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.

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A002212 Number of restricted hexagonal polyominoes with n cells.

Original entry on oeis.org

1, 1, 3, 10, 36, 137, 543, 2219, 9285, 39587, 171369, 751236, 3328218, 14878455, 67030785, 304036170, 1387247580, 6363044315, 29323149825, 135700543190, 630375241380, 2938391049395, 13739779184085, 64430797069375, 302934667061301, 1427763630578197
Offset: 0

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Author

N. J. A. Sloane, Ronald C. Read

Keywords

Comments

Number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0) with no peaks at odd level. Example: a(2)=3 because we have UUDD, UHD and HH. - Emeric Deutsch, Dec 06 2003
Number of 3-Motzkin paths of length n-1 (i.e., lattice paths from (0,0) to (n-1,0) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and three types of steps H=(1,0)). Example: a(4)=36 because we have 27 HHH paths, 3 HUD paths, 3 UHD paths and 3 UDH paths. - Emeric Deutsch, Jan 22 2004
Number of rooted, planar trees having edges weighted by strictly positive integers (multi-trees) with weight-sum n. - Roland Bacher, Feb 28 2005
Number of skew Dyck paths of semilength n. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. - Emeric Deutsch, May 10 2007
Equivalently, number of self-avoiding paths of semilength n in the first quadrant beginning at the origin, staying weakly above the diagonal, ending on the diagonal, and consisting of steps r=(+1,0) (right), U=(0,+1) (up), and D=(0,-1) (down). Self-avoidance implies that factors UD and DU and steps D reaching the diagonal before the end are forbidden. The a(3) = 10 such paths are UrUrUr, UrUUrD, UrUUrr, UUrrUr, UUrUrD, UUrUrr, UUUDrD, UUUrDD, UUUrrD, and UUUrrr. - Joerg Arndt, Jan 15 2024
Hankel transform of [1,3,10,36,137,543,...] is A000012 = [1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
From Gary W. Adamson, May 17 2009: (Start)
Convolved with A026375, (1, 3, 11, 45, 195, ...) = A026378: (1, 4, 17, 75, ...)
(1, 3, 10, 36, 137, ...) convolved with A026375 = A026376: (1, 6, 30, 144, ...).
Starting (1, 3, 10, 36, ...) = INVERT transform of A007317: (1, 2, 5, 15, 51, ...). (End)
Binomial transform of A032357. - Philippe Deléham, Sep 17 2009
a(n) = number of rooted trees with n vertices in which each vertex has at most 2 children and in case a vertex has exactly one child, it is labeled left, middle or right. These are the hex trees of the Deutsch, Munarini, Rinaldi link. This interpretation yields the second MATHEMATICA recurrence below. - David Callan, Oct 14 2012
The left shift (1,3,10,36,...) of this sequence is the binomial transform of the left-shifted Catalan numbers (1,2,5,14,...). Example: 36 =1*14 + 3*5 + 3*2 + 1*1. - David Callan, Feb 01 2014
Number of Schroeder paths from (0,0) to (2n,0) with no level steps H=(2,0) at even level. Example: a(2)=3 because we have UUDD, UHD and UDUD. - José Luis Ramírez Ramírez, Apr 27 2015
This is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the Catalan sequence A000108. See a Feb 17 2017 comment in A097805. - Wolfdieter Lang, Feb 17 2017
a(n) is the number of parking functions of size n avoiding the patterns 132 and 231. - Lara Pudwell, Apr 10 2023

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 36*x^4 + 137*x^5 + 543*x^6 + 2219*x^7 + 9285*x^8 + ...
		

References

  • J. Brunvoll, B. N. Cyvin, and S. J. Cyvin, Studies of some chemically relevant polygonal systems: mono-q-polyhexes, ACH Models in Chem., 133 (3) (1996), 277-298, Eq 14.
  • S. J. Cyvin, J. Brunvoll, G. Xiaofeng, and Z. Fuji, Number of perifusenes with one internal vertex, Rev. Roumaine Chem., 38(1) (1993), 65-78.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

First differences of A007317.
Row sums of triangle A104259.

Programs

  • Magma
    I:= [1,3]; [1] cat [n le 2 select I[n]  else ((6*n-3)*Self(n-1)-5*(n-2)*Self(n-2)) div (n+1): n in [1..30]]; // Vincenzo Librandi, Jun 15 2015
  • Maple
    t1 := series(1+ (1-3*x-(1-x)^(1/2)*(1-5*x)^(1/2))/(2*x), x, 50):
    A002212_list := len -> seq(coeff(t1,x,n),n=0..len): A002212_list(40);
    a[0] := 1: a[1] := 1: for n from 2 to 50 do a[n] := (3*(2*n-1)*a[n-1]-5*(n-2)*a[n-2])/(n+1) od: print(convert(a,list)); # Zerinvary Lajos, Jan 01 2007
    a := n -> `if`(n=0,1,simplify(GegenbauerC(n-1, -n, -3/2)/n)):
    seq(a(n), n=0..23); # Peter Luschny, May 09 2016
  • Mathematica
    InverseSeries[Series[(y)/(1+3*y+y^2), {y, 0, 24}], x] (* then A(x)=1+y(x) *) (* Len Smiley, Apr 14 2000 *)
    (* faster *)
    a[0]=1;a[1]=1;
    a[n_]/;n>=2 := a[n] = a[n-1] +  Sum[a[i]a[n-1-i],{i,0,n-1}];
    Table[a[n],{n,0,14}] (* See COMMENTS above, [David Callan, Oct 14 2012] *)
    (* fastest *)
    s[0]=s[1]=1;
    s[n_]/;n>=2 := s[n] = (3(2n-1)s[n-1]-5(n-2)s[n-2])/(n+1);
    Table[s[n],{n,0,14 }] (* See Deutsch, Munarini, Rinaldi link, [David Callan, Oct 14 2012] *)
    (* 2nd fastest *)
    a[n_] := Hypergeometric2F1[3/2, 1-n, 3, -4]; a[0]=1; Table[a[n], {n, 0, 14}]  (* Jean-François Alcover, May 16 2013 *)
    CoefficientList[Series[(1 - x - Sqrt[1 - 6x + 5x^2])/(2x), {x, 0, 20}], x] (* Nikolaos Pantelidis, Jan 30 2023 *)
  • Maxima
    makelist(sum(binomial(n,k)*binomial(n-k,k)*3^(n-2*k)/(k+1),k,0,n/2),n,0,24); /* for a(n+1) */ /* Emanuele Munarini, May 18 2011 */
    
