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%I #18 Dec 07 2021 03:59:35
%S 0,1,6,41,306,2426,20076,171481,1500666,13386206,121267476,1112674026,
%T 10318939956,96572168916,910896992856,8650566601401,82644968321226,
%U 793753763514806,7659535707782916,74225795172589006,722042370787826076
%N Series reversion of x/(1+6x+5x^2).
%C Hankel transform is -A127849(n)=-5^C(n,2)*(5^n-1)/4; a(n+1) counts (6,5)-Motzkin paths of length n, where there are 6 colors available for the H(1,0) steps and 5 for the U(1,1) steps. See A078009 for more information.
%H Vincenzo Librandi, <a href="/A127848/b127848.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: (1-6x-sqrt(1-12x+16x^2))/(10x); a(n)=sum{k=0..n-1, (1/n)*C(n,k)C(n,k+1)5^k}; a(n+1)=sum{k=0..floor(n/2), C(n, 2k)C(k)6^(n-2k)*5^k};
%F Recurrence: (n+1)*a(n) = 6*(2*n-1)*a(n-1) - 16*(n-2)*a(n-2). - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) ~ sqrt(10+6*sqrt(5))*(6+2*sqrt(5))^n/(10*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012. Equivalently, a(n) ~ 2^(2*n) * phi^(2*n + 1) / (5^(3/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Dec 07 2021
%F a(n) = A078009(n) for n>0. - _Philippe Deléham_, Apr 03 2013
%t CoefficientList[ InverseSeries[ Series[ x/(1+6x+5x^2), {x, 0, 20}], x], x] (* _Jean-François Alcover_, May 24 2012 *)
%Y Cf. A078009.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Feb 02 2007
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