A349844 Expansion of -(1 - 16*x)^(1/2) / (1 - 8*x)^(1/4).

-1, 6, 38, 340, 3482, 38740, 457500, 5654440, 72412410, 953696900, 12844323828, 176130113432, 2450987760676, 34524885571400, 491309242342264, 7052495781361488, 101992452504973882, 1484590294804096356, 21732695236734410500, 319745609409940857144

Offset

0,2

Comments

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.

Links

Table of n, a(n) for n=0..19.

Wikipedia, Cauchy product

Formula

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

Example

Let C(n) denote the Catalan numbers, P = A004981.
a(0) = -P(0) = -1;
a(1) = 2^3 * C(0) * P(0) - P(1) = 6;
a(2) = 2^3 * C(0) * P(1) + 2^5 * C(1) * P(0) - P(2) = 38;
a(3) = 2^3 * C(0) * P(2) + 2^5 * C(1) * P(1) + 2^7 * C(2) * P(0) - P(3) = 340;
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.

Prog

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

Crossrefs

Cf. A000108, A004981, A349845, A349846.

Sequence in context: A356458 A276473 A062814 * A367838 A319194 A303865

Adjacent sequences: A349841 A349842 A349843 * A349845 A349846 A349847

Keyword

sign

Author

Jianing Song, Dec 01 2021

Status

approved