  • PARI
    {a(n) = polcoeff( (1 - x - sqrt(1 - 6*x + 5*x^2 + x^2 * O(x^n))) / 2, n+1)};
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + 3*x + x^2) + x * O(x^n)), n))}; /* Michael Somos */
    
  • PARI
    my(N=66,x='x+O('x^N)); Vec((1 - x - sqrt(1-6*x+5*x^2))/(2*x)) \\ Joerg Arndt, Jan 13 2024
    
  • Sage
    def A002212():
        x, y, n = 1, 1, 1
        while True:
            yield x
            n += 1
            x, y = y, ((6*n - 3)*y - (5*n - 10)*x) / (n + 1)
    a = A002212()
    [next(a) for i in range(24)]  # Peter Luschny, Oct 12 2013
    

Formula

a(0)=1, for n > 0: a(n) = Sum_{j=0..n-1} Sum_{i=0..j} a(i)*a(j-i). G.f.: A(x) = 1 + x*A(x)^2/(1-x). - Mario Catalani (mario.catalani(AT)unito.it), Jun 19 2003
a(n) = Sum_{i=ceiling((n-1)/2)..n-1} (3^(2i+1-n)*binomial(n, i)*binomial(i, n-i-1))/n. - Emeric Deutsch, Jul 23 2002
a(n) = Sum_{k=1..n} binomial(2k, k)*binomial(n-1, k-1)/(k+1), i.e., binomial transform of the Catalan numbers 1, 2, 5, 14, 42, ... (A000108). a(n) = Sum_{k=0..floor((n-1)/2)} 3^(n-1-2*k)*binomial(2k, k)*binomial(n-1, 2k)/(k+1). - Emeric Deutsch, Aug 05 2002
D-finite with recurrence: a(1)=1, a(n) = (3(2n-1)*a(n-1)-5(n-2)*a(n-2))/(n+1) for n > 1. - Emeric Deutsch, Dec 18 2002
a(n) is asymptotic to c*5^n/n^(3/2) with c=0.63.... - Benoit Cloitre, Jun 23 2003
In closed form, c = (1/2)*sqrt(5/Pi) = 0.63078313050504... - Vaclav Kotesovec, Oct 04 2012
Reversion of Sum_{n>0} a(n)x^n = -Sum_{n>0} A001906(n)(-x)^n.
G.f. A(x) satisfies xA(x)^2 + (1-x)(1-A(x)) = 0.
G.f.: (1 - x - sqrt(1 - 6x + 5x^2))/(2x). For n > 1, a(n) = 3*a(n-1) + Sum_{k=1..n-2} a(k)*a(n-k-1). - John W. Layman, Feb 22 2001
The Hankel transform of this sequence gives A001519 = 1, 2, 5, 13, 34, 89, ... E.g., Det([1, 1, 3, 10, 36; 1, 3, 10, 36, 137; 3, 10, 36, 137, 543; 10, 36, 137, 543, 2219; 36, 137, 543, 2219, 9285 ])= 34. - Philippe Deléham, Jan 25 2004
a(m+n+1) = Sum_{k>=0} A091965(m, k)*A091965(n, k) = A091965(m+n, 0). - Philippe Deléham, Sep 14 2005
a(n+1) = Sum_{k=0..n} 2^(n-k)*M(k)*binomial(n,k), where M(k) = A001006(k) is the k-th Motzkin number (from here it follows that a(n+1) and M(n) have the same parity). - Emeric Deutsch, May 10 2007
a(n+1) = Sum_{k=0..n} A097610(n,k)*3^k. - Philippe Deléham, Oct 02 2007
G.f.: 1/(1-x/(1-x-x/(1-x/(1-x-x/(1-x/(1-x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: (1-x)/(1-2x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-.... (continued fraction). - Paul Barry, Oct 17 2009
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-x) (continued fraction); more generally g.f. C(x/(1-x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = -5^(1/2)/(10*(n+1)) * (5*hypergeom([1/2, n], [1], 4/5) -3*hypergeom([1/2, n+1], [1], 4/5)) (for n>0). - Mark van Hoeij, Nov 12 2009
For n >= 1, a(n) = (1/(2*Pi))*Integral_{x=1..5} x^(n-1)*sqrt((x-1)*(5-x)) dx. - Groux Roland, Mar 16 2011
a(n+1) = [x^n](1-x^2)(1+3*x+x^2)^n. - Emanuele Munarini, May 18 2011
From Gary W. Adamson, Jul 21 2011: (Start)
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows (with 3,2,2,2,... as the main diagonal):
3, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 1, 2, 1, 0, 0, ...
1, 1, 1, 2, 1, 0, ...
1, 1, 1, 1, 2, 0, ...
...
Alternatively, let M = the previous matrix but change the 3 to a 2. Then a(n) = sum of top row terms of M^(n-1). (End)
a(n) = hypergeometric([1-n,3/2],[3],-4), for n>0. - Peter Luschny, Aug 15 2012
a(n) = GegenbauerC(n-1, -n, -3/2)/n for n >= 1. - Peter Luschny, May 09 2016
E.g.f.: 1 + Integral (exp(3*x) * BesselI(1,2*x) / x) dx. - Ilya Gutkovskiy, Jun 01 2020
G.f.: 1 + x/G(0) with G(k) = (1 - 3*x - x^2/G(k+1)) (continued fraction). - Nikolaos Pantelidis, Dec 12 2022
From Peter Bala, Feb 03 2024: (Start)
G.f.: 1 + x/(1 - x) * c(x/(1 - x))^2 = 1 + x/(1 - 5*x) * c(-x/(1 - 5*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
a(n+1) = Sum_{k = 0..n} binomial(n, k)*Catalan(k+1).
a(n+1) = hypergeom([-n, 3/2], [3], -4).
a(n+1) = 5^n * Sum_{k = 0..n} (-5)^(-k)*binomial(n, k)*Catalan(k+1).
a(n+1) = 5^n * hypergeom([-n, 3/2], [3], 4/5). (End)

