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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

N. J. A. Sloane

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.

Links

Crossrefs

Columns give: A000108, A000245, A000344, A000588, A001392, A000589, A000590, A000012, A005408, A014107(n>1).
Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).
Cf. A008313, A039598, A084938, A000007, A067311, A321620.

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

A118973 Number of hill-free Dyck paths of semilength n+2 and having length of first descent equal to 2 (a hill in a Dyck path is a peak at level 1).

Original entry on oeis.org

1, 0, 2, 5, 16, 51, 168, 565, 1934, 6716, 23604, 83806, 300154, 1083137, 3934404, 14374413, 52787766, 194746632, 721435884, 2682522918, 10008240456, 37455101382, 140569122624, 528926230530, 1994980278636, 7541234323096
Offset: 0

Views

Author

Emeric Deutsch, May 08 2006

Keywords

Comments

Also, for a given j>=2, number of hill-free Dyck paths of semilength n+j and having length of first descent equal to j. a(n)=A000108(n+1)-A000108(n)-[A000957(n+2)-A000957(n+1)]. Columns 2,3,4,... of A118972 (without the initial 0's).

Examples

			a(2)=2 because we have uu(dd)uudd and uuu(dd)udd, where u=(1,1),d=(1,-1) (the first descents are shown between parentheses).

Links

Crossrefs

Cf. A000108, A000957, A118972, A001558.

Programs

  • Maple
    F:=(1-sqrt(1-4*z))/z/(3-sqrt(1-4*z)): C:=(1-sqrt(1-4*z))/2/z: g:=(1-z)*C*F: gser:=series(g,z=0,33): seq(coeff(gser,z,n),n=0..28);
A118973List := proc(m) local A, P, n; A := [1,0]; P := [1,0];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-2]]);
A := [op(A), P[-1]] od; A end: A118973List(26); # Peter Luschny, Mar 26 2022
  • Mathematica
    CoefficientList[Series[(1-x)*(1-Sqrt[1-4*x])/x/(3-Sqrt[1-4*x])*(1-Sqrt[1-4*x])/2/x, {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 20 2014 *)
  • Maxima
    a(n):=(sum((k+2)*(-1)^k*(binomial(2*n-k+1,n-k)/(n+2)-binomial(2*n-k-1,n-k-1)/(n+1)),k,0,n-1))+(-1)^(n); /* Vladimir Kruchinin, Mar 06 2016 */
  • PARI
    my(x='x+O('x^25)); Vec((1-x)*(1-sqrt(1-4*x))/x/(3-sqrt(1-4*x))*(1-sqrt(1-4*x))/2/x) \\ G. C. Greubel, Feb 08 2017

Formula

G.f.: (1-x)*C*F, where F = (1-sqrt(1-4*x))/(x*(3-sqrt(1-4*x))) and C = (1-sqrt(1-4*x))/(2*x) is the Catalan function.
a(n) ~ 5*4^n/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 20 2014
a(n) = (Sum_{k=0..n-1}((k+2)*(-1)^k*(binomial(2*n-k+1,n-k)/(n+2)-binomial(2*n-k-1,n-k-1)/(n+1))))+(-1)^(n). - Vladimir Kruchinin. Mar 06 2016
D-finite with recurrence +2*(n+2)*a(n) +(-7*n-2)*a(n-1) +2*(-3*n+1)*a(n-2) +(7*n-26)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Jul 26 2022

A167772 Riordan array (c(x)/(1+x*c(x)), x*c(x)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 6, 8, 6, 3, 1, 18, 24, 18, 10, 4, 1, 57, 75, 57, 33, 15, 5, 1, 186, 243, 186, 111, 54, 21, 6, 1, 622, 808, 622, 379, 193, 82, 28, 7, 1, 2120, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1, 7338, 9458, 7338, 4596, 2476, 1164, 474, 163, 45, 9, 1
Offset: 0

Views

Author

Philippe Deléham, Nov 11 2009, corrected Nov 12 2009

Keywords

Examples

			Triangle begins:
     1;
     0,    1;
     1,    1,    1;
     2,    3,    2,    1;
     6,    8,    6,    3,   1;
    18,   24,   18,   10,   4,   1;
    57,   75,   57,   33,  15,   5,   1;
   186,  243,  186,  111,  54,  21,   6,  1;
   622,  808,  622,  379, 193,  82,  28,  7,  1;
  2120, 2742, 2120, 1312, 690, 311, 118, 36,  8,  1;
Production matrix begins:
  0, 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, 1, 1;
  1, 1, 1, 1, 1, 1, 1, 1, 1;
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
  ... - _Philippe Deléham_, Mar 03 2013

Links

Crossrefs

Cf. A000957, A000958 (row sums), A001558, A001559, A104629, A167769.
Diagonals: A000012, A000217, A001477, A166830.

