A349844 - OEIS

%I #18 Dec 04 2021 06:40:47

%S -1,6,38,340,3482,38740,457500,5654440,72412410,953696900,12844323828,

%T 176130113432,2450987760676,34524885571400,491309242342264,

%U 7052495781361488,101992452504973882,1484590294804096356,21732695236734410500,319745609409940857144

%N Expansion of -(1 - 16*x)^(1/2) / (1 - 8*x)^(1/4).

%C Let b(n) = -a(n)/(-8)^n, {b(n)} = {1, 3/4, -19/32, 85/128, -1741/2048, 9685/8192, -114375/65536, ...}, then Sum_{n>=0} b(n) is clearly divergent since Sum_{n>=0} a(n)*x^n has radius of convergence 1/16. Let c(n) = -A349846(n)/(-4)^n, {c(n)} = {1, 3/2, -5/8, 7/16, -45/128, 77/256, -273/1024, ...}, then Sum_{n>=1} c(n) is the Cauchy product of Sum_{n>=0} b(n) with itself. Since |c(n)| ~ 1/sqrt(Pi*n) and |c(n+1)|/|c(n)| = ((2*n-1)*(2*n+3)) / ((2*n+1)*(2*n+2)) < 1, Sum_{n>=0} c(n) is conditionally convergent by Leibniz's criterion. {b(n)} serves as an example such that the Cauchy product of a divergent series with itself can be conditionally convergent.

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

%F a(n) = (Sum_{k=0..n-1} 2^(2*k+3) * CatalanNumber(k) * A004981(n-1-k)) - A004981(n).

%e Let C(n) denote the Catalan numbers, P = A004981.

%e a(0) = -P(0) = -1;

%e a(1) = 2^3 * C(0) * P(0) - P(1) = 6;

%e a(2) = 2^3 * C(0) * P(1) + 2^5 * C(1) * P(0) - P(2) = 38;

%e a(3) = 2^3 * C(0) * P(2) + 2^5 * C(1) * P(1) + 2^7 * C(2) * P(0) - P(3) = 340;

%e a(4) = 2^3 * C(0) * P(3) + 2^5 * C(1) * P(2) + 2^7 * C(2) * P(1) + 2^9 * C(3) * P(0) - P(4) = 3482.

%o (PARI) C(n) = binomial(2*n,n)/(n+1)

%o a(n) = sum(k=0, n-1, 2^(2*k+3) * C(k) * A004981(n-1-k)) - A004981(n) \\ See A004981 for its program

%Y Cf. A000108, A004981, A349845, A349846.

%K sign

%O 0,2

%A _Jianing Song_, Dec 01 2021