A005572 Number of walks on cubic lattice starting and finishing on the xy plane and never going below it.

Original entry on oeis.org

1, 4, 17, 76, 354, 1704, 8421, 42508, 218318, 1137400, 5996938, 31940792, 171605956, 928931280, 5061593709, 27739833228, 152809506582, 845646470616, 4699126915422, 26209721959656, 146681521121244, 823429928805936
Offset: 0

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Author

Keywords

Comments

Also number of paths from (0,0) to (n,0) in an n X n grid using only Northeast, East and Southeast steps and the East steps come in four colors. - Emeric Deutsch, Nov 03 2002
Number of skew Dyck paths of semilength n+1 with the left steps coming in two colors. - David Scambler, Jun 21 2013
Number of 2-colored Schroeder paths from (0,0) to (2n+2,0) with no level steps H=(2,0) at an even level. There are two ways to color an H-step at an odd level. Example: a(1)=4 because we have UUDD, UHD (2 choices) and UDUD. - José Luis Ramírez Ramírez, Apr 27 2015

Examples

			a(3) = 76 = sum of top row terms of M^3; i.e., (37 + 29 + 9 + 1).
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Binomial transform of A002212. Sequence shifted right twice is A025228.

Programs

  • Maple
    a := n -> simplify(2^n*hypergeom([3/2, -n], [3], -2)):
    seq(a(n), n=0..21); # Peter Luschny, Feb 03 2015
    a := n -> simplify(GegenbauerC(n, -n-1, -2))/(n+1):
    seq(a(n), n=0..21); # Peter Luschny, May 09 2016
  • Mathematica
    RecurrenceTable[{a[0]==1,a[1]==4,a[n]==((2n+1)a[n-1]-3(n-1)a[n-2]) 4/(n+2)}, a[n],{n,30}] (* Harvey P. Dale, Oct 04 2011 *)
    a[n_]:=If[n==0,1,Coefficient[(1+4x+x^2)^(n+1),x^n]/(n+1)]
    Table[a[n],{n,0,40}] (* Emanuele Munarini, Apr 06 2012 *)
  • Maxima
    a(n):=coeff(expand((1+4*x+x^2)^(n+1)),x^n)/(n+1); makelist(a(n),n,0,12); /* Emanuele Munarini, Apr 06 2012 */
    
  • PARI
    a(n)=polcoeff((1-4*x-sqrt(1-8*x+12*x^2+x^3*O(x^n)))/2,n+2)
    
  • PARI
    { A005572(n) = sum(k=0,n\2, binomial(n,2*k) * binomial(2*k,k) * 4^(n-2*k) / (k+1) ) } /* Max Alekseyev, Feb 02 2015 */
    
  • PARI
    {a(n)=sum(k=0,n, binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1) )}
    for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Feb 02 2015
    
  • Sage
    def A005572(n):
        A108198 = lambda n,k: (-1)^k*catalan_number(k+1)*rising_factorial(-n,k)/factorial(k)
        return sum(A108198(n,k)*2^(n-k) for k in (0..n))
    [A005572(n) for n in range(22)] # Peter Luschny, Feb 05 2015

