cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A108557 Row sums of triangle A108556, in which row n equals the inverse binomial transform of the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 3, 9, 63, 433, 2823, 17657, 107439, 642529, 3802167, 22357097, 130970271, 765564049, 4469342439, 26073165401, 152043343119, 886424978881, 5167271805207, 30119654732489, 175558462395135, 1023255914549617
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Limit a(n+1)/a(n) = 3+sqrt(8) = 5.82842712...

Crossrefs

Programs

  • PARI
    a(n)=local(A=vector(n+1,r,vector(n+1,c,if(r-1==0 || c-1==0,1,if(r-1==1,2*c-1, sum(j=0,c-1,binomial(r+c-j-2,c-j-1)*(binomial(2*r-2,2*j)-2*(r-1)*binomial(r-3,j-1)))))))); sum(k=0,n,polcoeff(subst(Ser(A[n+1]),x, x/(1+x))/(1+x),k))

Formula

G.f.: (1-9*x+19*x^2+33*x^3-80*x^4+12*x^5)/(1-12*x+46*x^2-60*x^3+9*x^4).

A063007 T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 12, 30, 20, 1, 20, 90, 140, 70, 1, 30, 210, 560, 630, 252, 1, 42, 420, 1680, 3150, 2772, 924, 1, 56, 756, 4200, 11550, 16632, 12012, 3432, 1, 72, 1260, 9240, 34650, 72072, 84084, 51480, 12870, 1, 90, 1980, 18480, 90090, 252252, 420420, 411840, 218790, 48620
Offset: 0

Views

Author

Henry Bottomley, Jul 02 2001

Keywords

Comments

T(n,k) is the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's Cluster algebra of finite type B_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of triangle A008459 (squares of binomial coefficients). For example, x^2+6*x+6 = y^2+4*y+1. - Paul Boddington, Mar 07 2003
T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k N=(0,1) steps. E.g. T(2,0)=1 because we have DD; T(2,1) = 6 because we have NED, NDE, EDN, END, DEN and DNE; T(2,2)=6 because we have NNEE, NENE, NEEN, EENN, ENEN and ENNE. - Emeric Deutsch, Apr 20 2004
Another version of [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 2, 0; 1, 6, 6, 0; 1, 12, 30, 20, 0; ..., where DELTA is the operator defined in A084938. - Philippe Deléham Apr 15 2005
Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with increasing powers of x.
From Peter Bala, Oct 28 2008: (Start)
Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p.60]. See A008459 for the corresponding h-vectors for associahedra of type B_n and A001263 and A033282 respectively for the h-vectors and f-vectors for associahedra of type A_n.
An alternative description of this triangle in terms of f-vectors is as follows. Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i,j <= n+1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the f-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A008459 is the corresponding array of h-vectors for these type A_n polytopes. See A127674 (without the signs) for the array of f-vectors for type C_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes.
The S-transform on the ring of polynomials is the linear transformation of polynomials that is defined on the basis monomials x^k by S(x^k) = binomial(x,k) = x(x-1)...(x-k+1)/k!. Let P_n(x) denote the S-transform of the n-th row polynomial of this array. In the notation of [Hetyei] these are the Stirling polynomials of the type B associahedra. The first few values are P_1(x) = 2*x + 1, P_2(x) = 3*x^2 + 3*x + 1 and P_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials P_n(-x) satisfy a Riemann hypothesis. See A142995 for further details. The sequence of values P_n(k) for k = 0,1,2,3, ... produces the n-th row of A108625. (End)
This is the row reversed version of triangle A104684. - Wolfdieter Lang, Sep 12 2016
T(n, k) is also the number of (n-k)-dimensional faces of a convex n-dimensional Lipschitz polytope of real functions f defined on the set X = {1, 2, ..., n+1} which satisfy the condition f(n+1) = 0 (see Gordon and Petrov). - Stefano Spezia, Sep 25 2021
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial ((x+1)*(x+2)*(x+3)*...*(x+n) / n!)^2 in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022
Chapoton's observation above is correct: the precise expansion is ((x+1)*(x+2)*(x+3)*...*(x+n)/ n!)^2 = Sum_{k = 0..n} (-1)^k*T(n,n-k)*binomial(x+2*n-k, 2*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives ((x+1)*(x+2)*(x+3)/3!)^2 = 20*binomial(x+6,6) - 30*binomial(x+5,5) + 12*binomial(x+4,4) - binomial(x+3,3). - Peter Bala, Jun 24 2023

