A082367 - OEIS

%I #43 Sep 08 2022 08:45:10

%S 1,9,90,981,11430,140058,1782900,23369805,313426350,4281280686,

%T 59360821740,833312907522,11820849447420,169182862497108,

%U 2440064033240040,35428651752626109,517446157031236350

%N G.f.: (1-8*x-sqrt(64*x^2-20*x+1))/(2*x).

%C More generally coefficients of (1-m*x-sqrt(m^2*x^2-(2*m+4)*x+1))/(2*x) are given by a(0)=1 and n>0 a(n)=(1/n)*Sum_{k=0..n} (m+1)^k*C(n,k)*C(n,k-1).

%C Hankel transform is 9^C(n+1,2). - _Philippe Deléham_, Feb 11 2009

%H Vincenzo Librandi, <a href="/A082367/b082367.txt">Table of n, a(n) for n = 0..200</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html">On Integer-Sequence-Based Constructions of Generalized Pascal Triangles</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

%F a(0)=1; for n > 0, a(n) = (1/n)*Sum_{k=0..n} 9^k*C(n, k)*C(n, k-1).

%F D-finite with recurrence: (n+1)*a(n) + 10*(1-2n)*a(n-1) + 64*(n-2)*a(n-2) = 0. - _R. J. Mathar_, Dec 08 2011 Recurrence follows from the D.E. (x-20*x^2+64*x^3)*y' + (1-10*x)*y - 1 - 8*x = 0 satisfied by the g.f.. - _Robert Israel_, Mar 16 2018

%F a(n) ~ sqrt(3)*2^(4*n+1)/(sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 14 2012

%F G.f.: 1/(1 - 8*x - x/(1 - 8*x - x/(1 - 8*x - x/(1 - 8*x - x/(1 - ...))))), a continued fraction. - _Ilya Gutkovskiy_, Apr 04 2018

%p f:= gfun:-rectoproc({64*n*a(n)+(-30-20*n)*a(1+n)+(3+n)*a(n+2), a(0) = 1, a(1) = 9}, a(n), remember):

%p map(f, [$0..30]); # _Robert Israel_, Mar 16 2018

%t Table[SeriesCoefficient[(1-8*x-Sqrt[64*x^2-20*x+1])/(2*x),{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)

%o (PARI) a(n)=if(n<1,1,sum(k=0,n,9^k*binomial(n,k)*binomial(n,k-1))/n)

%o (PARI) x='x+O('x^99); Vec((1-8*x-(64*x^2-20*x+1)^(1/2))/(2*x)) \\ _Altug Alkan_, Apr 04 2018

%o (Magma) m:=25; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-8*x-Sqrt(64*x^2-20*x+1))/(2*x))); // _G. C. Greubel_, Sep 16 2018

%Y Cf. A006318, A047891.

%K nonn

%O 0,2

%A _Benoit Cloitre_, May 10 2003