Formula

Generating function A(x) satisfies 1 + (xA)^2 = A - 4xA.
a(0) = 1 and, for n > 0, a(n) = 4a(n-1) + Sum_{i=1..n-1} a(i-1)*a(n-i-1). - John W. Layman, Jan 07 2000
G.f.: (1 - 4*x - sqrt(1 - 8*x + 12*x^2))/(2*x^2).
D-finite with recurrence: a(n) = ((2*n+1)*a(n-1) - 3*(n-1)*a(n-2))*4/(n+2), n > 0.
a(m+n) = Sum_{k>=0} A052179(m, k)*A052179(n, k) = A052179(m+n, 0). - Philippe Deléham, Sep 15 2005
a(n) = 4*a(n-1) + A052177(n-1) = A052179(n, 0) = 6*A005573(n)-A005573(n-1) = Sum_{j=0..floor(n/2)} 4^(n-2*j)*C(n, 2*j)*C(2*j, j)/(j+1). - Henry Bottomley, Aug 23 2001
a(n) = Sum_{k=0..n} A097610(n,k)*4^k. - Philippe Deléham, Dec 03 2009
Let A(x) be the g.f., then B(x) = 1 + x*A(x) = 1 + 1*x + 4*x^2 + 17*x^3 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-2*x) (continued fraction); more generally B(x) = C(x/(1-2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
From Gary W. Adamson, Jul 21 2011: (Start)
a(n) = sum of top row terms of M^n, M = an infinite square production matrix as follows:
3, 1, 0, 0, ...
1, 3, 1, 0, ...
1, 1, 3, 1, ...
1, 1, 1, 3, ...
... (End)
a(n) ~ 3*6^(n+1/2)/(n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Oct 05 2012
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k) * binomial(2k,k) * 4^(n-2k) / (k+1). - Max Alekseyev, Feb 02 2015
From Paul D. Hanna, Feb 02 2015: (Start)
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A000108(k+1).
a(n) = [x^n] (1 + 4*x + x^2)^(n+1) / (n+1).
G.f.: (1/x) * Series_Reversion( x/(1 + 4*x + x^2) ). (End)
a(n) = 2^n*hypergeom([3/2, -n], [3], -2). - Peter Luschny, Feb 03 2015
a(n) = 4^n*hypergeom([-n/2, (1-n)/2], [2], 1/4). - Robert Israel, Feb 04 2015
a(n) = Sum_{k=0..n} A108198(n,k)*2^(n-k). - Peter Luschny, Feb 05 2015
a(n) = 2*(12^(n/2))*(n!/(n+2)!)*GegenbauerC(n, 3/2,2/sqrt(3)), where GegenbauerC are Gegenbauer polynomials in Maple notation. This is a consequence of Robert Israel's formula. - Karol A. Penson, Feb 20 2015
a(n) = (2^(n+1)*3^((n+1)/2)*P(n+1,1,2/sqrt(3)))/((n+1)*(n+2)) where P(n,u,x) are the associated Legendre polynomials of the first kind. - Peter Luschny, Feb 24 2015
a(n) = -6^(n+1)*sqrt(3)*Integral{t=0..Pi}(cos(t)*(2+cos(t))^(-n-2))/(Pi*(n+2)). - Peter Luschny, Feb 24 2015
From Karol A. Penson and Wojciech Mlotkowski, Mar 16 2015: (Start)
Integral representation as the n-th moment of a positive function defined on a segment x=[2, 6]. This function is the Wigner's semicircle distribution shifted to the right by 4. This representation is unique. In Maple notation,
a(n) = int(x^n*sqrt(4-(x-4)^2)/(2*Pi), x=2..6),
a(n) = 2*6^n*Pochhammer(3/2, n)*hypergeom([-n, 3/2], [-n-1/2], 1/3)/(n+2)!
(End)
a(n) = GegenbauerC(n, -n-1, -2)/(n+1). - Peter Luschny, May 09 2016
E.g.f.: exp(4*x) * BesselI(1,2*x) / x. - Ilya Gutkovskiy, Jun 01 2020
From Peter Bala, Aug 18 2021: (Start)
G.f. A(x) = 1/(1 - 2*x)*c(x/(1 - 2*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Cf. A129400.
Conjecture: a(n) is even except for n of the form 2*(2^k - 1). [added Feb 03: the conjecture follows from the formula a(n) = Sum_{k = 0..n} 2^(n-k)*binomial(n, k)*Catalan(k+1) given above.] (End)
From Peter Bala, Feb 03 2024: (Start)
G.f.: 1/(1 - 2*x) * c(x/(1 - 2*x))^2 = 1/(1 - 6*x) * c(-x/(1 - 6*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
a(n) = 6^n * Sum_{k = 0..n} (-6)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 6^n * hypergeom([-n, 3/2], [3], 2/3). (End)

Extensions

Additional comments from Michael Somos, Jun 10 2000

A098410 Expansion of 1/(sqrt(1-4*x)*sqrt(1-8*x)).

Original entry on oeis.org

1, 6, 38, 252, 1734, 12276, 88796, 652728, 4856902, 36478404, 275975028, 2099978568, 16054486044, 123213933576, 948713646072, 7325088811632, 56692748053062, 439689331938276, 3416328042565124, 26587566855421608, 207218159714453044, 1617124976299315224, 12634892752595949192
Offset: 0

Views

Author

Paul Barry, Sep 07 2004

Keywords

Comments

Convolution of A000984(n) and 2^n*A000984(n). Convolution of A000984(n) and A059304. 4th binomial transform of A000984.
Largest coefficient of (1 + 6*x + x^2)^n. - Philippe Deléham, Oct 02 2007
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H steps can have 6 colors. - N-E. Fahssi, Mar 31 2008
Self-convolution of a(n)/4^n gives A126646. - Vladimir Reshetnikov, Oct 10 2016
Diagonal of rational function 1/(1 - (x^2 + 6*x*y + y^2)). - Gheorghe Coserea, Aug 03 2018

Examples

			G.f. = 1 + 6*x + 38*x^2 + 252*x^3 + 1734*x^4 + 12276*x^5 + 88796*x^6 + ...
		

Crossrefs

Column 6 of A292627. Cf. A025230, A104454 (binomial transf.)

Programs

  • Mathematica
    Table[SeriesCoefficient[1/(Sqrt[1-4*x]*Sqrt[1-8*x]),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 15 2012 *)
    a[ n_] := If[n < 0, 0, 4^n Hypergeometric2F1[-n, 1/2, 1, -1]]; (* Michael Somos, May 06 2017 *)
    a[ n_] := SeriesCoefficient[ D[ InverseJacobiSD[2 x, -1] / 2, x], {x, 0, 2 n}]; (* Michael Somos, May 06 2017 *)
  • PARI
    x='x+O('x^66); Vec(1/sqrt(1-12*x+32*x^2)) \\ Joerg Arndt, May 11 2013
    
  • PARI
    {a(n) = sum(k=0, n, 8^(n-k)*(-1)^k*binomial(n, k)*binomial(2*k, k))} \\ Seiichi Manyama, Apr 22 2019
    
  • PARI
    {a(n) = sum(k=0, n\2, 6^(n-2*k)*binomial(n, 2*k)*binomial(2*k, k))} \\ Seiichi Manyama, May 04 2019
    
  • Sage
    a = lambda n: 4^n*hypergeometric([-n, 1/2], [1], -1)
    [simplify(a(n)) for n in range(23)] # Peter Luschny, May 19 2015

