A098614 -id:A098614 - cp's OEIS frontend

A098615 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098614(n) = Fibonacci(n+1)Catalan(n).

1, 1, 3, 5, 13, 25, 61, 125, 295, 625, 1447, 3125, 7151, 15625, 35491, 78125, 176597, 390625, 880125, 1953125, 4390901, 9765625, 21920913, 48828125, 109486993, 244140625, 547018941, 1220703125, 2733608905, 6103515625, 13662695645, 30517578125, 68294088535, 152587890625, 341399727335, 762939453125, 1706739347095, 3814697265625, 8532741458075, 19073486328125, 42660172763995, 95367431640625

Offset: 0

Paul D. Hanna, Oct 14 2004

nonn

G.f. satisfies: A(x) = x/(series reversion of x*G098614(x)), where G098614 is the g.f. for A098614 = {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, ...}.
Hankel transform is 2^n. Image of F(n+1) under the Riordan array (c(x^2),xc(x^2)), c(x) the g.f. of A000108. The sequence 0,1,1,3,5,... has general term Sum_{k=0..floor(n/2)} (C(n-1,k) - C(n-1,k-1))*F(n-2k). It is the image of the Fibonacci numbers under the transform of generating functions g(x)-> g(xc(x^2)), c(x) the g.f. of A000108. This sequence has Hankel transform -(-4)^((n-1)/2)(1-(-1)^n)/2. - Paul Barry, Oct 01 2007
The sequence of fractions 1, 1/2, 3/4, 5/8, 13/16, 25/32, ... or a(n)/2^n is the image of F(n+1) under the Chebyshev related (rational) Riordan array c((x/2)^2),(x/2)c((x/2)^2)) where c(x) is the g.f. of A000108. The Hankel transform of this fraction sequence is 1/(2^(n^2)). - Paul Barry, Jun 17 2008

Paul D. Hanna, Table of n, a(n) for n = 0..300
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, example 30.
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.

Cf. A000045, A000108, A046748, A053121, A054335, A098614.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (x+Sqrt(1-4*x^2))/(1-5*x^2) )); // G. C. Greubel, Jul 31 2024

Array[Sum[Binomial[(# - 1)/2, (# - k)/2]*2^(# - k - 1)*((-1)^(# - k) + 1), {k, 0, #}] &, 42, 0] (* or *)
CoefficientList[Series[(Sqrt[1 - 4 x^2] + x)/(1 - 5 x^2), {x, 0, 41}], x] (* Michael De Vlieger, May 20 2021 *)

a(n):=sum(binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*((-1)^(n-k)+1),k,0,n); /* Vladimir Kruchinin, Apr 16 2011 */

{ a(n) = polcoeff((sqrt(1-4*x^2+x^2*O(x^n))+x)/(1-5*x^2),n) }
for(n=0,50,print1(a(n),", "))

def A098615_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (x+sqrt(1-4*x^2))/(1-5*x^2) ).list()
A098615_list(30) # G. C. Greubel, Jul 31 2024

G.f.: (x + sqrt(1-4*x^2)) / (1-5*x^2).
G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x) + 5*x^2*A(x)^2). - Paul D. Hanna, Nov 18 2014
a(2*n) = A046748(n).
a(2*n+1) = 5^n.
a(n) = Sum_{k=0..floor((n+1)/2)} (C(n,k) - C(n,k-1))*Fibonacci(n-2k+1). - Paul Barry, Oct 01 2007
G.f.: 1/(1-x-2x^2/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 09 2009
a(n) = Sum_{k=0..n} binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*(1+(-1)^(n-k)). - Vladimir Kruchinin, Apr 16 2011
From Gary W. Adamson, Sep 22 2011: (Start)
a(n) is the upper left term in M^n, M = an infinite square production matrix as follows:
   1, 1, 1, 0, 0, 0, ...
   1, 0, 0, 1, 0, 0, ...
   1, 0, 0, 0, 1, 0, ...
   0, 1, 0, 0, 0, 1, ...
   0, 0, 1, 0, 0, 0, ...
   0, 0, 0, 1, 0, 0, ...
   0, 0, 0, 0, 1, 0, ...
   0, 0, 0, 0, 0, 1, ...
   ... (End)
a(n) = Sum_{k=0..floor(n/2)} A054335(n-k,n-2k). - Philippe Deléham, Feb 01 2012
a(n) = Sum_{k=0..n} A053121(n,k)*A000045(k+1). - Philippe Deléham, Feb 03 2012
n*a(n) +(n-1)*a(n-1) -3*(3*n-4)*a(n-2) -3*(3*n-7)*a(n-3) +20*(n-3)*a(n-4) +20*(n-4)*a(n-5) = 0. - R. J. Mathar, Jul 21 2017

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

Paul D. Hanna, Oct 09 2004

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.

