cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A349847 Expansion of (1 + 8*x) / sqrt(1 - 4*x).

Original entry on oeis.org

1, 10, 22, 68, 230, 812, 2940, 10824, 40326, 151580, 573716, 2183480, 8347612, 32033848, 123321400, 476050320, 1842020550, 7142249340, 27743985060, 107946346200, 420608639220, 1641030105000, 6410161959240, 25066222437360, 98115049503900, 384391435902552
Offset: 0

Views

Author

Jianing Song, Dec 01 2021

Keywords

Comments

Sum_{n>=0} (a(n)/(-4)^n) is the Cauchy product of Sum_{n>=0} (-A349845(n)/8^n) with itself.

Examples

			a(1) = binomial(0,0) * (12 - 2/1) = 10;
a(2) = binomial(2,1) * (12 - 2/2) = 22;
a(3) = binomial(4,2) * (12 - 2/3) = 68;
a(4) = binomial(6,3) * (12 - 2/4) = 230.

Links

Crossrefs

Cf. A000984, A349845, A349835, A349846.

Programs

  • Mathematica
    CoefficientList[Series[(1+8x)/Sqrt[1-4x],{x,0,30}],x] (* Harvey P. Dale, Jun 08 2023 *)
  • PARI
    a(n) = if(n, binomial(2*(n-1),n-1) * (12 - 2/n), 1)

Formula

For n > 0, a(n) = 8*binomial(2*(n-1),n-1) + binomial(2*n,n) = binomial(2*(n-1),n-1) * (12 - 2/n).
a(n) ~ 4^n * (3/sqrt(Pi*n)).

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

Original entry on oeis.org

-1, 6, 38, 340, 3482, 38740, 457500, 5654440, 72412410, 953696900, 12844323828, 176130113432, 2450987760676, 34524885571400, 491309242342264, 7052495781361488, 101992452504973882, 1484590294804096356, 21732695236734410500, 319745609409940857144
Offset: 0

Views

Author

Jianing Song, Dec 01 2021

Keywords

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.

Examples

			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.

Links

Crossrefs

Cf. A000108, A004981, A349845, A349846.

Programs

  • 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

Formula

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

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