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 A210736 Expansion of (1 + sqrt( (1 + 2*x) / (1 - 2*x))) / 2 in powers of x. 7

%I

%S 1,1,1,2,3,6,10,20,35,70,126,252,462,924,1716,3432,6435,12870,24310,

%T 48620,92378,184756,352716,705432,1352078,2704156,5200300,10400600,

%U 20058300,40116600,77558760,155117520,300540195,601080390,1166803110,2333606220,4537567650

%N Expansion of (1 + sqrt( (1 + 2*x) / (1 - 2*x))) / 2 in powers of x.

%C Hankel transform is period 4 sequence [ 1, 0, -1, 0, ...] A056594 and the Hankel transform of sequence omitting a(0) is the all 1s sequence A000012. This is the unique sequence with that property.

%C Series reversion of x*A(x) apparently yields x*A036765(-x). - _R. J. Mathar_, Sep 24 2012

%C a(n) is the number of length n words on {-1,1} such that the sum of any of its prefixes is always positive. Cf. A001405 where the sum of all prefixes is nonnegative. - _Geoffrey Critzer_, Jul 08 2013

%H Vincenzo Librandi, <a href="/A210736/b210736.txt">Table of n, a(n) for n = 0..200</a>

%H P. Flajolet and R. Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/books.html">Analytic Combinatorics</a>, 2009; page 77.

%F G.f.: 2 * x / (-1 + 2*x + sqrt(1 - 4*x^2)).

%F G.f. A(x) satisfies A(x) = A(x)^2 - x / (1 - 2*x).

%F G.f. A(x) satisfies A( x / (1 + x^2) ) = 1 / (1 - x).

%F G.f. A(x) satisfies A(1/3) = (1 + sqrt(5))/2.

%F G.f. A(x) = 1 + x / (1 - 2*x + x / A(x)).

%F G.f. A(x) = 1 + x / (1 - x / (1 - x / (1 + x / A(x)))).

%F G.f. A(x) = 1 + x * A001405(x). a(n+1) = A001405(n).

%F Convolution inverse is A210628. Partial sums is A072100.

%F Binomial transform with offset 1 is A211278 with offset 1. a(n+2) * a(n) - a(n+1)^2 = A138350(n-1).

%F a(n) = (-1)^floor(n/2)*hypergeom2F1([1-n, -n],[1],-1). - _Peter Luschny_, Sep 01 2012

%F Conjecture: n*a(n) -2*a(n-1) +4*(2-n)*a(n-2)=0. - _R. J. Mathar_, Sep 14 2012

%F G.f. A(x) = 1 / (1 - x / (1 - x^2 / (1 - x^2 / (1 - x^2 / ...)))). - _Michael Somos_, Jan 02 2013

%F G.f.: 1/(1 - x*C(x)) where C(x) is the o.g.f. for A126120. - _Geoffrey Critzer_, Jul 08 2013

%F a(n) ~ 2^(n-1/2) / sqrt(Pi*n). - _Vaclav Kotesovec_, Feb 01 2014

%F G.f.: A(x) = 1 - x/(- 1 + x/A(-x)). - _Arkadiusz Wesolowski_, Feb 28 2014

%F From _Tom Copeland_, Nov 07 2014: (Start)

%F Setting a(0)=0 here, we have a signed version in A126930 and

%F O.g.f. G(x)=[-1+sqrt(1+4*x/(1-2x))]/2 = x + x^2 + 2 x^3 + ... = -C[-P(P(x,-1),-1)]= -C[-P(x,-2)] where C(x)= [1-sqrt(1-4*x)]/2= x + x^2 + 2 x^3 + ... = A000108(x) with inverse Cinv(x)=x*(1-x), and P(x,t)= x/(1 + t*x) with inverse P(x,-t).

%F These types of arrays are from linear fractional transformations of C(x). See A091867.

%F Ginv(x) = P[-Cinv(-x),2] = x*(1+x)/(1+2*x*(1-x))= (x+x^2)/(1+2(x+x^2)) (see A146559). (End)

%e G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 10*x^6 + 20*x^7 + 35*x^8 + 70*x^9 + ...

%t nn=36; d=(1-(1-4x^2)^(1/2))/(2x^2);CoefficientList[Series[1/(1-x d),{x,0,nn}],x] (* _Geoffrey Critzer_, Jul 08 2013 *)

%t CoefficientList[Series[2 x / (-1 + 2 x + Sqrt[1 - 4 x^2]), {x, 0, 40}], x] (* _Vincenzo Librandi_, Nov 10 2014 *)

%o (PARI) {a(n) = if( n<1, n==0, binomial( n - 1, (n - 1)\2))};

%o (PARI) {a(n) = polcoeff( (1 + sqrt( (1 + 2*x) / (1 - 2*x) + x * O(x^n))) / 2, n)};

%Y Cf. A001405, A056594, A138350, A210628, A211278.

%Y Cf. A000108, A091867, A146559, A126930.

%K nonn

%O 0,4

%A _Michael Somos_, May 10 2012

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Last modified February 16 09:04 EST 2019. Contains 320159 sequences. (Running on oeis4.)