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 1-3 of 3 results.

A098619 G.f. A(x) satisfies: A(x*G098618(x)) = G098618(x), where G098618 is the g.f. for A098618(n) = A007482(n)*Catalan(n).

Original entry on oeis.org

1, 3, 13, 51, 213, 867, 3589, 14739, 60853, 250563, 1033605, 4259571, 17565909, 72412707, 298586661, 1231016019, 5075753589, 20927272323, 86286346693, 355763629491, 1466857936405, 6047981701347, 24936516122469, 102815688922899, 423920292507061, 1747866711689283, 7206641564551429
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2004

Keywords

Comments

G.f. satisfies: A(x) = x/(series reversion of x*G098618(x)), where G098618 is the g.f. for A098618 = {1*1,3*1,11*2,39*5,139*14,495*42,1763*132,...}.

Links

Crossrefs

Cf. A098618, A098615, A098617, A007482, A000108.

Programs

  • Mathematica
    Flatten[{1,3,13,51,Table[17^(n/2)*(1/2+1/2*(-1)^n + 3/34*Sqrt[17]*(1-(-1)^n) + Sum[(-1)^j*(4/17 + Sum[Binomial[2*k-1,k-1]*2^(k+3)/ ((k+1)*17^(k+1)), {k,1,Floor[(j-1)/2]}]),{j,3,n-1}]),{n,4,20}]}] (* Vaclav Kotesovec, Oct 29 2012 *)
  • PARI
    a(n)=polcoeff((sqrt(1-8*x^2+x^2*O(x^n))+3*x)/(1-17*x^2),n);
  • PARI
    x='x+O('x^66); Vec((sqrt(1-8*x^2) + 3*x)/(1-17*x^2)) \\ Joerg Arndt, May 12 2013

Formula

G.f.: (sqrt(1-8*x^2) + 3*x)/(1-17*x^2).
a(2*n+1) = 3*17^n.
Recurrence: n*a(n) = (25*n-24)*a(n-2) - 136*(n-3)*a(n-4). - Vaclav Kotesovec, Oct 29 2012

A098614 Product of Fibonacci and Catalan numbers: a(n) = A000045(n+1)*A000108(n).

Original entry on oeis.org

1, 1, 4, 15, 70, 336, 1716, 9009, 48620, 267410, 1494844, 8465184, 48466796, 280073300, 1631408400, 9568812015, 56466198990, 335002137360, 1997007404700, 11955535480350, 71850862117320, 433322055191220, 2621615826231480, 15906988165723200, 96775058652983100
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2004

Keywords

Comments

Radius of convergence: r = (sqrt(5)-1)/8; A(r) = sqrt(2+2/sqrt(5)). More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
a(n) is also the number of nonnesting permutations of {1,1,2,2,...,n,n} that avoid the patterns 1223, 1332, 2113, or the patterns 1123, 1132, 2133. - Amya Luo, Dec 11 2024

Examples

			Sequence has the factored form: {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, 21*429, ...}.

Links

Crossrefs

Cf. A000045, A000108, A098615, A098616, A098618, A249925.

Programs

  • Magma
    [Fibonacci(n+1)*Catalan(n): n in [0..40]]; // G. C. Greubel, Jul 31 2024
  • Mathematica
    With[{nn=30},Times@@@Thread[{Fibonacci[Range[nn]],CatalanNumber[ Range[ 0,nn-1]]}]] (* Harvey P. Dale, Nov 14 2011 *)
  • PARI
    {a(n)=local(X=x+O(x^(n+3)), A); A = sqrt( (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n)=binomial(2*n,n)/(n+1)*round(((1+sqrt(5))^(n+1)-(1-sqrt(5))^(n+1))/(2^(n+1)*sqrt(5)))}
  • SageMath
    [fibonacci(n+1)*catalan_number(n) for n in range(41)] # G. C. Greubel, Jul 31 2024

Formula

G.f.: A(x) = sqrt( (1-2*x - sqrt(1-4*x-16*x^2))/10 )/x.
G.f. satisfies: A(x) = sqrt( 1 + 2*x*A(x)^2 + 5*x^2*A(x)^4 ).
a(n) == 1 (mod 2) iff n = 2^k - 1 for k>=0.
n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/8^n = 2*sqrt(2/5). - Amiram Eldar, May 06 2023

A098616 Product of Pell and Catalan numbers: a(n) = A000129(n+1)*A000108(n).

Original entry on oeis.org

1, 2, 10, 60, 406, 2940, 22308, 175032, 1408550, 11561836, 96425836, 814773960, 6960289532, 60012947800, 521582661000, 4564643261040, 40190674554630, 355772529165900, 3164408450118300, 28266363849505320, 253466716153665300, 2280803103062033160, 20588945107316958840
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2004

Keywords

Comments

Radius of convergence: r = (sqrt(2)-1)/4, where A(r) = sqrt(2+sqrt(2)).
More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

Examples

			Sequence begins: [1*1, 2*1, 5*2, 12*5, 29*14, 70*42, 169*132, 408*429,...].

Links

Crossrefs

Cf. A000129, A000108, A098614, A098617, A098618, A000984, A214377.

Programs

  • Mathematica
    With[{nn=30},Times@@@Thread[{LinearRecurrence[{2,1},{1,2},nn], CatalanNumber[ Range[0,nn-1]]}]] (* Harvey P. Dale, Jan 04 2012 *)
a[n_] := Fibonacci[n + 1, 2] * CatalanNumber[n]; Array[a, 25, 0] (* Amiram Eldar, May 05 2023 *)
  • PARI
    a(n) = binomial(2*n,n)/(n+1)*round(((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/(2*sqrt(2)))

Formula

G.f.: A(x) = sqrt( (1-4*x - sqrt(1-8*x-16*x^2))/16 )/x.
Run lengths of zeros (mod 10) equal (5^k - (-1)^k)/2 - 1 starting at index (5^k + (-1)^k)/2:
a(n) == 0 (mod 10) for n = (5^k + (-1)^k)/2 through n = 5^k - 1 when k>=1.
a(n) ~ 2^(2*n-3/2) * (1+sqrt(2))^(n+1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, May 09 2014
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). Compare with the o.g.f. B(x) of the central binomial numbers A000984, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/16^n = 2*sqrt(3-sqrt(7)). - Amiram Eldar, May 05 2023
G.f. A(x) satisfies A(x) = sqrt( 1 + 4*x*A(x)^2 + 8*x^2*A(x)^4 ). - Paul D. Hanna, Dec 14 2024
Showing 1-3 of 3 results.

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