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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A366694 G.f. satisfies A(x) = (1 + x)^2 + x*A(x)^2.

Original entry on oeis.org

1, 3, 7, 23, 88, 363, 1576, 7091, 32768, 154588, 741442, 3604495, 17721394, 87960004, 440165522, 2218289051, 11248850578, 57354875692, 293860786178, 1512169500356, 7811933144432, 40499933496818, 210643657689644, 1098802033533295, 5747266778089846
Offset: 0

Views

Author

Seiichi Manyama, Oct 16 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(2*(k+1), n-k)*binomial(2*k, k)/(k+1));

Formula

G.f.: A(x) = 2*(1+x)^2 / (1+sqrt(1-4*x*(1+x)^2)).
a(n) = Sum_{k=0..n} binomial(2*(k+1),n-k) * binomial(2*k,k)/(k+1).

A366695 G.f. satisfies A(x) = (1 + x)^3 + x*A(x)^2.

Original entry on oeis.org

1, 4, 11, 39, 166, 765, 3716, 18725, 96956, 512690, 2756806, 15027651, 82853678, 461215414, 2588619402, 14632777719, 83232244238, 476040155118, 2736005962314, 15793863291792, 91530881427964, 532343678619778, 3106141476531628, 18177446846299299
Offset: 0

Views

Author

Seiichi Manyama, Oct 16 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(3*(k+1), n-k)*binomial(2*k, k)/(k+1));

Formula

G.f.: A(x) = 2*(1+x)^3 / (1+sqrt(1-4*x*(1+x)^3)).
a(n) = Sum_{k=0..n} binomial(3*(k+1),n-k) * binomial(2*k,k)/(k+1).

A379171 G.f. A(x) satisfies A(x) = (1 + x)/(1 - x*A(x))^3.

Original entry on oeis.org

1, 4, 21, 139, 1021, 8010, 65708, 556751, 4834686, 42800265, 384832083, 3504693519, 32261240127, 299685628629, 2805773759322, 26448278629697, 250806022116194, 2390973659474304, 22901157688878983, 220279614235505630, 2126890041331033797, 20606993367985131716
Offset: 0

Views

Author

Seiichi Manyama, Dec 17 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n-k+1, k)*binomial(4*n-4*k+2, n-k)/(n-k+1));

Formula

a(n) = Sum_{k=0..n} binomial(n-k+1,k) * binomial(4*n-4*k+2,n-k)/(n-k+1).

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

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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

Crossrefs

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

Crossrefs

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

Crossrefs

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

A260772 Certain directed lattice paths.

Original entry on oeis.org

1, 3, 10, 41, 190, 946, 4940, 26693, 147990, 837102, 4811860, 28027210, 165057100, 981177060, 5879570200, 35478788269, 215398416870, 1314794380374, 8064119033220, 49673222082782, 307163049317540, 1906066361809148, 11865666767361960, 74081851132379426
Offset: 0

Views

Author

N. J. A. Sloane, Jul 30 2015

Keywords

Comments

See Dziemianczuk (2014) for precise definition.

Crossrefs

Programs

  • Maple
    # A260772 satisfies a 4th-order recurrence that can be reduced
    # to a 2nd-order recurrence given in this program t:
    t := proc(n) options remember;
    if n <= 1 then
        [-1/2, 0, 1, 4][2*n+2]
      else
        (16*(n-2)*(2*n-3)*(5*n-2)*t(n-2) + (440*n^3-1056*n^2+724*n-144)*t(n-1))
           /( n*(2*n+1)*(5*n-7) )
      fi
    end:
    A260772 := proc(n)
    t(n/2) + ( (2-2*n)*t((n-1)/2)+(n+2)*t((n+1)/2) ) / (1+5*n)
    end:
    seq(A260772(i),i=0..100);
    # Mark van Hoeij, Jul 14 2022
  • Maxima
    a(n):=if n=0 then 1 else sum((-1)^j*binomial(n,j)*binomial(3*n-4*j,n-4*j+1),j,0,(n+1)/4)/n; /* Vladimir Kruchinin, Apr 04 2019 */
    
  • PARI
    a(n) = if (n==0, 1, sum(j=0, (n+1)/4, (-1)^j*binomial(n,j)*binomial(3*n-4*j, n-4*j+1))/n); \\ Michel Marcus, Apr 05 2019

