%I #24 Jan 10 2022 11:44:00
%S 1,4,18,85,410,1999,9807,48304,238570,1180615,5851253,29033074,
%T 144190943,716652070,3564079250,17734184365,88280673770,439625873215,
%U 2189988826125,10912480440850,54389237971285,271142650382080
%N Row sums of Riordan array A110165.
%C Number of 5-ary words of length n in which the number of 1's does not exceed the number of 0's. - _David Scambler_, Aug 14 2012
%C From _Peter Bala_, Jan 09 2022: (Start)
%C Conjectures: for k >= 2, the number of k-ary words of length n such that the number of 1's <= the number of 0's is equal to the coefficient of x^n in the expansion of ( k*x + 1/(1 + x) )^n, and satisfies the recurrence u(0) = 1, u(1) = k-1 and n*u(n) = (k-2)*(2*n-1)*u(n-1) - k*(k-4)*(n-1)* u(n-2) + k^(n-1) for n >= 2.
%C For cases see A027306 (k = 2), A027914 (k = 3) and A032443 (k = 4). (End)
%H Vincenzo Librandi, <a href="/A110166/b110166.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: (1/sqrt(1-6*x+5*x^2))/(1-(1-3*x-sqrt(1-6*x+5*x^2))/(2*x)).
%F a(n) = Sum_{k = 0..n} Sum_{j = 0..n} C(n, j)*C(2*j, j+k).
%F Recurrence: n*a(n) = (11*n-8)*a(n-1) - 5*(7*n-10)*a(n-2) + 25*(n-2)*a(n-3). - _Vaclav Kotesovec_, Oct 18 2012
%F a(n) ~ 5^n/2*(1+sqrt(5)/(2*sqrt(Pi*n))). - _Vaclav Kotesovec_, Oct 18 2012
%F From _Peter Bala_, Jan 08 2022: (Start)
%F a(n) = (1/2)*(5^n + A026375(n)) = (1/2)*(5^n + Sum_{k = 0..n} binomial(n,k) *binomial(2*k,k)).
%F a(n) = (1/2)*(5^n)*(1 + Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(-1/5)^k).
%F a(n) = [x^n] ( 5*x + 1/(1 + x) )^n.
%F a(0) = 1, a(1) = 4 and n*a(n) = 3*(2*n-1)*a(n-1) - 5*(n-1)*a(n-2) + 5^(n-1) for n >= 2.
%F The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k.
%F Binomial transform of A032443. (End)
%p seq( (1/2)*(5^n + add(binomial(n,k)*binomial(2*k,k), k = 0..n)), n = 0..30); # _Peter Bala_, Jan 08 2022
%t Table[Sum[Sum[Binomial[n,j]Binomial[2j,j+k],{j,0,n}],{k,0,n}],{n,0,25}] (* _Harvey P. Dale_, Dec 16 2011 *)
%Y Row sums A110165. Cf. A026375, A032443, A246437, A242586.
%K nonn,easy
%O 0,2
%A _Paul Barry_, Jul 14 2005