Formula

G.f.: 1/sqrt(1 - 12*x + 32*x^2).
E.g.f.: exp(6*x)*BesselI(0, 2*x).
a(n) = Sum_{k=0..n} 2^k*binomial(2*k, k)*binomial(2*(n-k), n-k).
a(n) = Sum_{k=0..n} 4^(n-k)*binomial(n,k)*binomial(2k,k). - Paul Barry, Mar 08 2005
D-finite with recurrence: n*a(n) = 6*(2*n-1)*a(n-1) - 32*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ 2^(3*n+1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 15 2012
a(n) = 4^n*hypergeometric([-n, 1/2], [1], -1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..n} 8^(n-k) * (-1)^k * binomial(n,k) * binomial(2*k,k). - Seiichi Manyama, Apr 22 2019
a(n) = Sum_{k=0..floor(n/2)} 6^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, May 04 2019
From Peter Bala, Jan 10 2022: (Start)
3*x + x^2*exp(Sum_{n >= 1} a(n)*x^n/n) = 3*x + x^2 + 6*x^3 + 37*x^4 + 234*x^5 + 1514*x^6 + ... is the o.g.f. of A025230.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k.
a(n) = (1/Pi) * Integral_{x = -1..1} (4 + 4*x^2)^n/sqrt(1 - x^2) dx = (1/Pi) * Integral_{x = -1..1} (8 - 4*x^2)^n/sqrt(1 - x^2) dx. (End)

A068764 Generalized Catalan numbers 2*x*A(x)^2 -A(x) +1 -x =0.

Original entry on oeis.org

1, 1, 4, 18, 88, 456, 2464, 13736, 78432, 456416, 2697088, 16141120, 97632000, 595912960, 3665728512, 22703097472, 141448381952, 885934151168, 5575020435456, 35230798994432, 223485795258368, 1422572226146304, 9083682419818496, 58169612565614592, 373486362257899520, 2403850703479816192
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(2,2; n)/2 with K(a,b; n) defined in a comment to A068763.
Hankel transform is A166232(n+1). - Paul Barry, Oct 09 2009

Examples

			G.f. = 1 + x + 4*x^2 + 18*x^3 + 88*x^4 + 456*x^5 + 2464*x^6 + 13736*x^7 + ...
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(1-Sqrt[1-8*x*(1-x)])/(4*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 13 2012 *)
    Round@Table[4^(n-1) Hypergeometric2F1[(1-n)/2, 1-n/2, 2, 1/2] + KroneckerDelta[n]/Sqrt[2], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 07 2015 *)
    a[ n_] := If[ n < 1, Boole[n == 0], 4^(n - 1) Hypergeometric2F1[ (1 - n)/2, (2 - n)/2, 2, 1/2]]; (* Michael Somos, Nov 08 2015 *)
  • Maxima
    a(n):=sum(binomial(n-1,k-1)*1/k*sum(binomial(k,j)*binomial(k+j,j-1),j,1,k),k,1,n); /* Vladimir Kruchinin, Aug 11 2010 */
    
  • PARI
    {a(n) = my(A); if( n<1, n==0, n--;  A = x * O(x^n); n! * simplify( polcoeff( exp(4*x + A) * besseli(1, 2*x * quadgen(8) + A), n)))}; /* Michael Somos, Mar 31 2007 */
    
  • PARI
    x='x+O('x^66); Vec((1-sqrt(1-8*x*(1-x)))/(4*x)) \\ Joerg Arndt, May 06 2013

Formula

G.f.: (1-sqrt(1-8*x*(1-x)))/(4*x).
a(n+1) = 2*sum(a(k)*a(n-k), k=0..n), n>=1, a(0) = 1 = a(1).
a(n) = (2^n)*p(n, -1/2) with the row polynomials p(n, x) defined from array A068763.
E.g.f. (offset -1) is exp(4*x)*BesselI(1, 2*sqrt(2)*x)/(sqrt(2)*x). - Vladeta Jovovic, Mar 31 2004
The o.g.f. satisfies A(x) = 1 + x*(2*A(x)^2 - 1), A(0) = 1. - Wolfdieter Lang, Nov 13 2007
a(n) = subs(t=1,(d^(n-1)/dt^(n-1))(-1+2*t^2)^n)/n!, n >= 2, due to the Lagrange series for the given implicit o.g.f. equation. This formula holds also for n=1 if no differentiation is used. - Wolfdieter Lang, Nov 13 2007, Feb 22 2008
1/(1-x/(1-x-2x/(1-x/(1-x-2x/(1-x/(1-x-2x/(1-..... (continued fraction). - Paul Barry, Jan 29 2009
a(n) = A166229(n)/(2-0^n). - Paul Barry, Oct 09 2009
a(n) = sum(binomial(n-1,k-1)*1/k*sum(binomial(k,j)*binomial(k+j,j-1),j,1,k),k,1,n), n>0. - Vladimir Kruchinin, Aug 11 2010
D-finite with recurrence: (n+1)*a(n) = 4*(2*n-1)*a(n-1) - 8*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 13 2012
a(n) ~ sqrt(1+sqrt(2))*(4+2*sqrt(2))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 13 2012
a(n) = 4^(n-1)*hypergeom([(1-n)/2,1-n/2], [2], 1/2) + 0^n/sqrt(2). - Vladimir Reshetnikov, Nov 07 2015
0 = a(n)*(+64*a(n+1) - 160*a(n+2) + 32*a(n+3)) + a(n+1)*(+32*a(n+1) + 48*a(n+2) - 20*a(n+3)) + a(n+2)*(+4*a(n+2) + a(n+3)) for all n>=0. - Michael Somos, Nov 08 2015
a(n) = (-1)^n * Sum_{k=0..n} (-2)^k * binomial(n,k) * binomial(2*k+1,n) / (2*k+1). - Seiichi Manyama, Jul 24 2023

A068763 Irregular triangle of the Fibonacci polynomials of A011973 multiplied diagonally by the Catalan numbers.