Examples

			The triangle T(n, k) starts:
  n\k 0  1    2     3     4      5      6      7      8     9
  0:  1
  1:  1  2
  2:  1  6    6
  3:  1 12   30    20
  4:  1 20   90   140    70
  5:  1 30  210   560   630    252
  6:  1 42  420  1680  3150   2772    924
  7:  1 56  756  4200 11550  16632  12012   3432
  8:  1 72 1260  9240 34650  72072  84084  51480  12870
  9:  1 90 1980 18480 90090 252252 420420 411840 218790 48620
... reformatted by _Wolfdieter Lang_, Sep 12 2016
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92, in Ser's book) is
   1;
  -1/2,  1/2;
   1/3, -1/2,    1/6;
  -1/4,  9/20,  -1/4,   1/20;
   1/5, -2/5,    2/7,  -1/10,  1/70;
  -1/6,  5/14, -25/84,  5/36, -1/28,  1/252;
   1/7, -9/28,  25/84, -1/6,   9/154, -1/84, 1/924;
   ... (End)
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 366.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, Table I, p. 92.
  • D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: from Neolithic Scots to John McKay, CRM Proc. Lecture Notes 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349-366.

Crossrefs

See A331430 for an essentially identical triangle, except with signed entries.
Columns include A000012, A002378, A033487 on the left and A000984, A002457, A002544 on the right.
Main diagonal is A006480.
Row sums are A001850. Alternating row sums are A033999.
Cf. A033282 (f-vectors type A associahedra), A108625, A080721 (f-vectors type D associahedra).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Haskell
    a063007 n k = a063007_tabl !! n !! k
    a063007_row n = a063007_tabl !! n
    a063007_tabl = zipWith (zipWith (*)) a007318_tabl a046899_tabl
    -- Reinhard Zumkeller, Nov 18 2014
    
  • Magma
    /* As triangle: */ [[Binomial(n,k)*Binomial(n+k,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 03 2015
  • Maple
    p := (n,x) -> orthopoly[P](n,1+2*x): seq(seq(coeff(p(n,x),x,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Binomial[n, k]Binomial[n + k, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Dec 24 2011 *)
    Table[CoefficientList[Hypergeometric2F1[-n, n + 1, 1, -x], x], {n, 0, 9}] // Flatten
    (* Peter Luschny, Mar 09 2018 *)
  • PARI
    {T(n, k) = local(t); if( n<0, 0, t = (x + x^2)^n; for( k=1, n, t=t'); polcoeff(t, k) / n!)} /* Michael Somos, Dec 19 2002 */
    
  • PARI
    {T(n, k) = binomial(n, k) * binomial(n+k, k)} /* Michael Somos, Sep 22 2013 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n+k)! / (k!^2 * (n-k)!))} /* Michael Somos, Sep 22 2013 */
    

