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A352971
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Moments of the distribution of position of the first occurrence of pattern aa in a random ternary word.
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0
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1, 12, 258, 8274, 353742, 18904602, 1212354798, 90706565514, 7756033173342, 746093257148442, 79745110236049038, 9375786203927344554, 1202540991574287431742, 167091435183140588426682, 25003060551369349424359278, 4008624526767825553573112394
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OFFSET
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0,2
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COMMENTS
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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).
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).
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LINKS
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FORMULA
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E.g.f.: exp(2*t)/(9 - 6*exp(t) - 2*exp(2*t)).
a(n) ~ n! * (3 - sqrt(3)) / (12 * (log(3*(sqrt(3) - 1)/2))^(n+1)). - Vaclav Kotesovec, Apr 13 2022
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MAPLE
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g := exp(2*x)/(9-6*exp(x)-2*exp(2*x)) ;
taylor(g, x=0, 40) ;
L := gfun[seriestolist](%) ;
seq( op(i, L)*(i-1)!, i=1..nops(L)) ; # R. J. Mathar, Mar 02 2023
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MATHEMATICA
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nn = 15; c[z_] := (1 - z^k)/(1 - z);
T[z_] := z^k/(z^k + (1 - m z) c[z]); G[t_] := T[t/m];
Range[0, nn]! CoefficientList[Series[G[Exp[t]] /. {k -> 2, m -> 3}, {t, 0, nn}], t]
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PROG
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(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
(SageMath) # uses[egfExpand from A362718]
def egf(x): return exp(2*x)/(9 - 6*exp(x) - 2*exp(2*x))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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