Original entry on oeis.org

1, 1, 1, 2, 2, 5, 6, 1, 14, 20, 6, 42, 70, 30, 2, 132, 252, 140, 20, 429, 924, 630, 140, 5, 1430, 3432, 2772, 840, 70, 4862, 12870, 12012, 4620, 630, 14, 16796, 48620, 51480, 24024, 4620, 252, 58786, 184756, 218790
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

The row length sequence of this array is [1,2,2,3,3,4,4,5,5,...] = A008619(n+2), n>=0.
The row polynomials p(n,x) := Sum_{m=0..floor((n+1)/2)} a(n,m)*x^m produce, for x = (b-a^2)/a^2 (not 0), the two parameter family of sequences K(a,b; n) := (a^(n+1))*p(n,(b-a^2)/a^2) with g.f. K(a,b; x) := (1-sqrt(1-4*x*(a+x*(b-a^2))))/(2*x).
Some members are: K(1,1; n)=A000108(n) (Catalan), K(1,2; n)=A025227(n-1), K(2,1; n)=A025228(n-1), K(1,3; n)=A025229(n-1), K(3,1; n)=A025230(n-1). For a=b=2..10 the sequences K(a,a; n)/a are A068764-A068772.
The column sequences (without leading 0's) are: A000108 (Catalan), A000984 (central binomial), A002457, 2*A002802, 5*A020918, 14*A020920, 42*A020922, ...
a(n,m) is the number of ways to designate exactly m cherries over all binary trees with n internal nodes. A cherry is an internal node whose descendants are both external nodes. Cf. A091894 which gives the number of binary trees with m cherries. - Geoffrey Critzer, Jul 24 2020
This irregular triangle is essentially that of A011973 with its diagonals multiplied by the Catalan numbers of A000108. The diagonals of this triangle are then rows of the Pascal matrix A007318 multiplied by the Catalan numbers. - Tom Copeland, Dec 23 2023

Examples

			The irregular triangle begins:
   n\m    0     1     2     3    4   5
   0:     1
   1:     1     1
   2:     2     2
   3:     5     6     1
   4:    14    20     6
   5:    42    70    30     2
   6:   132   252   140    20
   7:   429   924   630   140    5
   8:  1430  3432  2772   840   70
   9:  4862 12870 12012  4620  630  14
  10: 16796 48620 51480 24024 4620 252
  ...
p(3,x) = 5 + 6*x + x^2.
		

Crossrefs

Cf. A025227(n-1) (row sums).
Cf. A000007(n) (alternating row sums).

Programs

  • Mathematica
    nn = 10; b[z_] := (1 - Sqrt[1 - 4 z])/(2 z);Map[Select[#, # > 0 &] &,
    CoefficientList[Series[v b[v z] /. v -> (1 + u z ), {z, 0, nn}], {z, u}]] // Grid (* Geoffrey Critzer, Jul 24 2020 *)

Formula

a(n, m) = binomial(n+1-m, m)*C(n-m) if 0 <= m <= floor((n+1)/2), otherwise 0, with C(n) := A000108(n) (Catalan).
G.f. for column m=1, 2, ...: (x^(2*m-1))*C(m-1)/(1-4*x)^((2*m-1)/2); m=0: c(x), g.f. for A000108 (Catalan).
G.f. for row polynomials p(n, x): c(z) + x*z*c(x*(z^2)/(1-4*z))/sqrt(1-4*z) = (1-sqrt(1-4*z*(1+x*z)))/(2*z), where c(x) is the g.f. of A000108 (Catalan).
G.f. for triangle: (1 - sqrt(1 - 4*x (1 + y*x)))/(2*x). - Geoffrey Critzer, Jul 24 2020
The series expansion of f(x) = (1 + 2sx - sqrt(1 + 4sx + 4d^2x^2))/(2x) at x = 0 is (s^2 - d^2) x + (2 d^2s - 2 s^3) x^2 + (d^4 - 6 d^2 s^2 + 5 s^4) x^3 + (-6 d^4 s + 20 d^2 s^3 - 14 s^5) x^4 + ..., containing the coefficients of this array. With s = (a+b)/2 and d = (a-b)/2, then f(x)/ab = g(x) = (1 + (a+b)x - sqrt((1+(a+b)x)^2 - 4abx^2))/(2abx) = x - (a + b) x^2 + (a^2 + 3 a b + b^2) x^3 - (a^3 + 6 a^2 b + 6 a b^2 + b^3) x^4 + ..., containing the Narayana polynomials of A001263, which can be simply transformed into A033282. The compositional inverse about the origin of g(x) is g^(-1)(x) = x/((1-ax)(1-bx)) = x/((1-(s+d)x)(1-(s-d)x)) = x + (a + b) x^2 + (a^2 + a b + b^2) x^3 + (a^3 + a^2 b + a b^2 + b^3) x^4 + ..., containing the complete homogeneous symmetric polynomials h_n(a,b) = (a^n - b^n)/(a-b), which are the polynomials of A034867 when expressed in s and d, e.g., ((s + d)^7 - (s - d)^7)/(2 d) = d^6 + 21 d^4 s^2 + 35 d^2 s^4 + 7 s^6. A133437 and A134264 for compositional inversion of o.g.f.s can be used to relate the sets of polynomials above. - Tom Copeland, Nov 28 2023

Extensions

Title changed by Tom Copeland, Dec 23 2023

A182401 Number of paths from (0,0) to (n,0), never going below the x-axis, using steps U=(1,1), H=(1,0) and D=(1,-1), where the H steps come in five colors.