Programs

  • Haskell
    import Data.List (genericIndex)
a167772 n k = genericIndex (a167772_row n) k
a167772_row n = genericIndex a167772_tabl n
a167772_tabl = [1] : [0, 1] :
               map (\xs@(:x:) -> x : xs) (tail a065602_tabl)
-- Reinhard Zumkeller, May 15 2014
  • Mathematica
    A065602[n_, k_]:= A065602[n,k]= Sum[(k-1+2*j)*Binomial[2*(n-j)-k-1, n-1]/(2*(n-j) -k-1), {j, 0, (n-k)/2}];
T[n_, k_]:= If[k==0, A065602[n+1,3] + Boole[n==0], A065602[n+1, k+1]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 26 2022 *)
  • SageMath
    def A065602(n,k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
def A167772(n,k):
    if (k==0): return A065602(n+1,3) + bool(n==0)
    else: return A065602(n+1,k+1)
flatten([[A167772(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2022

Formula

Sum_{k=0..n} T(n, k) = A000958(n+1).
From Philippe Deléham, Nov 12 2009: (Start)
Sum_{k=0..n} T(n,k)*2^k = A014300(n).
Sum_{k=0..n} T(n,k)*2^(n-k) = A064306(n). (End)
For n > 0: T(n,0) = A065602(n+1,3), T(n,k) = A065602(n+1,k+1), k = 1..n. - Reinhard Zumkeller, May 15 2014

A118972 Triangle read by rows: T(n,k) is the number of hill-free Dyck paths of semilength n and having length of first descent equal to k (1<=k<=n; n>=1). A hill in a Dyck path is a peak at level 1.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 3, 2, 0, 1, 10, 5, 2, 0, 1, 33, 16, 5, 2, 0, 1, 111, 51, 16, 5, 2, 0, 1, 379, 168, 51, 16, 5, 2, 0, 1, 1312, 565, 168, 51, 16, 5, 2, 0, 1, 4596, 1934, 565, 168, 51, 16, 5, 2, 0, 1, 16266, 6716, 1934, 565, 168, 51, 16, 5, 2, 0, 1, 58082, 23604, 6716, 1934, 565, 168
Offset: 1

Views

Author

Emeric Deutsch, May 08 2006

Keywords

Comments

Row sums are the Fine numbers (A000957).
T(n,1) = A001558(n-3) for n>=3.
T(n,k) = A118973(n-k) for n>=k>=2.
Sum_{k=1..n} k*T(n,k) = A118974(n).

Examples

			T(5,2)=5 because we have uu(dd)uududd, uu(dd)uuuddd,uuu(dd)uuddd,uuu(dd)ududd and uuuu(dd)uddd, where u=(1,1), d=(1,-1) (the first descents are shown between parentheses).
Triangle starts:
  0;
  0,1;
  1,0,1;
  3,2,0,1;
  10,5,2,0,1;
  33,16,5,2,0,1;
  ...

Links

Crossrefs

Cf. A000957, A001558, A118973, A118974.

Programs

  • Maple
    F:=(1-sqrt(1-4*z))/z/(3-sqrt(1-4*z)): C:=(1-sqrt(1-4*z))/2/z: G:=t*z^2*C*F*(C-(1-t)/(1-t*z)): Gser:=simplify(series(G,z=0,15)): for n from 1 to 12 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 1 to 12 do seq(coeff(P[n],t,j),j=1..n) od; # yields sequence in triangular form

Formula

G.f.: t*z^2*C*F*(C-(1-t)/(1-t*z)), where F = (1-sqrt(1-4*z))/(z*(3-sqrt(1-4*z))) and C = (1-sqrt(1-4*z))/(2*z) is the Catalan function.

A167769 Pendular trinomial triangle (p=0), read by rows of 2n+1 terms (n>=0), defined by the recurrence : if 0

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 6, 8, 6, 3, 1, 0, 0, 1, 4, 10, 18, 24, 18, 10, 4, 1, 0, 0, 1, 5, 15, 33, 57, 75, 57, 33, 15, 5, 1, 0, 0, 1, 6, 21, 54, 111, 186, 243, 186, 111, 54, 21, 6, 1, 0, 0, 1, 7, 28, 82, 193, 379, 622, 808, 622, 379, 193, 82, 28, 7, 1, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Nov 11 2009

Keywords

Comments

See A119369 for p=1 and A122445 for p=2. The diagonals may be generated by iterated convolutions of a base sequence B (A000108(n)) with the sequence C (A000957(n+1)) of central terms.

