%I #14 Dec 04 2021 06:40:41

%S 1,6,14,44,150,532,1932,7128,26598,100100,379236,1444456,5525884,

%T 21217224,81719000,315583920,1221550470,4737927780,18409560180,

%U 71645805000,279227584020,1089643989720,4257130461480,16649826582480,65181326593500,255401021170152

%N Expansion of (1 + 4*x)/sqrt(1 - 4*x).

%C Let b(n) = A349834(n)/4^n, {b(n)} = {1, 3/2, 11/8, 23/16, 179/128, 365/256, 1439/1024, ...}. Since A349834(n) >= 4^n, Sum_{n>=0} b(n) is divergent. Let c(n) = a(n)/(-4)^n, {c(n)} = {1, -3/2, 7/8, -11/16, 75/128, -133/256, 483/1024, ...}. Since |c(n)| ~ 2/sqrt(Pi*n) and |c(n+1)|/|c(n)| = ((4*n+3)*(2*n-1)) / ((4*n-1)*(2*n+2)) < 1, Sum_{n>=0} c(n) is conditionally convergent by Leibniz's criterion. Note that Sum_{n>=0} b(n)*x^n = sqrt(1 + x)/(1 - x), Sum_{n>=0} c(n)*x^n = (1 - x)/sqrt(1 + x), hence the Cauchy product of Sum_{n>=0} b(n) and Sum_{n>=0} c(n) is 1 + 0 + 0 + .... {b(n)} and {c(n)} serve as an example such that the Cauchy product of a divergent series and a conditionally convergent series can be absolutely convergent.

%H Wikipedia, <a href="https://en.m.wikipedia.org/wiki/Cauchy_product">Cauchy product</a>

%F For n > 0, a(n) = binomial(2*n,n) + 4*binomial(2*(n-1),n-1) = binomial(2*(n-1),n-1) * (8 - 2/n).

%F a(n) ~ 4^n * (2/sqrt(Pi*n)).

%e a(1) = binomial(0,0) * (8 - 2/1) = 6;

%e a(2) = binomial(2,1) * (8 - 2/2) = 14;

%e a(3) = binomial(4,2) * (8 - 2/3) = 44;

%e a(4) = binomial(6,3) * (8 - 2/4) = 150.

%o (PARI) a(n) = if(n, binomial(2*(n-1),n-1) * (8 - 2/n), 1)

%Y Cf. A000984, A349834, A349846, A349847.

%K nonn,easy

%O 0,2

%A _Jianing Song_, Dec 01 2021