Formula

G.f.: P1(x) = (2*(1-x)/3)/x - ((2*sqrt(1-5*x-2*x^2)/3)/x)*sin((Pi/6 + arccos(((20*x^3-6*x^2+15*x-2)/2)/(1-5*x-2*x^2)^(3/2))/3)). - See Dziemianczuk (2014), Proposition 11.
a(n) = (1/n)*Sum_{j=0..(n+1)/4} (-1)^j*C(n,j)*C(3*n-4*j,n-4*j+1), a(0)=1. - Vladimir Kruchinin, Apr 04 2019
n*(n+1)*(25*n^2-70*n+21)*a(n) - 30*(7*n-15)*n*a(n-1) + (-1100*n^4+5280*n^3-6424*n^2-1188*n+3816)*a(n-2) + 120*(n+2)*(n-3)*a(n-3) - 16*(n-3)*(n-4)*(25*n^2-20*n-24)*a(n-4) = 0. - Mark van Hoeij, Jul 14 2022
a(n) ~ 2^(n - 1/2) * phi^((10*n - 1)/4) / (sqrt(Pi) * 5^(1/4) * sqrt(phi^(3/2) - 2) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Jul 15 2022

Extensions

More terms from Lars Blomberg, Aug 01 2015

A367042 G.f. satisfies A(x) = 1 + x^3 + x*A(x)^2.

Original entry on oeis.org

1, 1, 2, 6, 16, 48, 152, 500, 1688, 5816, 20368, 72288, 259424, 939808, 3432192, 12622416, 46706144, 173762016, 649569216, 2438748864, 9191656192, 34765298944, 131912452864, 501987944832, 1915417307392, 7326620001536, 28088736525824, 107913607531520
Offset: 0

Views

Author

Seiichi Manyama, Nov 03 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\3, binomial(n-3*k+1, k)*binomial(2*(n-3*k), n-3*k)/(n-3*k+1));

Formula

G.f.: A(x) = 2*(1+x^3) / (1+sqrt(1-4*x*(1+x^3))).
a(n) = Sum_{k=0..floor(n/3)} binomial(n-3*k+1,k) * binomial(2*(n-3*k),n-3*k)/(n-3*k+1).

A367639 G.f. A(x) satisfies A(x) = (1 + x)^2 + x*A(x)^2 / (1 + x).

Original entry on oeis.org

1, 3, 6, 16, 52, 184, 688, 2672, 10672, 43552, 180800, 761088, 3241088, 13937408, 60435968, 263962880, 1160188672, 5127762432, 22775636992, 101608357888, 455105255424, 2045751037952, 9225923895296, 41731062358016, 189275050729472, 860630181167104
Offset: 0

Views

Author

Seiichi Manyama, Nov 25 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(k+2, n-k)*binomial(2*k, k)/(k+1));

Formula

G.f.: A(x) = 2*(1+x)^2 / (1+sqrt(1-4*x*(1+x))).
a(n) = Sum_{k=0..n} binomial(k+2,n-k) * binomial(2*k,k)/(k+1).
a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(n + 3/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 25 2023
D-finite with recurrence (n+1)*a(n) +(-3*n+1)*a(n-1) +2*(-4*n+9)*a(n-2) +4*(-n+4)*a(n-3)=0. - R. J. Mathar, Dec 04 2023
From Peter Bala, May 05 2024: (Start)
A(x) = (1 + x)*S(x/(1 + x)), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the g.f. of the large Schröder numbers A006318. Cf. A025227.
A333090(n) = [x^n] A(x)^n. (End)

A378317 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(2*r+k,n)/(2*r+k) for k > 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 4, 0, 1, 6, 12, 12, 0, 1, 8, 24, 40, 40, 0, 1, 10, 40, 92, 144, 144, 0, 1, 12, 60, 176, 360, 544, 544, 0, 1, 14, 84, 300, 752, 1440, 2128, 2128, 0, 1, 16, 112, 472, 1400, 3200, 5872, 8544, 8544, 0, 1, 18, 144, 700, 2400, 6352, 13664, 24336, 35008, 35008, 0
Offset: 0

Views

Author

Seiichi Manyama, Nov 23 2024

Keywords

Examples

			Square array begins:
  1,   1,    1,    1,     1,     1,     1, ...
  0,   2,    4,    6,     8,    10,    12, ...
  0,   4,   12,   24,    40,    60,    84, ...
  0,  12,   40,   92,   176,   300,   472, ...
  0,  40,  144,  360,   752,  1400,  2400, ...
  0, 144,  544, 1440,  3200,  6352, 11616, ...
  0, 544, 2128, 5872, 13664, 28480, 54768, ...
		

Crossrefs

Columns k=0..1 give A000007, A025227(n+1).
Main diagonal gives A333473.

Programs

  • PARI
    T(n, k, t=0, u=2) = if(k==0, 0^n, k*sum(r=0, n, binomial(n, r)*binomial(t*n+u*r+k, n)/(t*n+u*r+k)));
    matrix(7, 7, n, k, T(n-1, k-1))

Formula

G.f. A_k(x) of column k satisfies A_k(x) = ( 1 + x + x * A_k(x)^(2/k) )^k for k > 0.
G.f. of column k: (B(x)/x)^k where B(x) is the g.f. of A025227.
B(x)^k = B(x)^(k-1) + x * B(x)^(k-1) + x * B(x)^(k+1). So T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k+1) for n > 0.
Previous Showing 21-30 of 41 results. Next