Original entry on oeis.org

1, 5, 26, 140, 777, 4425, 25755, 152675, 919139, 5606255, 34578292, 215322310, 1351978807, 8550394455, 54419811354, 348309105300, 2240486766555, 14476490777175, 93914850905862, 611489638708140, 3994697746533171, 26175407271617955, 171991872078871311
Offset: 0

Views

Author

Emanuele Munarini, Apr 27 2012

Keywords

Comments

Number of 3-colored Schroeder paths from (0,0) to (2n+2,0) with no level steps H=(2,0) at even level. H-steps at odd levels are colored with one of the three colors. Example: a(2)=5 because we have UUDD, UHD (3 choices) and UDUD. - José Luis Ramírez Ramírez, Apr 27 2015

Examples

			seq(3^n * simplify(hypergeom([3/2, -n], [3], -4/3)), n = 0..20); # _Peter Bala_, Feb 04 2024
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-5*x-Sqrt[1-10*x+21*x^2])/(2*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)
    a[n_] := 5^n*Hypergeometric2F1[(1-n)/2, -n/2, 2, 4/25]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 22 2013, after 2nd formula *)
  • Maxima
    a(n):=coeff(expand((1+5*x+x^2)^(n+1)),x^n)/(n+1);
    makelist(a(n),n,0,30);
    
  • PARI
    x='x+O('x^66); Vec((1-5*x-sqrt(1-10*x+21*x^2))/(2*x^2)) \\ Joerg Arndt, Jun 02 2013

Formula

a(n) = [x^n] (1+5*x+x^2)^(n+1)/(n+1).
a(n) = Sum_{k=0..floor(n/2)} (binomial(n,2*k)*binomial(2*k,k)/(k+1))*5^(n-2*k).
G.f.: (1-5*x-sqrt(1-10*x+21*x^2))/(2*x^2).
Conjecture: (n+2)*a(n) +5*(-2*n-1)*a(n-1) +21*(n-1)*a(n-2)=0. - R. J. Mathar, Jul 24 2012
a(n) ~ 7^(n+3/2)/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
a(n) = A125906(n,0). - Philippe Deléham, Mar 04 2013
G.f.: 1/(1 - 5*x - x^2/(1 - 5*x - x^2/(1 - 5*x - x^2/(1 - 5*x - x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
From Seiichi Manyama, Jan 15 2024: (Start)
G.f.: (1/x) * Series_Reversion( x / (1+5*x+x^2) ).
a(n) = (1/(n+1)) * Sum_{k=0..n} 3^(n-k) * binomial(n+1,n-k) * binomial(2*k+2,k). (End)
From Peter Bala, Feb 03 2024: (Start)
G.f: 1/(1 - 3*x)*c(x/(1 - 3*x))^2 = 1/(1 - 7*x)*c(-x/(1 - 7*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
a(n) = Sum_{k = 0..n} 3^(n-k)*binomial(n, k)*Catalan(k+1).
a(n) = 3^n * hypergeom([3/2, -n], [3], -4/3).
a(n) = 7^n * Sum_{k = 0..n} (-7)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 7^n * hypergeom([3/2, -n], [3], 4/7). (End)

A068772 Generalized Catalan numbers 10*x*A(x)^2 -A(x) +1 -9*x =0.

Original entry on oeis.org

1, 1, 20, 410, 8600, 184200, 4020000, 89205000, 2008700000, 45816140000, 1056825200000, 24618524200000, 578457724000000, 13695679012000000, 326448619920000000, 7827776361090000000, 188701194087000000000
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

This is the tenth member in the a-family of sequences K(a,a; n), a=1,2,3,...,n>=0, defined in a comment to the array A068763.

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[1] = 1; a[n_] := (360 (2 - n) a[n - 2] + 20 (2 n - 1) a[n - 1])/(n + 1); Table[a[n], {n, 0, 20}] (* Wesley Ivan Hurt, Mar 04 2014 *)
    CoefficientList[Series[(1-Sqrt[1-40*x*(1-9*x)])/(20*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 06 2014 *)

Formula

a(n) = (10^n) * p(n, -9/10) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 10*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-40*x*(1-9*x)))/(20*x).
Recurrence: (n+1)*a(n) = 360*(2-n)*a(n-2) + 20*(2*n-1)*a(n-1). - Fung Lam, Mar 05 2014
a(n) ~ sqrt(5+5*sqrt(10)) * (20+2*sqrt(10))^n / (10*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 06 2014

A247495 Generalized Motzkin numbers: Square array read by descending antidiagonals, T(n, k) = k!*[x^k](exp(n*x)* BesselI_{1}(2*x)/x), n>=0, k>=0.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 2, 4, 5, 3, 1, 0, 9, 14, 10, 4, 1, 5, 21, 42, 36, 17, 5, 1, 0, 51, 132, 137, 76, 26, 6, 1, 14, 127, 429, 543, 354, 140, 37, 7, 1, 0, 323, 1430, 2219, 1704, 777, 234, 50, 8, 1, 42, 835, 4862, 9285, 8421, 4425, 1514, 364, 65, 9, 1
Offset: 0

Views

Author

Peter Luschny, Dec 11 2014

Keywords

Comments

This two-dimensional array of numbers can be seen as a generalization of the Motzkin numbers A001006 for two reasons: The case n=1 reduces to the Motzkin numbers and the columns are the values of the Motzkin polynomials M_{k}(x) = sum_{j=0..k} A097610(k,j)*x^j evaluated at the nonnegative integers.

Examples

			Square array starts:
[n\k][0][1] [2]  [3]   [4]   [5]    [6]     [7]      [8]
[0]   1, 0,  1,   0,    2,    0,     5,      0,      14, ...  A126120
[1]   1, 1,  2,   4,    9,   21,    51,    127,     323, ...  A001006
[2]   1, 2,  5,  14,   42,  132,   429,   1430,    4862, ...  A000108
[3]   1, 3, 10,  36,  137,  543,  2219,   9285,   39587, ...  A002212
[4]   1, 4, 17,  76,  354, 1704,  8421,  42508,  218318, ...  A005572
[5]   1, 5, 26, 140,  777, 4425, 25755, 152675,  919139, ...  A182401
[6]   1, 6, 37, 234, 1514, 9996, 67181, 458562, 3172478, ...  A025230
A000012,A001477,A002522,A079908, ...
.
Triangular array starts:
              1,
             0, 1,
           1, 1, 1,
          0, 2, 2, 1,
        2, 4, 5, 3, 1,
      0, 9, 14, 10, 4, 1,
   5, 21, 42, 36, 17, 5, 1,
0, 51, 132, 137, 76, 26, 6, 1.
		