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

Harvey P. Dale, Table of n, a(n) for n = 0..1000

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

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 *)

a(n) = binomial(2*n,n)/(n+1)*round(((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/(2*sqrt(2)))

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

A200375 Product of Catalan and Jacobsthal numbers: a(n) = A000108(n)*A001045(n+1).

Original entry on oeis.org

1, 1, 6, 25, 154, 882, 5676, 36465, 244530, 1657942, 11471668, 80242890, 568080772, 4056976900, 29212908120, 211783889025, 1544811959970, 11328491394990, 83473572128100, 617702666484750, 4588654943721420, 34206312386929020, 255803818897858920, 1918528298674328250, 14427334095935095764

Offset: 0

Paul D. Hanna, Nov 16 2011

nonn

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.

			G.f.: A(x) = 1 + x + 2*3*x^2 + 5*5*x^3 + 14*11*x^4 + 42*21*x^5 + 132*43*x^6 + 429*85*x^7 + 1430*171*x^8 +...+ A000108(n)*A001045(n)*x^n +...
The g.f. of the Jacobsthal sequence A001045, F(x) = 1/(1-x-2*x^2), begins:
F(x) = 1 + x + 3*x^2 + 5*x^3 + 11*x^4 + 21*x^5 + 43*x^6 + 85*x^7 + 171*x^8 +...
The g.f. of A200376, where G(x) =  A(x/G(x)), begins:
G(x) = 1 + x + 5*x^2 + 9*x^3 + 37*x^4 + 81*x^5 + 301*x^6 + 729*x^7 +...
in which the odd-indexed coefficients are powers of 9.

Michael De Vlieger, Table of n, a(n) for n = 0..1112
Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.
Paul Barry and Arnauld Mesinga Mwafise, Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017).

Cf. A200376, A098614, A098616, A200312.

Array[CatalanNumber[# - 1] (2^# - (-1)^#)/3 &, 25] (* Michael De Vlieger, Apr 24 2018 *)

{a(n) = binomial(2*n, n)/(n+1) * (2^(n+1) + (-1)^n)/3}

{a(n) = polcoef(sqrt((1-2*x - sqrt(1-4*x-32*x^2 +O(x^(n+3))))/2)/(3*x), n)}

{a(n) = polcoef((1/x)*serreverse(x-x^2 - 4*x^3*sum(m=0,n\2,binomial(2*m,m)/(m+1)*3^m*x^(2*m)) +x^3*O(x^n)), n)}

G.f.: sqrt( (1-2*x - sqrt(1-4*x-32*x^2))/2 )/(3*x).
G.f.: (1/x)*Series_Reversion(x-x^2 - 4*x^3*Sum_{n>=0} A000108(n)*3^n*x^(2*n) ).
G.f. satisfies: A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where G(x) is the g.f. of A200376: G(x) = 1/sqrt(1-10*x^2 + x^4/(1-8*x^2)) + x/(1-9*x^2).
n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -8*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2011
a(n) = binomial(2*n,n)/(n+1) * (2^(n+1) + (-1)^n)/3.
From Peter Bala, Aug 17 2021: (Start)
G.f.: A(x) = (sqrt(1 + 4*x) - sqrt(1 - 8*x))/(6*x).
A(x) = 1/sqrt(1 + 4*x)*c( 3*x/(1 + 4*x) ), where c(x) = (1 - sqrt(1- 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Cf. A151374.
In general, [x^n] ( 1/sqrt(1 + 4*x)*c( k*x/(1 + 4*x) ) ) = Catalan(n)*((k-1)^(n+1) + (-1)^(n+1))/k.
A(x) = 1/sqrt(1 - 8*x)*c( -3*x/(1 - 8*x) ). (End)
G.f. A(x) satisfies A(x) = sqrt( 1 + 2*x*A(x)^2 + 9*x^2*A(x)^4 ). - Paul D. Hanna, Dec 14 2024

Typo in Name corrected by Peter Bala, Aug 17 2021

A098618 Products of A007482 and Catalan numbers: a(n) = A007482(n)*A000108(n).

