%I #30 Sep 20 2024 02:10:21
%S 1,6,42,300,2154,15492,111492,802584,5778090,41600532,299517996,
%T 2156509416,15526797252,111792690600,804906480840,5795323452720,
%U 41726317225770,300429441596340,2163091823919900,15574260559056840,112134673904493420,807369644235408120
%N G.f. 1/(1-6*x*c(x)) where c(x) is the g.f. of A000108.
%C Image of 6^n under the Catalan transform g(x)->g(xc(x)). The Hankel transform of this sequence and of the aerated version with g.f. 1/(1-6*x^2*c(x^2)) is 6^n. In general, the expansions of 1/(1-k*x*c(x)) and 1/(1-k*x^2*c(x^2)) have Hankel transform k^n.
%H Robert Israel, <a href="/A127628/b127628.txt">Table of n, a(n) for n = 0..1165</a>
%F a(n) = 1 if n=0, Sum_{k=1..n} C(2*n-k-1,n-k)*k*6^k/n otherwise;
%F a(n) = Sum_{k=0..n} C(2*n,n-k)*(2*k+1)*5^k/(n+k+1).
%F a(n) = Sum_{k=0..n} A106566(n,k)*6^k. - _Philippe Deléham_, Feb 04 2007
%F a(n) = Sum_{k=0..n} A039599(n,k)*5^k. - _Philippe Deléham_, Sep 08 2007
%F a(n) = (36*a(n-1) - 6*A000108(n-1))/5 for n >= 1, a(0) = 1. - _Philippe Deléham_, Nov 27 2007
%F Conjecture: 5*n*a(n) + 2*(15-28*n)*a(n-1) + 72*(2*n-3)*a(n-2) = 0. - _R. J. Mathar_, Nov 14 2011
%F G.f.: (2+3*sqrt(1-4*x))/(5-36*x). Mathar's conjecture verified using the differential equation (144*x^2-56*x+5)*y'+(72*x-26)*y = 4 satisfied by the g.f. - _Robert Israel_, Aug 28 2020
%p f:= gfun:-rectoproc({(72 + 144*n)*a(n) + (-82 - 56*n)*a(n + 1) + (5*n + 10)*a(n + 2), a(0) = 1, a(1) = 6},a(n),remember):
%p map(f, [$0..50]); # _Robert Israel_, Aug 28 2020
%t RecurrenceTable[{(-56*n - 82)*a[n + 1] + (5*n + 10)*a[n + 2] + (144*n + 72)*a[n] == 0, a[0] == 1, a[1] == 6}, a, {n, 0, 50}] (* _Jean-François Alcover_, Sep 15 2022, after _Robert Israel_ *)
%o (PARI) my(x='x+O('x^25)); Vec(1/(1-6*x*(1-sqrt(1-4*x))/(2*x))) \\ _Michel Marcus_, Sep 15 2022
%Y Cf. A000108, A039599, A106566.
%K nonn
%O 0,2
%A _Paul Barry_, Jan 20 2007