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).

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

Author

Paul D. Hanna, Oct 14 2004

Keywords

• nonn

Comments

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

Links

• 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.

Crossrefs

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

Programs

Magma

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

Mathematica

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 *)

Maxima

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 */

PARI

{ 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),", "))

SageMath

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

Formula

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