A349834 Expansion of sqrt(1 + 4*x)/(1 - 4*x).

1, 6, 22, 92, 358, 1460, 5756, 23288, 92294, 372036, 1478420, 5947272, 23671516, 95102088, 378922552, 1521039088, 6064766662, 24329781988, 97059838372, 389194630888, 1553243997172, 6226104229528, 24855484384072, 99604902663568, 397733491426972

Offset: 0

Author

Jianing Song, Dec 01 2021

Keywords

• nonn

Comments

Let b(n) = a(n)/4^n, {b(n)} = {1, 3/2, 11/8, 23/16, 179/128, 365/256, 1439/1024, ...}. Since a(n) >= 4^n, Sum_{n>=0} b(n) is divergent. Let c(n) = A349835(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.

Examples

			Let C(n) denote the Catalan numbers.
a(0) = 2^0 = 1;
a(1) = 2^2 + 2^1 * C(0) = 6;
a(2) = 2^4 + 2^3 * C(0) - 2^1 * C(1) = 22;
a(3) = 2^6 + 2^5 * C(0) - 2^3 * C(1) + 2^1 * C(2) = 92;
a(4) = 2^8 + 2^7 * C(0) - 2^5 * C(1) + 2^3 * C(2) - 2^1 * C(3) = 358.
		

Links

• Wikipedia, Cauchy product

Crossrefs

Cf. A000108, A349835.

Programs

PARI

C(n) = binomial(2*n, n)/(n+1)
a(n) = 2^(2*n) + sum(k=0, n-1, (-1)^k * 2^(2*n-1-2*k) * C(k))

Formula

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

a(n) = 2^(2*n + 1/2) - ((-1)^n * CatalanNumber(n) * hypergeom([1, n + 1/2], [n + 2], -1)) / 2. - Peter Luschny, Dec 03 2021

D-finite with recurrence n*a(n) -6*a(n-1) +8*(-2*n+3)*a(n-2)=0. - R. J. Mathar, Jul 27 2022