Formula

T(n, k) = (n+k)!/(k!^2*(n-k)!) = T(n-1, k)*(n+k)/(n-k) = T(n, k-1)*(n+k)*(n-k+1)/k^2 = T(n-1, k-1)*(n+k)*(n+k-1)/k^2.
binomial(x, n)^2 = Sum_{k>=0} T(n,k) * binomial(x, n+k). - Michael Somos, May 11 2012
T(n, k) = A109983(n, k+n). - Michael Somos, Sep 22 2013
G.f.: G(t, z) = 1/sqrt(1-2*z-4*t*z+z^2). Row generating polynomials = P_n(1+2z), i.e., T(n, k) = [z^k] P_n(1+2*z), where P_n are the Legendre polynomials. - Emeric Deutsch, Apr 20 2004
Sum_{k>=0} T(n, k)*A000172(k) = Sum_{k>=0} T(n, k)^2 = A005259(n). - Philippe Deléham, Jun 08 2005
1 + z*d/dz(log(G(t,z))) = 1 + (1 + 2*t)*z + (1 + 8*t + 8*t^2)*z^2 + ... is the o.g.f. for a signed version of A127674. - Peter Bala, Sep 02 2015
If R(n,t) denotes the n-th row polynomial then x^3 * exp( Sum_{n >= 1} R(n,t)*x^n/n ) = x^3 + (1 + 2*t)*x^4 + (1 + 5*t + 5*t^2)*x^5 + (1 + 9*t + 21*t^2 + 14*t^3)*x^6 + ... is an o.g.f for A033282. - Peter Bala, Oct 19 2015
P(n,x) := 1/(1 + x)*Integral_{t = 0..x} R(n,t) dt are (modulo differences of offset) the row polynomials of A033282. - Peter Bala, Jun 23 2016
From Peter Bala, Mar 09 2018: (Start)
R(n,x) = Sum_{k = 0..n} binomial(2*k,k)*binomial(n+k,n-k)*x^k.
R(n,x) = Sum_{k = 0..n} binomial(n,k)^2*x^k*(1 + x)^(n-k).
n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x).
R(n,x) = (-1)^n*R(n,-1 - x).
R(n,x) = 1/n! * (d/dx)^n ((x^2 + x)^n). (End)
The row polynomials are R(n,x) = hypergeom([-n, n + 1], [1], -x). - Peter Luschny, Mar 09 2018
T(n,k) = C(n+1,k)*A009766(n,k). - Bob Selcoe, Jan 18 2020 (Connects this triangle with the Catalan triangle. - N. J. A. Sloane, Jan 18 2020)
If we let A(n,k) = (-1)^(n+k)*(2*k+1)*(n*(n-1)*...*(n-(k-1)))/((n+1)*...*(n+(k+1))) for n >= 0 and k = 0..n, and we consider both T(n,k) and A(n,k) as infinite lower triangular arrays, then they are inverses of one another. (Empty products are by definition 1.) See the example below. The rational numbers |A(n,k)| appear in Table II on p. 92 in Ser's (1933) book. - Petros Hadjicostas, Jul 11 2020
From Peter Bala, Nov 28 2021: (Start)
Row polynomial R(n,x) = Sum_{k >= n} binomial(k,n)^2 * x^(k-n)/(1+x)^(k+1) for x > -1/2.
R(n,x) = 1/(1 + x)^(n+1) * hypergeom([n+1, n+1], [1], x/(1 + x)).
R(n,x) = (1 + x)^n * hypergeom([-n, -n], [1], x/(1 + x)).
R(n,x) = hypergeom([(n+1)/2, -n/2], [1], -4*x*(1 + x)).
If we set R(-1,x) = 1, we can run the recurrence n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x) backwards to give R(-n,x) = R(n-1,x).
R(n,x) = [t^n] ( (1 + t)*(1 + x*(1 + t)) )^n. (End)
n*T(n,k) = (2*n-1)*T(n-1,k) + (4*n-2)*T(n-1,k-1) - (n-1)*T(n-2,k). - Fabián Pereyra, Jun 30 2022
From Peter Bala, Oct 07 2024: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n, k) * x^k o (1 + x)^(n-k), where o denotes the black diamond product of power series as defined by Dukes and White (see Bala, Section 4.4, exercise 3).
Denote this triangle by T. Then T * transpose(T) = A143007, the square array of crystal ball sequences for the A_n X A_n lattices.
Let S denote the triangle ((-1)^(n+k)*T(n, k))n,k >= 0, a signed version of this triangle. Then S^(-1) * T = A007318, Pascal's triangle; it appears that T * S^(-1) = A110098.
T = A007318 * A115951. (End)

A108553 Square array, read by antidiagonals, where row n equals the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 13, 13, 7, 1, 1, 25, 55, 25, 9, 1, 1, 41, 169, 147, 41, 11, 1, 1, 61, 411, 625, 309, 61, 13, 1, 1, 85, 853, 2051, 1681, 561, 85, 15, 1, 1, 113, 1583, 5577, 6981, 3721, 923, 113, 17, 1, 1, 145, 2705, 13203, 23673, 18733, 7225, 1415, 145, 19, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Rows 0 and 2 are included by extension since they fit the formula. Row 1 equals the odd numbers in order that triangle A108556 maintains that A108556(n,n-1) = (n/2)*A108556(n,n) for all n>=1, where row n of triangle A108556 equals the inverse binomial transform of row n of this square array.

