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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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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

A052701 a(0) = 0; for n >= 1, a(n) = 2^(n-1)*C(n-1), where C(n) = A000108(n) Catalan numbers, n>0.

Original entry on oeis.org

0, 1, 2, 8, 40, 224, 1344, 8448, 54912, 366080, 2489344, 17199104, 120393728, 852017152, 6085836800, 43818024960, 317680680960, 2317200261120, 16992801914880, 125210119372800, 926554883358720, 6882979133521920
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

A151374 shifted one place right. - Joerg Arndt, Mar 17 2011
The number of rooted Eulerian n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005
This is also the number of strings of length 2n-2 of two different types of balanced parentheses. For example, a(2) = 2, since the two possible strings of length 2 are [] and (), a(3) = 8, since the 8 possible strings of length 4 are (()), [()], ([]), [[]], ()(), [](), ()[], and [][]. - Jeffrey Shallit, Jun 03 2006
Row sums of number triangle A110506. - Paul Barry, Jul 24 2005
Also row sums of triangle in A085880. - Philippe Deléham, Aug 01 2005
Row sums of number triangle A114608. - Philippe Deléham, Oct 15 2008

References

  • V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.

Crossrefs

Limit of array A102544.

Programs

  • Maple
    spec := [S,{B=Union(C,Z),S=Union(B,C),C=Prod(S,S)},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    InverseSeries[Series[y-2*y^2, {y, 0, 24}], x] (* then A(x)=y(x) *) (* Len Smiley, Apr 07 2000 *)
    Join[{0},Table[2^n CatalanNumber[n],{n,0,30}]] (* Harvey P. Dale, Aug 29 2015 *)
  • PARI
    a(n)=if(n<1,0,2^(n-1)*(2*n-2)!/(n-1)!/n!)
    
  • PARI
    a(n)=if(n<1,0,polcoeff(serreverse(x-2*x^2+x*O(x^n)),n))
    
  • PARI
    a(n)=if(n<1,0,polcoeff(2*x/(1+sqrt(1-8*x+O(x^n))),n))

Formula

a(n) = A052714(n)/n!.
a(n) = A003645(n-2)*2, n>1.
a(n) = 8^(n-1)*GAMMA(n-1/2)/GAMMA(n+1)/Pi^(1/2), n>0.
D-finite with recurrence: a(1)=1, (-4+8*n)*a(n) - (n+1)*a(n+1) = 0.
G.f.: (1-sqrt(1-8*x))/4 = x*C(2*x) where C(x) is g.f. for Catalan numbers, A000108.
G.f. A(x) satisfies 2*A(x)^2-A(x)+x=0, A(0)=0 and A(x)=x+2*A(x)^2=x/(1-2*A(x)). Series reversion of x-2*x^2. - Michael Somos, Sep 06 2003
a(0)=0, a(1)=1; a(n) = 2*Sum_{i=1..n-1} a(i)*a(n-i). - Benoit Cloitre, Mar 16 2004
With a different offset, a(0)=1, a(n) = Sum_{k=0..n} Sum_{j=0..n} (j*C(2n-j-1, n-j)*C(j, k)*2^(n-j)/n), n>0. - Paul Barry, Jul 24 2005
The Hankel transform of a(n+1) = [1,2,8,40,224,1344,...] is 4^C(n+1,2). - Philippe Deléham, Nov 06 2007
G.f.: x + 4*x^2/(G(0)-4*x) where G(k) = k*(8*x+1) + 4*x + 2 - 2*x*(2*k+3)*(2*k+4)/G(k+1) ; (continued fraction ). - Sergei N. Gladkovskii, Apr 05 2013
a(n) ~ 8^(n-1)/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Dec 04 2016
From Amiram Eldar, Mar 06 2022: (Start)
Sum_{n>=1} 1/a(n) = 68/49 + 96*arcsin(sqrt(1/8))/(49*sqrt(7)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 20/27 - 16*log(2)/81. (End)
a(n) = A025225(n)/2 for n>=1. - Alois P. Heinz, Feb 16 2025

Extensions

Better description from Claude Lenormand (claude.lenormand(AT)free.fr), Mar 19 2001
Additional comments from Michael Somos, Feb 24 2002

A110511 Riordan array (1/(1+x), x(1-x)/(1+x)^2).

Original entry on oeis.org

1, -1, 1, 1, -4, 1, -1, 9, -7, 1, 1, -16, 26, -10, 1, -1, 25, -70, 52, -13, 1, 1, -36, 155, -190, 87, -16, 1, -1, 49, -301, 553, -403, 131, -19, 1, 1, -64, 532, -1372, 1462, -736, 184, -22, 1, -1, 81, -876, 3024, -4446, 3206, -1216, 246, -25, 1, 1, -100, 1365, -6084, 11826, -11584, 6190, -1870, 317, -28, 1, -1, 121, -2035
Offset: 0

Views

Author

Paul Barry, Jul 24 2005

Keywords

Comments

Inverse of number triangle A110506. Row sums are A110512. Diagonal sums are A110513. Product of (1/(1+x), x/(1+x)) (inverse binomial transform matrix) and (1, x(1-2x)) (A110509).

