A158500 Expansion of (1+sqrt(1+4x))*(1+2x)/(2*sqrt(1+4x)).

1, 1, 1, -4, 15, -56, 210, -792, 3003, -11440, 43758, -167960, 646646, -2496144, 9657700, -37442160, 145422675, -565722720, 2203961430, -8597496600, 33578000610, -131282408400, 513791607420, -2012616400080

Offset

0,4

Comments

Hankel transform is A158501. Row sums of the Riordan array

((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.

With the proviso that the negative signs be ignored,

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

Links

Michael De Vlieger, Table of n, a(n) for n = 0..1665

Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.

Formula

a(n)=C(1,n)+(-1)^n*C(2n-2,n-2).

n*(n-2)*a(n) +2*(n-1)*(2*n-3)*a(n-1)=0. - R. J. Mathar, Oct 25 2012

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

Mathematica

{1}~Join~Array[Binomial[1, #] + (-1)^#*Binomial[2 # - 2, # - 2] &,
  24] (* Michael De Vlieger, Jul 23 2020 *)

Crossrefs

Cf. A001791.

Sequence in context: A010905 A026030 A047038 * A001791 A047128 A087438

Adjacent sequences: A158497 A158498 A158499 * A158501 A158502 A158503

Keyword

easy,sign

Author

Paul Barry, Mar 20 2009

Status

approved