Examples

			Square array begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,5,13,25,41,61,85,113,145,181,221,265,...
1,13,55,147,309,561,923,1415,2057,2869,...
1,25,169,625,1681,3721,7225,12769,21025,...
1,41,411,2051,6981,18733,42783,86983,...
1,61,853,5577,23673,76389,204205,476113,...
1,85,1583,13203,68853,264825,824083,...
Inverse binomial transform of rows gives
rows of triangle A108556:
1,
1,2,
1,4,4,
1,12,30,20,
1,24,120,192,96,
1,40,330,940,1080,432, ...
Product of the g.f. of row n and (1-x)^(n+1)
generates the symmetric triangle A108558:
1;
1,1;
1,2,1;
1,9,9,1;
1,20,54,20,1;
1,35,180,180,35,1; ...
The row sums of triangle A108558 equals the
main diagonal of triangle A108556.
		

Crossrefs

Cf. A108554 (diagonal), A108555 (antidiagonal sums), A108556, A108558, A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).

Programs

  • PARI
    T(n,k)=if(n<0 || k<0,0,if(n==0 || k==0,1,if(n==1,2*k+1, sum(j=0,k,binomial(n+k-j,k-j)*(binomial(2*n,2*j)-2*n*binomial(n-2,j-1))))))

Formula

T(n, k) = Sum_{j=0..n} C(n+k-j, k-j)*[C(2*n, 2*j) - 2*j*(n-j)*C(n, j)/(n-1)] for n>1, with T(0, k)=1, T(1, k)=2*k+1.

A127674 Coefficient table for Chebyshev polynomials T(2*n,x) (increasing even powers x, without zeros).

Original entry on oeis.org

1, -1, 2, 1, -8, 8, -1, 18, -48, 32, 1, -32, 160, -256, 128, -1, 50, -400, 1120, -1280, 512, 1, -72, 840, -3584, 6912, -6144, 2048, -1, 98, -1568, 9408, -26880, 39424, -28672, 8192, 1, -128, 2688, -21504, 84480, -180224, 212992, -131072, 32768, -1, 162, -4320, 44352, -228096, 658944, -1118208
Offset: 0

Views

Author

Wolfdieter Lang Mar 07 2007

Keywords

Comments

Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of (the signless version of) this array are the f-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A086645 for the corresponding array of h-vectors for these type C_n polytopes. See A063007 for the array of f-vectors for type A_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes. - Peter Bala, Oct 23 2008

Examples

			[1];
[-1,2];
[1,-8,8];
[-1,18,-48,32];
[1,-32,160,-256,128];
...
See a link for the row polynomials.
The T-polynomial for row n=3, [-1,18,-48,32], is T(2*3,x) =  -1*x^0 + 18*x^2 - 48*x^4 + 32*x^6.
		

References

  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. p. 37, eq.(1.96) and p. 4. eq.(1.10).

Crossrefs

Cf. A075733 (different signs and offset). A084930 (coefficients of odd-indexed T-polynomials).
Cf. A053120 (coefficients of T-polynomials, with interspersed zeros).