Examples

			Rows begin
   1;
  -1,   1;
   1,  -4,   1;
  -1,   9,  -7,   1;
   1, -16,  26, -10,   1;
  -1,  25, -70,  52, -13,   1;
		

Programs

  • Mathematica
    T[n_, k_] := Sum[(-1)^(n - j)*Binomial[n, j]*(-2)^(j - k)*Binomial[k, j - k], {j, 0, n}]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 29 2017 *)
  • PARI
    for(n=0,20, for(k=0,n, print1(sum(j=0,n, (-1)^(n-j)*binomial(n, j)*(-2)^(j-k)*binomial(k, j-k)), ", "))) \\ G. C. Greubel, Aug 29 2017

Formula

Number triangle: T(n, k) = Sum_{j=0..n} (-1)^(n-j)*C(n, j)*(-2)^(j-k)*C(k, j-k).
T(n, k) = Sum_{j=0..n} Sum_{i=0..k} C(k, i)*C(n+k-i-j-1, n-k-i-j)*(-1)^(n-k).
T(n,k) = T(n-1,k-1) - 2*T(n-1,k) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=-1, T(1,1)=1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Jan 12 2014

A114189 Riordan array (1/(1+xc(-2x)), xc(-2x)/(1+xc(-2x))), c(x) the g.f. of A000108.

Original entry on oeis.org

1, -1, 1, 3, -4, 1, -13, 19, -7, 1, 67, -102, 44, -10, 1, -381, 593, -278, 78, -13, 1, 2307, -3640, 1795, -568, 121, -16, 1, -14589, 23231, -11849, 4051, -999, 173, -19, 1, 95235, -152650, 79750, -28770, 7820, -1598, 234, -22, 1, -636925, 1025965, -545680, 204760, -59650, 13642, -2392, 304, -25, 1
Offset: 0

Views

Author

Paul Barry, Nov 16 2005

Keywords

Comments

Inverse of A114188. Factors as (1,xc(-2x))*(1/(1+x), x/(1+x)). Row sums are 0^n. Diagonal sums are A114190. First column is A114191. A signed version of A110506.
Triangle T(n,k), 0 <= k <= n, read by rows, given by [ -1,-2,-2,-2,-2,-2,-2,...] DELTA [1,0,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 01 2007

Examples

			Triangle begins
     1;
    -1,    1;
     3,   -4,    1;
   -13,   19,   -7,   1;
    67, -102,   44, -10,   1;
  -381,  593, -278,  78, -13, 1;
		

Programs

  • Mathematica
    c[x_] := (1 - Sqrt[1 - 4x])/(2x);
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1/(1 + # c[-2#])&, # c[-2#]/(1 + # c[-2#])&, 10] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Formula

Riordan array ((3-sqrt(1+8x))/(2(1-x)), (sqrt(1+8x)-2x-1)/(2(1-x))).
T(n,k) = (-1)^(n-k)*A110506(n,k). - Philippe Deléham, Mar 24 2007

A110508 Expansion of 1/(1-(x+x^2)c(2x)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 1, 4, 17, 87, 490, 2945, 18517, 120340, 802005, 5451651, 37652546, 263480357, 1864065017, 13311094644, 95816113129, 694511157535, 5064818563258, 37135165923801, 273581694291309, 2024194855052180, 15034769479254861
Offset: 0

Views

Author

Paul Barry, Jul 24 2005

Keywords

Comments

Diagonal sums of A110506.

Programs

  • Mathematica
    CoefficientList[Series[1/(1-(x+x^2)*(1-Sqrt[1-8*x])/(4*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)
  • PARI
    x='x+O('x^50); Vec(1/(1-(x+x^2)*(1-sqrt(1-8*x))/(4*x))) \\ G. C. Greubel, Aug 29 2017

Formula

a(0)=1; for n>0, a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..(n-k)} j*C(2n-2k-j-1, n-k-j)*C(j, k)*2^(n-k-j)/(n-k).
Conjecture: n*(3*n-7)*a(n) -4*(3*n-4)*(2*n-5)*a(n-1) +2*n*(3*n-7) +(-45*n^2+177*n-160)*a(n-3) -4*(3*n-4)*(2*n-5)*a(n-4) = 0. - R. J. Mathar, Nov 15 2011
a(n) ~ 9 * 2^(3*n+4) / (529 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 08 2014
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