Original entry on oeis.org

1, 3, 22, 195, 1946, 20790, 232716, 2693691, 31979090, 387243714, 4764470932, 59391201870, 748472730628, 9520446996300, 122067269204760, 1575965219205195, 20470515781159170, 267325017886787850

Offset: 0

Paul D. Hanna, Oct 09 2004

nonn

Radius of convergence: r = (sqrt(17)-3)/16; A(r) = sqrt(2+6/sqrt(17)). Recurrence of A007482 is A007482(n) = 3*A007482(n-1) + 2*A007482(n-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.

			Begins: {1*1, 3*1, 11*2, 39*5, 139*14, 495*42, 1763*132, 6279*429,...}.

Cf. A007482, A000108, A098614, A098616, A098619.

{a(n)=binomial(2*n,n)/(n+1)*((3+sqrt(17))^(n+1)-(3-sqrt(17))^(n+1))/2^(n+1)/sqrt(17)}

G.f.: A(x) = sqrt((1-6*x - sqrt(1-12*x-32*x^2))/34 )/x.

n*(n+1)*a(n) -6*n*(2*n-1)*a(n-1) -8*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2018

A200312 a(n) = A000108(n)A006130(n), where A000108 is the Catalan numbers and A006130(n) = A006130(n-1) + 3A006130(n-2).

Original entry on oeis.org

1, 1, 8, 35, 266, 1680, 12804, 93093, 726440, 5635058, 45063668, 362121760, 2955642508, 24284658100, 201428123040, 1680921310635, 14119413718770, 119205791509200, 1011387051005100, 8617021562542470, 73704123363739440, 632601537174078420

Offset: 0

Paul D. Hanna, Nov 16 2011

nonn

More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), S(0)=1, |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.

			G.f.: A(x) = 1 + x + 2*4*x^2 + 5*7*x^3 + 14*19*x^4 + 42*40*x^5 + 132*97*x^6 + 429*217*x^7 + ... + A000108(n)*A006130(n)*x^n + ...
where the g.f. of A006130, F(x) = 1/(1-x-3*x^2), begins:
F(x) = 1 + x + 4*x^2 + 7*x^3 + 19*x^4 + 40*x^5 + 97*x^6 + 217*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..450
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017)

Cf. A098614, A098616, A200375.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!(Sqrt((1-2*x - Sqrt(1-4*x-48*x^2))/26)/x)); // G. C. Greubel, Jul 27 2018

CoefficientList[Series[Sqrt[(1 - 2*x - Sqrt[1 - 4*x - 48*x^2])/26]/x, {x, 0, 30}], x] (* G. C. Greubel, Jul 27 2018 *)

{a(n)=binomial(2*n, n)/(n+1)*polcoeff(1/(1-x-3*x^2+x*O(x^n)),n)}

{a(n)=polcoeff(sqrt((1-2*x - sqrt(1-4*x-48*x^2+x^3*O(x^n)))/26)/x,n)}

{a(n)=polcoeff(serreverse(x*sqrt(1-12*x^2+x^2*O(x^n)) - x^2)/x,n)}

{a(n)=polcoeff((1/x)*serreverse(x-x^2 - 6*x^3*sum(m=0,n\2,binomial(2*m,m)/(m+1)*3^m*x^(2*m))+x^3*O(x^n)),n)}

G.f.: sqrt( (1-2*x - sqrt(1-4*x-48*x^2))/26 )/x.
G.f.: (1/x)*Series_Reversion( x*sqrt(1-12*x^2) - x^2 ).
G.f.: (1/x)*Series_Reversion( x-x^2 - 6*x^3*Sum_{n>=0} A000108(n)*3^n*x^(2*n) ).
G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x)^2 + 13*x^2*A(x)^4).
Conjecture: n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -12*(2*n-1)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 17 2011
a(n) = ( ((1+sqrt(13))/2)^(n+1) - ((1-sqrt(13))/2)^(n+1) )/sqrt(13) * binomial(2*n+1,n)/(2*n+1). - Paul D. Hanna, Sep 25 2012
0 = +a(n)*(+110592*a(n+3) -9216*a(n+4) -7392*a(n+5) +858*a(n+6)) +a(n+1)*(+6912*a(n+3) -1968*a(n+4) -910*a(n+5) +154*a(n+6)) +a(n+2)*(-240*a(n+3) -2*a(n+4) +41*a(n+5) -4*a(n+6)) +a(n+3)*(+6*a(n+3) +5*a(n+4) +3*a(n+5) -a(n+6)) for all n in Z. - Michael Somos, Jul 28 2018

A200539 Product of Fibonacci and Motzkin numbers: a(n) = A000045(n+1)*A001006(n).

