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Number of ternary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.
8

%I #45 Sep 08 2022 08:45:58

%S 1,1,5,29,181,1181,7941,54573,381333,2699837,19319845,139480397,

%T 1014536117,7426790749,54669443141,404388938349,3004139083221,

%U 22402851226749,167640057210981,1258340276153229,9471952718661621,71481616200910749,540715584181142661

%N Number of ternary words either empty or beginning with the first character of the alphabet, that can be built by inserting n doublets into the initially empty word.

%H Alois P. Heinz, <a href="/A194723/b194723.txt">Table of n, a(n) for n = 0..1000</a>

%H C. Kassel and C. Reutenauer, <a href="https://arxiv.org/abs/1303.3481">Algebraicity of the zeta function associated to a matrix over a free group algebra</a>, arXiv preprint arXiv:1303.3481 [math.CO], 2013-2014.

%F G.f.: 2/3 + 4/(3*(1+3*sqrt(1-8*x))).

%F a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*2^j for n>0.

%F D-finite with recurrence: n*a(n) = (17*n-12)*a(n-1) - 36*(2*n-3)*a(n-2). - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) ~ 2^(3*n+1)/(sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012

%F G.f.: 2-4/( Q(0) + 3), where Q(k) = 1 + 8*x*(4*k+1)/( 4*k+2 - 8*x*(4*k+2)*(4*k+3)/( 8*x*(4*k+3) + 4*(k+1)/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Nov 20 2013

%F From _Karol A. Penson_, Jul 13 2015: (Start)

%F Special values of the hypergeometric function 2F1, in Maple notation:

%F a(n+1) = (16/9)*8^n*GAMMA(n+3/2)*hypergeom([1, n+3/2], [n+3],8/9)/(sqrt(Pi)*(n+2)!), n=0,1,... .

%F Integral representation as the n-th moment of a positive function W(x) = sqrt((8-x)*x)*(1/(9-x))/(2*Pi) on (0,8): a(n+1) = int(x^n*W(x),x=0..8), n=0,1,... . This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)

%F a(n) = 2^(n+1)*binomial(2*n,n)*hypergeom([2,1-n],[n+2],-2)/(n+1) - 3^(2*n-1) for n>=1. - _Peter Luschny_, Apr 07 2018

%e a(2) = 5: aaaa, aabb, aacc, abba, acca (with ternary alphabet {a,b,c}).

%p a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *2^j, j=0..n-1)/n):

%p seq(a(n), n=0..25);

%t CoefficientList[Series[2/3+4/(3*(1+3*Sqrt[1-8*x])), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 20 2012 *)

%t a[n_] := 2^(n+1) CatalanNumber[n] Hypergeometric2F1[2, 1-n, n+2, -2] - 3^(2n - 1);

%t Table[If[n == 0, 1, a[n]], {n, 0, 22}] (* _Peter Luschny_, Apr 08 2018 *)

%o (PARI) a(n) = if (n==0, 1, sum(j=0, n-1, binomial(2*n,j)*(n-j)*2^j)/n); \\ _Michel Marcus_, Apr 07 2018

%o (PARI) x='x+O('x^99); Vec(4/(3*(1+3*(1-8*x)^(1/2)))+2/3) \\ _Altug Alkan_, Apr 07 2018

%o (Magma) [1] cat [&+[(Binomial(2*n, k)*(n-k)*2^k)/n: k in [0..n]]: n in [1..25]]; // _Vincenzo Librandi_, Apr 08 2018

%Y Column k=3 of A183134.

%Y Cf. A194726.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Sep 02 2011