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Dot product of the rows of triangle A046899 with vector (1,2,4,8,...) (= A000079).
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%I #61 Jan 21 2023 01:38:04

%S 1,5,31,209,1471,10625,78079,580865,4361215,32978945,250806271,

%T 1916280833,14698053631,113104519169,872801042431,6751535300609,

%U 52337071357951,406468580343809,3162019821780991,24634626678980609,192179216026959871

%N Dot product of the rows of triangle A046899 with vector (1,2,4,8,...) (= A000079).

%C Hankel transform is A133460.

%H Vincenzo Librandi, <a href="/A178792/b178792.txt">Table of n, a(n) for n = 0..200</a>

%H J. Abate and W. Whitt, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Whitt/whitt6.html">Brownian motion and generalized Catalan numbers</a>, Journal of Integer Sequences, Vol. 14 (2011).

%H Karl Dilcher and Maciej Ulas, <a href="https://arxiv.org/abs/1909.11222">Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1) = 1</a>, arXiv:1909.11222 [math.NT], 2019.

%H Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, <a href="https://arxiv.org/abs/2301.02628">Pinnacle sets of signed permutations</a>, arXiv:2301.02628 [math.CO], 2023.

%F a(n) = Sum_{k = 0..n} A046899(n,k)*2^k = Sum_{k = 0..n} 2^k*binomial(n+k,k).

%F G.f.: (1/3)*(4/sqrt(1 - 8*x) - 1/(1 - x*c(2*x))) with c(x) the g.f. of the Catalan numbers A000108.

%F a(n) = (1/3)*(4*2^n*A000984(n) - A064062(n)).

%F a(n) + a(n+1) = 6*2^n*A001700(n).

%F O.g.f.: (3 - sqrt(1 - 8*x))/(2*(1 + x)*sqrt(1 - 8*x)). - _Peter Bala_, Apr 10 2012

%F a(n) = 2^n *binomial(2+2*n,1+n)*2F1(1, 2+2*n; 2+n;-1). - _Olivier Gérard_, Aug 19 2012

%F D-finite with recurrence n*a(n) = (7*n - 4)*a(n-1) + 4*(2*n - 1)*a(n-2). - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) ~ 2^(3n+2)/(3*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) = Sum_{k = 0..n} binomial(k+n,n)*binomial(2*n+1,n-k). - _Vladimir Kruchinin_, Oct 28 2016

%F a(n) = 1/2*(n + 1)*binomial(2*n+2,n+1)*Sum_{k = 0..n} binomial(n,k)/(n + k + 1). - _Peter Bala_, Feb 21 2017

%F a(n) = binomial(2*n+2,n+1)*hypergeom([-n, n+1], [n+2], -1)/2. - _Peter Luschny_, Feb 21 2017

%F a(n) = (-1)^(n+1) - 2^(n+1)*(2*n+1)*binomial(2*n,n)*hypergeom([1, 2*n+2], [n+2], 2)/(n+1). - _John M. Campbell_, Jul 14 2018

%e a(3) = (1,4,10,20)dot(1,2,4,8) = 209.

%p a := n -> binomial(2*n+2,n+1)*hypergeom([-n, n + 1], [n + 2], -1)/2:

%p seq(simplify(a(n)), n=0..20); # _Peter Luschny_, Feb 21 2017

%t CoefficientList[Series[(3-Sqrt[1-8*x])/(2*(1+x)*Sqrt[1-8*x]), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 20 2012 *)

%t Table[Sum[2^k*Binomial[n + k, k], {k, 0, n}], {n, 0, 20}] (* _Michael De Vlieger_, Oct 28 2016 *)

%t a[n_] := (-1)^(n + 1) - 2^(n + 1) (2n + 1) Binomial[2n, n] Hypergeometric2F1[1, 2n + 2, n + 2, 2]/(n + 1); Array[a, 22, 0] (* _Robert G. Wilson v_, Jul 21 2018 *)

%Y Row sums of A091811.

%K nonn,easy

%O 0,2

%A _Joseph Abate_, Jun 15 2010