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 1-6 of 6 results.

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

Views

Author

N. J. A. Sloane

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).

Links

Crossrefs

Binomial transform of A002212. Sequence shifted right twice is A025228.
Cf. A001006, A025228, A025230, A108198, A129400, A182401.

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

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 + ...

Links

Crossrefs

Cf. A025227, A025228, A025229, A025230.
Cf. A071356, A001003, A025235.

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.

Links

Crossrefs

Cf. A025227(n-1) (row sums).
Cf. A000007(n) (alternating row sums).
Cf. A000108, A001263, A007318, A011973, A033832, A034867, A133437 and A134264.

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 alternating row sums are {1,repeat 0} = {A000007(n)}{n>=0}. From the second comment p(n, x=-1) = K(1,0; n), with g.f. K(1,0; x) = 1. Or from the g.f. of the triangle with y = -1. Thanks to Aaron Li for asking for a proof. - _Wolfdieter Lang, Jan 21 2023
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

A103970 Expansion of (1 - sqrt(1 - 4*x - 12*x^2))/(2*x).

Original entry on oeis.org

1, 4, 8, 32, 128, 576, 2688, 13056, 65024, 330752, 1710080, 8962048, 47497216, 254132224, 1370849280, 7447117824, 40707293184, 223731253248, 1235630948352, 6853893292032, 38166664839168, 213288826699776, 1195775593807872, 6723691157127168, 37908469021409280, 214260335517892608, 1213784937073737728, 6890689428042285056
Offset: 0

Views

Author

Paul Barry, Feb 23 2005

Keywords

Comments

Image of c(x), the g.f. of the Catalan numbers A000108 under the mapping g(x) -> (1+3x)g(x(1+3x)). In general, the image of the Catalan numbers under the mapping g(x) -> (1+i*x)g(x(1+i*x)) is given by a(n) = Sum_{k=0..n} i^(n-k)*C(k)*C(k+1,n-k).
Hankel transform is 4^C(n+1,2)*A128018(n). [Paul Barry, Nov 20 2009]
By following L. Comtet [Analyse Combinatoire Tomes 1 et 2, PUF, Paris 1970], we also obtain (n+1)*C(n) - 2*a*(2*n-1)*C(n-1) + 4*(n-2)*(a^2-b)*C(n-2) = 0. In the present case, we also have the asymptotic result: a(n) ~ sqrt(4/3)*2^(n-1)*3^(n+1)/sqrt(Pi*n^3) for large n. - Richard Choulet, Dec 17 2009

Links

Crossrefs

Cf. A025227, A025229, A103971, A103972.
Cf. A000108, A025228, A025230, A025231, A025232. [Richard Choulet, Dec 17 2009]

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( (1-Sqrt(1-4*x-12*x^2))/(2*x) )); // G. C. Greubel, Mar 16 2019
  • Maple
    n:=30:a(0):=1:a(1):=4: k:=1: for k from 1 to n do a(k+1):=sum('a(p)*a(k-p)','p'=0..k):od:seq(a(k),k=0..n); # Richard Choulet, Dec 17 2009
taylor(((1-(1-4*z-12*z^2)^0.5)/(2*z)),z=0,32); # Richard Choulet, Dec 17 2009
  • Mathematica
    CoefficientList[Series[(1 - Sqrt[1-4x-12x^2])/(2x), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 18 2017 *)
  • PARI
    my(x='x+O('x^35)); Vec((1-sqrt(1-4*x-12*x^2))/(2*x)) \\ G. C. Greubel, Mar 16 2019
  • Sage
    ((1-sqrt(1-4*x-12*x^2))/(2*x)).series(x, 35).coefficients(x, sparse=False) # G. C. Greubel, Mar 16 2019

Formula

G.f.: (1 - sqrt(1-4*x*(1+3*x)))/(2*x).
a(n) = Sum_{k=0..n} 3^(n-k)*C(k)*C(k+1, n-k).
D-finite with recurrence: (n+1)*a(n) = 2*(2*n-1)*a(n-1) + 12*(n-2)*a(n-2). - Richard Choulet, Dec 17 2009

A103971 Expansion of (1 - sqrt(1 - 4*x - 16*x^2))/(2*x).