Original entry on oeis.org

1, 1, 4, 12, 45, 168, 663, 2667, 10982, 45925, 194732, 834912, 3614063, 15771795, 69316740, 306534564, 1362986799, 6089916936, 27328613142, 123118156260, 556626199974, 2524659817449, 11484671681511, 52384730922720, 239534402969925, 1097805759803893, 5042014405418968

Offset: 0

Paul D. Hanna, Nov 18 2011

nonn

The g.f. for the Fibonacci numbers is 1/(1-x-x^2) and the g.f. M(x) for the Motzkin numbers satisfies: M(x) = 1 + x*M(x) + x^2*M(x)^2.

			G.f.: A(x) = 1 + x + 4*x^2 + 12*x^3 + 45*x^4 + 168*x^5 + 663*x^6 +...
where A(x) = 1*1 + 1*1*x + 2*2*x^2 + 3*4*x^3 + 5*9*x^4 + 8*21*x^5 + 13*51*x^6 + 21*127*x^7 + 34*323*x^8 +...+ A000045(n+1)*A001006(n)*x^n +...

Cf. A098614, A200538, A200540, A098616.

{A001006(n)=polcoeff((1-x-sqrt((1-x)^2-4*x^2+x^3*O(x^n)))/(2*x^2),n)}
{a(n)=fibonacci(n+1)*A001006(n)}

A200540 Product of Pell and Motzkin numbers: a(n) = A000129(n+1)*A001006(n).

Original entry on oeis.org

1, 2, 10, 48, 261, 1470, 8619, 51816, 318155, 1985630, 12561308, 80360280, 519013571, 3379514970, 22161470850, 146227235904, 970126550763, 6467496499590, 43304243215638, 291087137589552, 1963598081845335, 13288619577124122, 90195242361688863, 613843707553183800

Offset: 0

Paul D. Hanna, Nov 18 2011

nonn

The g.f. for the Pell numbers is 1/(1-2*x-x^2) and the g.f. M(x) for the Motzkin numbers satisfy: M(x) = 1 + x*M(x) + x^2*M(x)^2.

			G.f.: A(x) = 1 + 2*x + 10*x^2 + 48*x^3 + 261*x^4 + 1470*x^5 + 8619*x^6 +...
where A(x) = 1*1 + 2*1*x + 5*2*x^2 + 12*4*x^3 + 29*9*x^4 + 70*21*x^5 + 169*51*x^6 + 408*127*x^7 + 985*323*x^8 +...+ A000129(n+1)*A001006(n)*x^n +...

Cf. A098616, A200538, A200539, A098614.

{A001006(n)=polcoeff((1-x-sqrt((1-x)^2-4*x^2+x^3*O(x^n)))/(2*x^2),n)}
{A000129(n)=polcoeff( x/(1-2*x-x^2+x*O(x^n)),n)}
{a(n)=A000129(n+1)*A001006(n)}

A215931 Product of Fibonacci and Catalan numbers: a(n) = A000045(2n+2)A000108(n).

Original entry on oeis.org

1, 3, 16, 105, 770, 6048, 49764, 423423, 3695120, 32891430, 297473956, 2725789248, 25251200716, 236101791900, 2225241057600, 21118368117105, 201640796593290, 1935642349666080, 18670022226540300, 180851385211254450, 1758621701183524320, 17160853351737885660

Offset: 0

Paul D. Hanna, Aug 27 2012

nonn

More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2) with S(0)=1, |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.

