A349845 - OEIS

%I #15 Dec 04 2021 06:40:53

%S -1,10,6,332,1498,29964,269660,4066456,48190842,679524828,8993585460,

%T 126419889960,1757062172580,25004701186680,356647387079160,

%U 5145713721249072,74607994412294970,1089344167433473788,15981504546211353156,235635552851036269704

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

%C Let b(n) = -a(n)/8^n, {b(n)} = {1, -5/4, -3/32, -83/128, -749/2048, -7491/8192, -67415/65536, ...}, 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) = A349847(n)/(-4)^n, {c(n)} = {1, -5/2, 11/8, -17/16, 115/128, -203/256, 735/1024, ...}, then Sum_{n>=0} c(n) is the Cauchy product of Sum_{n>=1} b(n) with itself. Since |c(n)| ~ 3/sqrt(Pi*n) and |c(n+1)|/|c(n)| = ((6*n+5)*(2*n-1)) / ((6*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} (-1)^(n-1-k) * 2^(2*k+3) * CatalanNumber(k) * A004981(n-1-k)) + (-1)^(n-1) * 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) = 10;

%e a(2) = -2^3 * C(0) * P(1) + 2^5 * C(1) * P(0) - P(2) = 6;

%e a(3) = 2^3 * C(0) * P(2) - 2^5 * C(1) * P(1) + 2^7 * C(2) * P(0) + P(3) = 332;

%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) = 1498.

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

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

%Y Cf. A000108, A004981, A349844, A349847.

%K sign

%O 0,2

%A _Jianing Song_, Dec 01 2021