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%I #18 Jul 23 2020 19:36:40
%S 1,1,1,-4,15,-56,210,-792,3003,-11440,43758,-167960,646646,-2496144,
%T 9657700,-37442160,145422675,-565722720,2203961430,-8597496600,
%U 33578000610,-131282408400,513791607420,-2012616400080
%N Expansion of (1+sqrt(1+4x))*(1+2x)/(2*sqrt(1+4x)).
%C Hankel transform is A158501. Row sums of the Riordan array
%C ((1+2x)/sqrt(1+4x), xc(-x^2))=((1-x^2)/(1+x^2),x/(1-x)^2)^{-1}, where c(x) is the g.f. of A000108.
%C With the proviso that the negative signs be ignored,
%C a(n)=the sum of the consecutive pairwise products of the terms in row(n) of Pascal's triangle. For example, the seventh row for row(6) has the terms 1,6,15,20,15,6,1 giving a sum of 2*(1*6+6*15+15*20)=792=a(6). For row(10) the terms are 1,9,36,84,126,126,84,36,9,1 giving 2*(1*9+9*36+36*84+84*126)+126*126=43758=a(10). - _J. M. Bergot_, Jul 26 2012
%H Michael De Vlieger, <a href="/A158500/b158500.txt">Table of n, a(n) for n = 0..1665</a>
%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.
%F a(n)=C(1,n)+(-1)^n*C(2n-2,n-2).
%F n*(n-2)*a(n) +2*(n-1)*(2*n-3)*a(n-1)=0. - _R. J. Mathar_, Oct 25 2012
%F E.g.f.: 1 + 2*x - x*Q(0), where Q(k)= 1 + 2*x/(k+2 - (k+2)*(2*k+3)/(2*k+3 - (k+2)/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Apr 28 2013
%t {1}~Join~Array[Binomial[1, #] + (-1)^#*Binomial[2 # - 2, # - 2] &,
%t 24] (* _Michael De Vlieger_, Jul 23 2020 *)
%Y Cf. A001791.
%K easy,sign
%O 0,4
%A _Paul Barry_, Mar 20 2009
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