			G.f.: A(x) = 1 + 3*x + 16*x^2 + 105*x^3 + 770*x^4 + 6048*x^5 + 49764*x^6 +...
such that the coefficients equal the term-wise products:
A = [1*1, 3*1, 8*2, 21*5, 55*14, 144*42, 377*132, 987*429, 2584*1430, ...].
Related expansions.
A(x)^2 = 1 + 6*x + 41*x^2 + 306*x^3 + 2426*x^4 + 20076*x^5 + 171481*x^6 +...
A(x)^3 = 1 + 9*x + 75*x^2 + 630*x^3 + 5400*x^4 + 47223*x^5 + 420277*x^6 +...
Incidentally, note that (2*n+1) divides [x^n] A(x)^3:
A^3 = [1*1, 3*3, 5*15, 7*90, 9*600, 11*4293, 13*32329, 15*253110, ...].

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A098614, A098616, A001906, A000045, A000032.

[Fibonacci(2*n+2)*Binomial(2*n, n)/(n+1): n in [0..22]] // Vincenzo Librandi, Aug 28 2012

Table[Fibonacci[2*n+2]*Binomial[2*n,n]/(n+1), {n,0,25}] (* Vincenzo Librandi, Aug 28 2012 *)

{a(n)=fibonacci(2*n+2)*binomial(2*n,n)/(n+1)}

{a(n)=fibonacci(n+1)*(2*fibonacci(n)+fibonacci(n+1))*binomial(2*n,n)/(n+1)}

{a(n)=polcoeff( sqrt( (1-6*x - sqrt(1-12*x+16*x^2 +x^3*O(x^n)))/10 )/x,n)}
for(n=0,21,print1(a(n),", "))

G.f.: sqrt( (1-6*x - sqrt(1-12*x+16*x^2))/10 )/x.
a(n) = Fibonacci(2*n+2) * binomial(2*n,n)/(n+1).
a(n) = Fibonacci(n+1) * Lucas(n+1) * binomial(2*n,n)/(n+1), where Lucas(n+1) = 2*Fibonacci(n) + Fibonacci(n+1) = A000032(n+1).
a(n) = A000032(n+1) * A098614(n).
n*(n+1)*a(n) -6*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 = 4*sqrt(1-2/sqrt(5)). - Amiram Eldar, May 06 2023

A249925 G.f. satisfies: A(x) = 1 + 2xA(x) + 5x^2A(x)^2.

Original entry on oeis.org

1, 2, 9, 38, 186, 932, 4889, 26238, 143966, 802652, 4536874, 25932348, 149650516, 870675912, 5101656889, 30078478318, 178309845686, 1062198928812, 6355149937934, 38172142221748, 230094601968876, 1391444403490552, 8439240940653834, 51323083138005388, 312896262064813036, 1911980839096481432

Offset: 0

Paul D. Hanna, Nov 22 2014

nonn

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 38*x^3 + 186*x^4 + 932*x^5 + 4889*x^6 +...
where the square-root of the g.f. yields
sqrt(A(x)) = 1 + x + 4*x^2 + 15*x^3 + 70*x^4 + 336*x^5 + 1716*x^6 + 9009*x^7 + 48620*x^8 +...+ Fibonacci(n+1)*A000108(n)*x^n + +...
Related expansions.
A(x)^2 = 1 + 4*x + 22*x^2 + 112*x^3 + 605*x^4 + 3292*x^5 + 18298*x^6 +...
which obeys A(x) = 1 + 2*x*A(x) + 5*x^2*A(x)^2.
Given series bisections A(x) = B0(x^2) + x*B1(x^2),
B0(x) = 1 + 9*x + 186*x^2 + 4889*x^3 + 143966*x^4 + 4536874*x^5 +...
B1(x) = 2 + 38*x + 932*x^2 + 26238*x^3 + 802652*x^4 + 25932348*x^5 +...
then B1(x)/B0(x) = 2 + 10*x*B1(x):
B1(x)/B0(x) = 2 + 20*x + 380*x^2 + 9320*x^3 + 262380*x^4 + 8026520*x^5 +...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.

Cf. A098614, A000045, A000108.

