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%I #45 Feb 06 2022 06:47:10
%S 1,1,3,5,13,25,61,125,295,625,1447,3125,7151,15625,35491,78125,176597,
%T 390625,880125,1953125,4390901,9765625,21920913,48828125,109486993,
%U 244140625,547018941,1220703125,2733608905,6103515625,13662695645,30517578125,68294088535,152587890625,341399727335,762939453125,1706739347095,3814697265625,8532741458075,19073486328125,42660172763995,95367431640625
%N 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).
%C 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, ...}.
%C 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
%C 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
%H Paul D. Hanna, <a href="/A098615/b098615.txt">Table of n, a(n) for n = 0..300</a>
%H Paul Barry and A. Hennessy, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barry5/barry96s.html">Meixner-Type Results for Riordan Arrays and Associated Integer Sequences</a>, J. Int. Seq. 13 (2010) # 10.9.4, example 30.
%H Cyril Banderier, Markus Kuba, and Michael Wallner, <a href="https://arxiv.org/abs/2103.03751">Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions</a>, arXiv:2103.03751 [math.PR], 2021.
%F G.f.: (sqrt(1-4*x^2) + x) / (1-5*x^2).
%F G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x) + 5*x^2*A(x)^2). - _Paul D. Hanna_, Nov 18 2014
%F a(2*n) = A046748(n);
%F a(2*n+1) = 5^n.
%F a(n) = Sum_{k=0..floor((n+1)/2)} (C(n,k) - C(n,k-1))*F(n-2k+1). - _Paul Barry_, Oct 01 2007
%F 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
%F a(n) = Sum_{k=0..n} binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*((-1)^(n-k)+1). - _Vladimir Kruchinin_, Apr 16 2011
%F From _Gary W. Adamson_, Sep 22 2011: (Start)
%F a(n) is the upper left term in M^n, M = an infinite square production matrix as follows:
%F 1, 1, 1, 0, 0, 0, ...
%F 1, 0, 0, 1, 0, 0, ...
%F 1, 0, 0, 0, 1, 0, ...
%F 0, 1, 0, 0, 0, 1, ...
%F 0, 0, 1, 0, 0, 0, ...
%F 0, 0, 0, 1, 0, 0, ...
%F 0, 0, 0, 0, 1, 0, ...
%F 0, 0, 0, 0, 0, 1, ...
%F ... (End)
%F a(n) = Sum_{k=0..floor(n/2)} A054335(n-k,n-2k). - _Philippe Deléham_, Feb 01 2012
%F a(n) = Sum_{k=0..n} A053121(n,k)*A000045(k+1). - _Philippe Deléham_, Feb 03 2012
%F 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
%t Array[Sum[Binomial[(# - 1)/2, (# - k)/2]*2^(# - k - 1)*((-1)^(# - k) + 1), {k, 0, #}] &, 42, 0] (* or *)
%t CoefficientList[Series[(Sqrt[1 - 4 x^2] + x)/(1 - 5 x^2), {x, 0, 41}], x] (* _Michael De Vlieger_, May 20 2021 *)
%o (PARI) { a(n) = polcoeff((sqrt(1-4*x^2+x^2*O(x^n))+x)/(1-5*x^2),n) }
%o for(n=0,50,print1(a(n),", "))
%o (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 */
%Y Cf. A098614, A046748.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 14 2004
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