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%I #24 May 06 2023 21:32:28
%S 1,12,258,8274,353742,18904602,1212354798,90706565514,7756033173342,
%T 746093257148442,79745110236049038,9375786203927344554,
%U 1202540991574287431742,167091435183140588426682,25003060551369349424359278,4008624526767825553573112394
%N Moments of the distribution of position of the first occurrence of pattern aa in a random ternary word.
%C Let X be the random variable that assigns to each word on alphabet {a,b,c} the number of letters required for the first occurrence of the pattern aa. Then a(n) = E(X^n).
%C Let X(m,k) be the random variable that assigns to each m-ary word the number of letters required for the first occurrence of the pattern aa...a (k copies of a). The moment generating function for X(m,k) is G(exp(t)) where G(t) = T(t/m), T(z) = z^k/(z^k + c(z)(1- m*z)), c(z) = (1-z^k)/(1-z).
%H Philippe Flajolet and Robert Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/AnaCombi/anacombi.html">Analytic Combinatorics</a>, Cambridge Univ. Press, 2009, page 59.
%F E.g.f.: exp(2*t)/(9 - 6*exp(t) - 2*exp(2*t)).
%F a(n) ~ n! * (3 - sqrt(3)) / (12 * (log(3*(sqrt(3) - 1)/2))^(n+1)). - _Vaclav Kotesovec_, Apr 13 2022
%p g := exp(2*x)/(9-6*exp(x)-2*exp(2*x)) ;
%p taylor(g,x=0,40) ;
%p L := gfun[seriestolist](%) ;
%p seq( op(i,L)*(i-1)!,i=1..nops(L)) ; # _R. J. Mathar_, Mar 02 2023
%t nn = 15; c[z_] := (1 - z^k)/(1 - z);
%t T[z_] := z^k/(z^k + (1 - m z) c[z]); G[t_] := T[t/m];
%t Range[0, nn]! CoefficientList[Series[G[Exp[t]] /. {k -> 2, m -> 3}, {t, 0, nn}],t]
%o (PARI) seq(n)=my(p=exp(x + O(x*x^n))); Vec(serlaplace(p^2/(9 - 6*p - 2*p^2))) \\ _Andrew Howroyd_, May 06 2023
%o (SageMath) # uses[egfExpand from A362718]
%o def egf(x): return exp(2*x)/(9 - 6*exp(x) - 2*exp(2*x))
%o print(egfExpand(egf, 1, 15)) # _Peter Luschny_, May 06 2023
%Y Cf. A302922.
%K nonn
%O 0,2
%A _Geoffrey Critzer_, Apr 12 2022
%E Typo in a(7) corrected by _Georg Fischer_, May 06 2023
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