Examples

			Triangle begins :
  1;
  1, 0,  0;
  1, 1,  1,  0,  0;
  1, 2,  3,  2,  1,  0,  0;
  1, 3,  6,  8,  6,  3,  1,  0,  0;
  1, 4, 10, 18, 24, 18, 10,  4,  1, 0, 0,
  1, 5, 15, 33, 57, 75, 57, 33, 15, 5, 1, 0, 0; ...

References

  • Kim, Ki Hang; Rogers, Douglas G.; Roush, Fred W. Similarity relations and semiorders. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 577--594, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561081 (81i:05013) - From N. J. A. Sloane, Jun 05 2012

Links

Crossrefs

Cf. A000108, A000957, A000958, A001558, A001559, A071724, A104629.

Programs

  • Maple
    T:= proc(n, k) option remember;
      if k=0 and n=0 then 1;
    elif k<0 or k>2*(n-1) then 0;
    elif n=2 and k<3 then 1;
    elif kG. C. Greubel, Mar 17 2021
  • Mathematica
    T[n_, k_]:= T[n, k]= If[k==0 && n==0, 1, If[k<0 || k>2*(n-1), 0, If[n==2 && k<3, 1, If[kG. C. Greubel, Mar 17 2021 *)
  • PARI
    T(n, k)=if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, if(kPaul D. Hanna, Nov 12 2009
  • Sage
    @CachedFunction
def T(n, k):
    if (k==0 and n==0): return 1
    elif (k<0 or k>2*(n-1)): return 0
    elif (n==2 and k<3): return 1
    elif (kG. C. Greubel, Mar 17 2021

Formula

Sum_{k=0..2*n} T(n,k) = A071724(n) = [n=0] + 3*binomial(2n,n-1)/(n+2) = [n=0] + n*C(n)/(n+2), where C(n) are the Catalan numbers (A000108). - G. C. Greubel, Mar 17 2021

A237619 Riordan array (1/(1+x*c(x)), x*c(x)) where c(x) is the g.f. of Catalan numbers (A000108).

Original entry on oeis.org

1, -1, 1, 0, 0, 1, -1, 1, 1, 1, -2, 2, 3, 2, 1, -6, 6, 8, 6, 3, 1, -18, 18, 24, 18, 10, 4, 1, -57, 57, 75, 57, 33, 15, 5, 1, -186, 186, 243, 186, 111, 54, 21, 6, 1, -622, 622, 808, 622, 379, 193, 82, 28, 7, 1, -2120, 2120, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1
Offset: 0

Views

Author

Philippe Deléham, Feb 10 2014

Keywords

Examples

			Triangle begins:
    1;
   -1,  1;
    0,  0,  1;
   -1,  1,  1,  1;
   -2,  2,  3,  2,  1;
   -6,  6,  8,  6,  3,  1;
  -18, 18, 24, 18, 10,  4, 1;
  -57, 57, 75, 57, 33, 15, 5, 1;
Production matrix begins:
  -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, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1, 1, 1;

Links

Crossrefs

Diagonals: A000012, A000217, A023443, A166830.
Cf. A000957, A000958, A001558, A001559, A104629, A126983.
Cf. A065602, A167772.

Programs

  • Mathematica
    A065602[n_, k_]:= A065602[n, k]= Sum[(k-1+2*j)*Binomial[2*(n-j)-k-1, n-1]/(2*(n - j) -k-1), {j,0,(n-k)/2}];
T[n_, k_]:= If[k==0, A065602[n, 0], If[n==1 && k==1, 1, A065602[n, k]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 27 2022 *)
  • SageMath
    def A065602(n, k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
def A237619(n, k):
    if (n<2): return (-1)^(n-k)
    elif (k==0): return A065602(n, 0)
    else: return A065602(n, k)
flatten([[A237619(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 27 2022

Formula

Sum_{k=0..n} T(n,k)*x^k = A126983(n), A000957(n+1), A026641(n) for x = 0, 1, 2 respectively.
T(n, k) = A167772(n-1, k-1) for k > 0, with T(n, 0) = A167772(n, 0).
T(n, 0) = A126983(n).
T(n+1, 1) = A000957(n+1).
T(n+2, 2) = A000958(n+1).
T(n+3, 3) = A104629(n) = A000957(n+3).
T(n+4, 4) = A001558(n).
T(n+5, 5) = A001559(n).
T(n, k) = A065602(n, k) for k > 0, with T(n, k) = (-1)^(n-k), for n < 2, and T(n, 0) = A065602(n, 0). - G. C. Greubel, May 27 2022
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