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.

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

A025239 a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ...+ a(n-2)*a(2) for n >= 3.

Original entry on oeis.org

2, 3, 6, 21, 78, 318, 1356, 5997, 27222, 126138, 594132, 2836290, 13692300, 66729180, 327855768, 1622216829, 8076311142, 40427919714, 203353800324, 1027318915254, 5210182030308, 26517609163812, 135397544040744, 693364054299474
Offset: 1

Views

Author

Clark Kimberling

Keywords

Links

Crossrefs

Essentially the same as A025229.

Programs

  • Mathematica
    Join[{2},Drop[CoefficientList[Series[(1-Sqrt[1-4x-8x^2])/2, {x,0,30}], x],2]] (* Harvey P. Dale, Nov 05 2011 *)
  • PARI
    a(n)=polcoeff((1-sqrt(1-4*x-8*x^2+x*O(x^n)))/2,n)

Formula

G.f.: (1-sqrt(1-4*x-8*x^2))/2. - Michael Somos, Jun 08 2000
Recurrence (for n>3): n*a(n) = 2*(2*n-3)*a(n-1) + 8*(n-3)*a(n-2). - Vaclav Kotesovec, Oct 07 2012
a(n) ~ sqrt(3-sqrt(3))*(2+2*sqrt(3))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012
Showing 1-6 of 6 results.

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