2n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n, s(0) = 3, s(2n) = 3. Also T(2n,n), where T is defined in A026022.

Original entry on oeis.org

1, 2, 6, 20, 69, 242, 858, 3068, 11050, 40052, 145996, 534888, 1968685, 7276050, 26993490, 100490220, 375287550, 1405622460, 5278838100, 19873977240, 74994427170, 283595947284, 1074568266756, 4079184055640, 15511924233204, 59083160374952, 225384613313944
Offset: 0

Views

Author

Keywords

Comments

Hankel transform is A008619(n+1). - Paul Barry, May 11 2009

Programs

  • Mathematica
    CoefficientList[Series[(1 - 2*x)*(-1 + Sqrt[1 - 4*x] + 2*x)^2 / (4*x^4), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 03 2019 *)

Formula

Expansion of (1+x^2*C^4)*C^2, where C = (1-(1-4*x)^(1/2))/(2*x) is g.f. for Catalan numbers, A000108.
a(n) = Sum_{k=0..n} C(n, k)*Sum_{i=0..k} C(k, 2i)*A000108(i+1). - Paul Barry, Jul 18 2003
a(n) = Sum_{k=0..3} A039599(n,k) = A000108(n) + A000245(n) + A000344(n) + A000588(n) = A026012(n) + A000588(n). - Philippe Deléham, Nov 12 2008
a(n) = C(2n,n) - C(2n,n-4). - Paul Barry, May 11 2009
Conjecture: (n+4)*a(n) + 6*(-n-2)*a(n-1) + 4*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 24 2012
a(n) ~ 4^(n+2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 03 2019
E.g.f.: exp(2*x)*(BesselI(0, 2*x) - BesselI(4, 2*x)). - Stefano Spezia, Jan 17 2024

A026020 a(n) = binomial(4n, n) - binomial(4n, n - 3).

Original entry on oeis.org

1, 4, 28, 219, 1804, 15314, 132572, 1163565, 10316924, 92195488, 829016968, 7492106505, 67991427828, 619193535380, 5655829748520, 51794730347745, 475390078267356, 4371917301657488, 40276635724273936
Offset: 0

Views

Author

Keywords

Crossrefs

a(n) = T(4n, n), where T is the array defined in A026009.
Bisections are A026012 and A026016.

Programs

  • Magma
    [Binomial(4*n, n) - Binomial(4*n, n-3): n in [0..20]]; // G. C. Greubel, Mar 22 2021
    
  • Maple
    A026020:= n-> binomial(4*n,n) - binomial(4*n,n-3); seq(A026020(n), n=0..20); # G. C. Greubel, Mar 22 2021
  • Mathematica
    Table[Binomial[4n, n] - Binomial[4n, n - 3], {n, 0, 19}] (* Alonso del Arte, Jun 06 2019 *)
  • PARI
    a(n) = binomial(4*n, n) - binomial(4*n, n-3) \\ Felix Fröhlich, Jun 06 2019
    
  • Sage
    [binomial(4*n, n) - binomial(4*n, n-3) for n in (0..20)] # G. C. Greubel, Mar 22 2021

Formula

G.f.: (g - 2)*(1 - g + g^2)*g/(3*g - 4) where g = 1 + x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
a(n) = A005810(n) - A004333(n) for n > 2 - Felix Fröhlich, Jun 06 2019

A127158 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and having k branches of length 1 starting from the root (0<=k<=n).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 3, 5, 5, 0, 1, 7, 18, 9, 7, 0, 1, 20, 52, 37, 13, 9, 0, 1, 59, 168, 113, 60, 17, 11, 0, 1, 184, 546, 388, 190, 87, 21, 13, 0, 1, 593, 1826, 1313, 688, 283, 118, 25, 15, 0, 1, 1964, 6211, 4545, 2408, 1076, 392, 153, 29, 17, 0, 1, 6642, 21459
Offset: 0

Views

Author

Emeric Deutsch, Mar 01 2007

Keywords

Comments

Row sums are the Catalan numbers (A000108). T(n,0)=A030238(n-2) for n>=2. Sum(k*T(n,k),k=0..n)=A026012(n-1) for n>=1.

Examples

			Triangle starts:
1;
0,1;
1,0,1;
1,3,0,1;
3,5,5,0,1;
7,18,9,7,0,1;
		

Crossrefs

Programs

  • Maple
    C:=(1-sqrt(1-4*z))/2/z: G:=1/(1-t*z*C+t*z^2*C-z^2*C): Gser:=simplify(series(G,z=0,15)): for n from 0 to 12 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 12 do seq(coeff(P[n],t,j),j=0..n) od; # yields sequence in triangular form

Formula

G.f.= 1/(1-tzC+tz^2*C-z^2*C), where C=[1-sqrt(1-4z)]/(2z) is the Catalan function.