CoefficientList[Series[(1-2*x-Sqrt[1-4*x-16*x^2]) / (10*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Nov 29 2014 *)

{a(n)=local(X=x+O(x^(n+3)),A); A = (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

{a(n) = sum(k=0,n,fibonacci(n-k+1)*fibonacci(k+1)*binomial(2*(n-k),n-k)*binomial(2*k,k)/((n-k+1)*(k+1)))}
for(n=0,30,print1(a(n),", "))

G.f.: (1-2*x - sqrt(1-4*x-16*x^2)) / (10*x^2).
Self-convolution square of A098614, where A098614(n) = A000045(n+1)*A000108(n), the term-wise product of Fibonacci and Catalan numbers.
a(n) = Sum_{k=0..n} A000045(k+1)*A000045(n-k+1) * A000108(k)*A000108(n-k).
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*Fibonacci(k+1) * C(2*(n-k),n-k)*C(2*k,k) / ((n-k+1)*(k+1)).
a(n) == 1 (mod 2) iff n = 2*(2^k - 1) for k>=0.
Given series bisections B0 and B1 such that A(x) = B0(x^2) + x*B1(x^2), then B1(x)/B0(x) = 2 + 10*x*B1(x), thus B1(x) = 2*B0(x)/(1 - 10*x*B0(x)).
a(n) ~ sqrt(5+2*sqrt(5)) * 2^(n+2) * (1+sqrt(5))^n / (5 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 29 2014. Equivalently, a(n) ~ 5^(1/4) * 2^(2*n+2) * phi^(n + 3/2) / (5 * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021
Recurrence: (n+2)*a(n) = 2*(2*n+1)*a(n-1) + 16*(n-1)*a(n-2). - Vaclav Kotesovec, Nov 29 2014

A216541 Product of Lucas and Catalan numbers: a(n) = A000032(n+1)*A000108(n).

Original entry on oeis.org

1, 3, 8, 35, 154, 756, 3828, 20163, 108680, 598026, 3342404, 18929092, 108374252, 626264700, 3647936160, 21396522915, 126262239570, 749087596620, 4465444206300, 26733390275130, 160663411399920, 968937572793060, 5862111195487560, 35569106862459300, 216395609659221564

Offset: 0

Paul D. Hanna, Sep 08 2012

nonn

			G.f.: A(x) = 1 + 3*x + 8*x^2 + 35*x^3 + 154*x^4 + 756*x^5 + 3828*x^6 +...
such that the coefficients equal the term-wise products:
A = [1*1, 3*1, 4*2, 7*5, 11*14, 18*42, 29*132, 47*429, 76*1430, ...].

Cf. A098614, A215931, A098616, A000032, A000108.

a[n_] := LucasL[n+1] * CatalanNumber[n]; Array[a, 25, 0] (* Amiram Eldar, May 05 2023 *)

{a(n)=(2*fibonacci(n)+fibonacci(n+1))*binomial(2*n,n)/(n+1)}

{a(n)=polcoeff( (1 - sqrt( (1-2*x + sqrt(1-4*x-16*x^2 +x^2*O(x^n)))/2 )) / x,n)}
for(n=0,25,print1(a(n),", "))

G.f.: (1 - sqrt( (1-2*x + sqrt(1-4*x-16*x^2))/2 )) / x.
G.f. satisfies: A(x) = (2+5*x) - (1+4*x)*A(x) + x*(5+2*x)*A(x)^2 - 4*x^2*A(x)^3 + x^3*A(x)^4.
n*(n+1)*a(n) -2*n*(2n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Sep 11 2012
Sum_{n>=0} a(n)/8^n = 8 - 2*sqrt(10). - Amiram Eldar, May 05 2023

cp's OEIS Frontend

A098615 G.f. A(x) satisfies: A(x*G(x)) = G(x), where G(x) is the g.f. for A098614(n) = Fibonacci(n+1)*Catalan(n).

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

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A200375 Product of Catalan and Jacobsthal numbers: a(n) = A000108(n)*A001045(n+1).

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A098618 Products of A007482 and Catalan numbers: a(n) = A007482(n)*A000108(n).

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A200312 a(n) = A000108(n)*A006130(n), where A000108 is the Catalan numbers and A006130(n) = A006130(n-1) + 3*A006130(n-2).

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A200539 Product of Fibonacci and Motzkin numbers: a(n) = A000045(n+1)*A001006(n).

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A200540 Product of Pell and Motzkin numbers: a(n) = A000129(n+1)*A001006(n).

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A098615 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098614(n) = Fibonacci(n+1)Catalan(n).

A200312 a(n) = A000108(n)A006130(n), where A000108 is the Catalan numbers and A006130(n) = A006130(n-1) + 3A006130(n-2).

A215931 Product of Fibonacci and Catalan numbers: a(n) = A000045(2n+2)A000108(n).

A249925 G.f. satisfies: A(x) = 1 + 2xA(x) + 5x^2A(x)^2.