Formula

a(n,m) = 0 if n < m, a(0,0) = 1; otherwise a(n,m) = ((-1)^(n-m))*(2^(2*m-1))*binomial(n+m,2*m)*(2*n)/(n+m).
O.g.f.: (1 + z*(1 - 2*x))/((1 + z)^2 - 4*x*z) = 1 + (-1 + 2*x)*z + (1 - 8*x + 8*x^2)*z^2 + ... . [Peter Bala, Oct 23 2008] For the t-polynomials actually with x -> x^2. - Wolfdieter Lang, Aug 02 2014
Denoting the row polynomials by R(n,x) we have exp( Sum_{n >= 1} R(n,x)*z^n/n ) = 1/sqrt( (1 + z)^2 - 4*x*z ) = 1 + (-1 + 2*x)*z + (1 - 6*x + 6*x^2)*z^2 + ..., the o.g.f. for a signed version of A063007. - Peter Bala, Sep 02 2015
The n-th row polynomial equals T(n, 2*x - 1). - Peter Bala, Jul 09 2023

A108558 Symmetric triangle, read by rows, where row n equals the (n+1)-th differences of the crystal ball sequence for D_n lattice, for n>=0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 9, 9, 1, 1, 20, 54, 20, 1, 1, 35, 180, 180, 35, 1, 1, 54, 447, 852, 447, 54, 1, 1, 77, 931, 2863, 2863, 931, 77, 1, 1, 104, 1724, 7768, 12550, 7768, 1724, 104, 1, 1, 135, 2934, 18186, 43128, 43128, 18186, 2934, 135, 1, 1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Row n equals the (n+1)-th differences of row n of the square array A108553. G.f. of row n equals: (1-x)^(n+1)*CBD_n(x), where CBD_n denotes the g.f. of the crystal ball sequence for D_n lattice.
From Peter Bala, Oct 23 2008: (Start)
Let D_n be the root lattice generated as a monoid by {+-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(D_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(D_n) [Ardila et al.]. See A108556 for the corresponding array of f-vectors for these type D_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A086645 for the array of h-vectors associated with type C_n polytopes.
The Hilbert transform of this array (as defined in A145905) equals A108553.
(End)

Examples

			G.f.s of initial rows of square array A108553 are:
  (1)/(1-x),
  (1 + x)/(1-x)^2,
  (1 + 2*x + x^2)/(1-x)^3,
  (1 + 9*x + 9*x^2 + x^3)/(1-x)^4,
  (1 + 20*x + 54*x^2 + 20*x^3 + x^4)/(1-x)^5,
  (1 + 35*x + 180*x^2 + 180*x^3 + 35*x^4 + x^5)/(1-x)^6.
Triangle begins:
  1;
  1,   1;
  1,   2,    1;
  1,   9,    9,     1;
  1,  20,   54,    20,      1;
  1,  35,  180,   180,     35,      1;
  1,  54,  447,   852,    447,     54,      1;
  1,  77,  931,  2863,   2863,    931,     77,     1;
  1, 104, 1724,  7768,  12550,   7768,   1724,   104,    1;
  1, 135, 2934, 18186,  43128,  43128,  18186,  2934,  135,   1;
  1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1;
  ...
		

Crossrefs

Cf. A108553, A008353, A108558, A008459, A086645, A108556. Row n equals (n+1)-th differences of: A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).
T(2n,n) gives A305693.

Programs

  • Mathematica
    T[1, 0] = T[1, 1]=1; T[n_, k_] := Binomial[2n, 2k] - 2n Binomial[n-2, k-1];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 25 2018 *)
  • PARI
    T(n,k)=if(n
    				

Formula

T(n, k) = C(2*n, 2*k) - 2*n*C(n-2, k-1) for n>1, with T(0, 0)=1, T(1, 0)=T(1, 1)=1. Row sums equal A008353: 2^(n-1)*(2^n-n) for n>1.
From Peter Bala, Oct 23 2008: (Start)
O.g.f. : rational function N(x,z)/D(x,z), where N(x,z) = 1 - 3*(1 + x)*z + (3 + 2*x + 3*x^2)*z^2 - (1 + x)*(1 - 8*x + x^2)z^3 - 8*x*(1 + x^2)*z^4 + 2*x*(1 + x)*(1 - x)^2*z^5 and D(x,z) = ((1 - z)^2 - 2*x*z*(1 + z) + x^2*z^2)*(1 - z*(1 + x))^2.
For n >= 2, the row n generating polynomial equals 1/2*[(1 + sqrt(x))^(2n) + (1 - sqrt(x))^(2n)] - 2*n*x*(1 + x)^(n-2).
(End)
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