A071738 Expansion of (1+x^3*C^4)*C, where C = (1-(1-4*x)^(1/2))/(2*x) is g.f. for Catalan numbers, A000108.

Original entry on oeis.org

1, 1, 2, 6, 19, 62, 207, 704, 2431, 8502, 30056, 107236, 385662, 1396652, 5088865, 18642420, 68624295, 253706790, 941630580, 3507232740, 13105289370, 49114150020, 184560753390, 695267483664, 2625197720454, 9933364416572
Offset: 0

Views

Author

N. J. A. Sloane, Jun 06 2002

Keywords

Crossrefs

Essentially the same as A026012.

A098977 Triangle read by rows: counts ordered trees by number of edges and position of first edge that terminates at a vertex of outdegree 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 5, 3, 2, 9, 14, 9, 6, 4, 21, 42, 28, 19, 13, 9, 51, 132, 90, 62, 43, 30, 21, 127, 429, 297, 207, 145, 102, 72, 51, 323, 1430, 1001, 704, 497, 352, 250, 178, 127, 835, 4862, 3432, 2431, 1727, 1230, 878, 628, 450, 323, 2188, 16796, 11934, 8502
Offset: 1

Views

Author

David Callan, Oct 24 2004

Keywords

Comments

T(n,k) = number of ordered trees on n edges whose k-th edge (in preorder or "walk around from root" order) is the first one that terminates at a vertex of outdegree 1 (k=0 if there is no such edge). The first column and the main diagonal (after initial entry) are Motzkin numbers (A001006). Each interior entry is the sum of its North and East neighbors.

Examples

			Table begins
\ k 0, 1, 2, ...
n
1 | 1
2 | 1, 1
3 | 2, 2, 1
4 | 4, 5, 3, 2
5 | 9, 14, 9, 6, 4
6 | 21, 42, 28, 19, 13, 9
7 | 51, 132, 90, 62, 43, 30, 21
8 |127, 429, 297, 207, 145, 102, 72, 51
T(4,2)=3 counts the following ordered trees (drawn down from root).
..|..../\..../|\..
./.\....|.....|...
.|......|.........
		

Crossrefs

Column k=1 is A000108 (apart from first term), k=2 is A000245, k=3 is A026012.

Programs

  • Mathematica
    Clear[v] MotzkinNumber[n_]/;IntegerQ[n] && n>=0 := If[0<=n<=1, 1, Module[{x = 1, y = 1}, Do[temp = ((2*i + 1)*y + 3*(i - 1)*x)/(i + 2); x = y; y = temp, {i, 2, n}]; y]]; v[n_, 0]/; n>=1 := MotzkinNumber[n-1]; v[n_, k_]/; k>=n := 0; v[n_, k_]/; n>=2 && k==n-1 := MotzkinNumber[n-2]; v[n_, k_]/; n>=3 && 1<=k<=n-2 := v[n, k] = v[n, k+1]+v[n-1, k]; TableForm[Table[v[n, k], {n, 10}, {k, 0, n-1}]]

Formula

G.f. for column k=0 is (1 - z - (1-2*z-3*z^2)^(1/2))/(2*z^2) = Sum_{n>=1}T(n, 0)z^n. G.f. for columns k>=1 is (t*(1 - (1 - 4*z)^(1/2) - 2*z))/ (1 - t + t*(1 - 4*z)^(1/2) + t*z + (1 - 2*t*z - 3*t^2*z^2)^(1/2)) = Sum_{n>=2, 1<=k<=n-1}T(n, k)z^n*t^k.

A123353 Triangle read by rows, giving Kekulé numbers for certain benzenoids (see the Cyvin-Gutman book for details).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 20, 19, 1, 1, 5, 50, 155, 62, 1, 1, 6, 105, 805, 1315, 207, 1, 1, 7, 196, 3136, 15218, 11957, 704, 1, 1, 8, 336, 9996, 118188, 326053, 114972, 2431, 1, 1, 9, 540, 27468, 690480, 5355756, 7736092, 1157650, 8502, 1
Offset: 0

Views

Author

N. J. A. Sloane, Oct 14 2006

Keywords

Examples

			Triangle begins:
  1
  1 1
  1 2 1
  1 3 6 1
  1 4 20 19 1
  1 5 50 155 62 1
  1 6 105 805 1315 207 1
  1 7 196 3136 15218 11957 704 1
  1 8 336 9996 118188 326053 114972 2431 1
  1 9 540 27468 690480 5355756 7736092 1157650 8502 1
  1 10 825 67320 3256308 60864012 282497568 199806100 12115220 30056 1
		

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 188).

Crossrefs

Diagonals give A026012, A123355, A123356.
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