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Number of 6-ary 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.
7

%I #23 Feb 08 2021 06:37:48

%S 1,1,11,146,2131,32966,530526,8786436,148733571,2561439806,

%T 44731364266,790211926076,14095578557486,253519929631996,

%U 4592415708939356,83709533881191816,1534227271236577251,28256420350942562286,522675506718404898546,9706083027629177910156

%N Number of 6-ary 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="/A194726/b194726.txt">Table of n, a(n) for n = 0..500</a>

%F G.f.: 5/6 + 5/(3*(4+6*sqrt(1-20*x))).

%F a(0) = 1, a(n) = 1/n * Sum_{j=0..n-1} C(2*n,j)*(n-j)*5^j for n>0.

%F a(n) ~ 5*20^n/(16*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Aug 13 2013

%F D-finite with recurrence: n*a(n) = 2*(28*n-15)*a(n-1) - 360*(2*n-3)*a(n-2). - _Vaclav Kotesovec_, Aug 13 2013

%F From _Karol A. Penson_, Jul 12 2015: (Start)

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

%F a(n+1) = (25/9)*20^n*GAMMA(n+3/2)*hypergeom([1, n+3/2], [n+3],5/9)/(sqrt(Pi)*(n+2)!), n=0,1,... .

%F Integral representation as the n-th moment of a positive function W(x) = sqrt((20-x)*x)*(1/(36-x))/(2*Pi) on (0,20): a(n+1) = int(x^n*W(x),x=0..20), n=0,1,... . This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)

%e a(2) = 11: aaaa, aabb, aacc, aadd, aaee, aaff, abba, acca, adda, aeea, affa (with 6-ary alphabet {a,b,c,d,e,f}).

%p a:= n-> `if`(n=0, 1, add(binomial(2*n, j) *(n-j) *5^j, j=0..n-1) /n):

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

%Y Column k=6 of A183134.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Sep 02 2011