Crossrefs

Programs

  • Maple
    # RECURRENCE
    T := proc(n,k) option remember; if k=0 then 1 elif k=1 then n else
    (n*(2*k+1)*T(n,k-1)-(n-2)*(n+2)*(k-1)*T(n,k-2))/(k+2) fi end:
    seq(print(seq(T(n,k),k=0..9)),n=0..6);
    # OGF (row)
    ogf := n -> (1-n*x-sqrt(((n-2)*x-1)*((n+2)*x-1)))/(2*x^2):
    seq(print(seq(coeff(series(ogf(n),x,12),x,k),k=0..9)),n=0..6);
    # EGF (row)
    egf := n -> exp(n*x)*hypergeom([],[2],x^2):
    seq(print(seq(k!*coeff(series(egf(n),x,k+2),x,k),k=0..9)),n=0..6);
    # MOTZKIN polynomial (column)
    A097610 := proc(n,k) if type(n-k,odd) then 0 else n!/(k!*((n-k)/2)!^2* ((n-k)/2+1)) fi end: M := (k,x) -> add(A097610(k,j)*x^j,j=0..k):
    seq(print(seq(M(k,n),n=0..9)),k=0..6);
    # OGF (column)
    col := proc(n, len) local G; G := A247497_row(n); (-1)^(n+1)* add(G[k+1]/(x-1)^(k+1), k=0..n); seq(coeff(series(%, x, len+1),x,j), j=0..len) end: seq(print(col(n,8)), n=0..6); # Peter Luschny, Dec 14 2014
  • Mathematica
    T[0, k_] := If[EvenQ[k], CatalanNumber[k/2], 0];
    T[n_, k_] := n^k*Hypergeometric2F1[(1 - k)/2, -k/2, 2, 4/n^2];
    Table[T[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 03 2017 *)
  • Sage
    def A247495(n,k):
        if n==0: return(k//2+1)*factorial(k)/factorial(k//2+1)^2 if is_even(k) else 0
        return n^k*hypergeometric([(1-k)/2,-k/2],[2],4/n^2).simplify()
    for n in (0..7): print([A247495(n,k) for k in range(11)])

Formula

T(n,k) = (n*(2*k+1)*T(n,k-1)-(n-2)*(n+2)*(k-1)*T(n,k-2))/(k+2) for k>=2.
T(n,k) = Sum_{j=0..floor(k/2)} n^(k-2*j)*binomial(k,2*j)*binomial(2*j,j)/(j+1).
T(n,k) = n^k*hypergeom([(1-k)/2,-k/2], [2], 4/n^2) for n>0.
T(n,n) = A247496(n).
O.g.f. for row n: (1-n*x-sqrt(((n-2)*x-1)*((n+2)*x-1)))/(2*x^2).
O.g.f. for row n: R(x)/x where R(x) is series reversion of x/(1+n*x+x^2).
E.g.f. for row n: exp(n*x)*hypergeom([],[2],x^2).
O.g.f. for column k: the k-th column consists of the values of the k-th Motzkin polynomial M_{k}(x) evaluated at x = 0,1,2,...; M_{k}(x) = sum_{j=0..k} A097610(k,j)*x^j = sum_{j=0..k} (-1)^j*binomial(k,j)*A001006(j)*(x+1)^(k-j).
O.g.f. for column k: sum_{j=0..k} (-1)^(k+1)*A247497(k,j)/(x-1)^(j+1). - Peter Luschny, Dec 14 2014
O.g.f. for row n: 1/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
T(n,k) is the coefficient of x^k in the expansion of 1/(k+1) * (1 + n*x + x^2)^(k+1). - Seiichi Manyama, May 07 2019

A068767 Generalized Catalan numbers 5*x*A(x)^2 -A(x) +1 -4*x=0.

Original entry on oeis.org

1, 1, 10, 105, 1150, 13050, 152500, 1825625, 22293750, 276758750, 3483287500, 44352006250, 570333187500, 7396680812500, 96638930625000, 1270796364765625, 16806545339843750, 223400240246093750
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(5,5; n)/5 with K(a,b; n) defined in a comment to A068763.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-20*x*(1-4*x)])/(10*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (5^n) * p(n, -4/5) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 5*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-20*x*(1-4*x)))/(10*x).
(n+1)*a(n) = 80*(2-n)*a(n-2) + 10*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(10+10*sqrt(5)) * (10+2*sqrt(5))^n / (10*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014
Equivalently, a(n) ~ 2^(2*n) * 5^((n-1)/2) * phi^(n + 1/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021

A068768 Generalized Catalan numbers 6*x*A(x)^2 -A(x) +1 -5*x =0.

Original entry on oeis.org

1, 1, 12, 150, 1944, 25992, 356832, 5008824, 71629920, 1040509152, 15315578496, 227981324736, 3426473187072, 51929043390720, 792725911280640, 12178706839758720, 188158789025809920, 2921622674591946240
Offset: 0

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Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(6,6; n)/6 with K(a,b; n) defined in a comment to A068763.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-24*x*(1-5*x)])/(12*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (6^n) * p(n, -5/6) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 6*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-24*x*(1-5*x)))/(12*x).
D-finite with recurrence: (n+1)*a(n) = 120*(2-n)*a(n-2) + 12*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(3+3*sqrt(6)) * (12+2*sqrt(6))^n / (6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014
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