Original entry on oeis.org

1, 5, 10, 45, 190, 930, 4660, 24445, 131190, 719830, 4013260, 22684370, 129661740, 748252580, 4353379560, 25508284445, 150392391590, 891549228430, 5310994644060, 31775749689670, 190860711108740, 1150473009844380
Offset: 0

Views

Author

Paul Barry, Feb 23 2005

Keywords

Comments

Image of c(x), the g.f. of the Catalan numbers A000108 under the mapping g(x) -> (1+4x)g(x(1+4x)). In general, the image of the Catalan numbers under the mapping g(x)->(1+i*x)g(x(1+i*x)) is given by a(n) = Sum_{k=0..n} i^(n-k)C(k)C(k+1,n-k).
More generally, the sequence C for which C(0)=a, C(1)=b and C(n+1) = sum(C(k)*C(n-k),k=0..n) has the following g.f. f: f(z) = (1-sqrt(1-4*z*(a-(a^2-b)*z)))/(2*z). We obtain: C(n)=(sum(-1)^(p-1)*2^{n-p}a^{n-2*p-1}*(a^2-b)^p*((2*n-2*p-1)*...*5*3*1/(p!*(n-2*p+1)!)),p=0..floor((n+1)/2)). By following Comtet [Analyse Combinatoire Tomes 1 et 2, PUF, Paris 1970], we obtain also: (n+1)*C(n) - 2*a*(2*n-1)*C(n-1) + 4*(n-2)*(a^2-b)*C(n-2) = 0. - Richard Choulet, Dec 17 2009

Links

Crossrefs

Cf. A000108, A025227, A025228, A025229, A025230, A025231, A025232. - Richard Choulet, Dec 17 2009

Programs

  • Maple
    n:=30:a(0):=1:a(1):=5: for k from 1 to n do a(k+1):=sum('a(p)*a(k-p)','p'=0..k):od:seq(a(k),k=0..n); # Richard Choulet, Dec 17 2009
  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-4x-16x^2])/(2x),{x,0,30}],x] (* Harvey P. Dale, Apr 02 2012 *)

Formula

G.f.: (1-sqrt(1-4*x*(1+4*x)))/(2*x).
a(n) = Sum_{k=0..n} 4^(n-k)*C(k)*C(k+1, n-k).
Another recurrence formula: (n+1)*a(n) = 2*(2*n-1)*a(n-1) + 16*(n-2)*a(n-2). - Richard Choulet, Dec 17 2009
a(n) ~ sqrt(10 + 2*sqrt(5))*(2 + 2*sqrt(5))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
Equivalently, a(n) ~ 5^(1/4) * 2^(2*n) * phi^(n + 1/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021

A103972 Expansion of (1-sqrt(1-4*x-20*x^2))/(2*x).

Original entry on oeis.org

1, 6, 12, 60, 264, 1392, 7392, 41424, 236640, 1384512, 8224896, 49554816, 301884672, 1856878080, 11514915840, 71915838720, 451938731520, 2855705994240, 18132621772800, 115637702461440, 740356410961920, 4756888756101120, 30662391191715840, 198229520200704000, 1285001080928845824
Offset: 0

Views

Author

Paul Barry, Feb 23 2005

Keywords

Comments

Image of c(x), the g.f. of the Catalan numbers A000108 under the mapping g(x)->(1+5x)g(x(1+5x)). In general, the image of the Catalan numbers under the mapping g(x)->(1+i*x)g(x(1+i*x)) is given by a(n)=sum{k=0..n, i^(n-k)C(k)C(k+1,n-k)}.
More generally, the sequence C for which C(0)=a, C(1)=b and C(n+1)=sum(C(k)*C(n-k),k=0..n) has the following G.f f: f(z)= (1-sqrt(1-4*z*(a-(a^2-b)*z)))/(2*z). We obtain: C(n)=(sum(-1)^(p-1)*2^{n-p}a^{n-2*p-1}*(a^2-b)^p*((2*n-2*p-1)*...*5*3*1/(p!*(n-2*p+1)!)),p=0..floor((n+1)/2)). By following L. Comtet [Analyse Combinatoire Tomes 1 et 2, PUF, Paris 1970], we obtain also: (n+1)*C(n)-2*a*(2*n-1)*C(n-1)+4*(n-2)*(a^2-b)*C(n-2)=0. - Richard Choulet, Dec 17 2009

Links

Crossrefs

Cf. A000108, A025227, A025228, A025229, A025230, A025231, A025232. [From Richard Choulet, Dec 17 2009]

Programs

  • Maple
    n:=30:a(0):=1:a(1):=6 :for k from 1 to n do a(k+1):=sum('a(p)*a(k-p)','p'=0..k):od:seq(a(k),k=0..n); # Richard Choulet, Dec 17 2009
  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-4*x-20*x^2])/(2*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)
  • PARI
    x='x+O('x^66); Vec((1-sqrt(1-4*x-20*x^2))/(2*x)) \\ Joerg Arndt, May 13 2013

Formula

G.f.: (1-sqrt(1-4*x*(1+5*x)))/(2*x).
a(n) = Sum_{k=0..n} 5^(n-k)*C(k)*C(k+1, n-k).
Another recurrence formula: (n+1)*a(n)=2*(2n-1)*a(n-1)+20*(n-2)*a(n-2). - Richard Choulet, Dec 17 2009
a(n) ~ sqrt(12+2*sqrt(6))*(2+2*sqrt(6))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
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