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.

Showing 1-2 of 2 results.

A155988 a(n) = (2*n + 1)*9^n.

Original entry on oeis.org

1, 27, 405, 5103, 59049, 649539, 6908733, 71744535, 731794257, 7360989291, 73222472421, 721764371007, 7060738412025, 68630377364883, 663426981193869, 6382625094934119, 61149666232110753, 583701359488329915, 5553501505988967477, 52683216989246691471, 498464283821334080841
Offset: 0

Views

Author

Jaume Oliver Lafont, Feb 01 2009

Keywords

Crossrefs

Cf. A058962 for the similar (2n+1)4^n.

Programs

  • Magma
    [(2*n+1)*9^n: n in [0..20]]; // Vincenzo Librandi, Jun 08 2011
    
  • Maxima
    makelist((2*n+1)*9^n, n, 0, 20); /* Martin Ettl, Nov 11 2012 */
  • PARI
    a(n)=(2*n+1)*9^n;
    

Formula

G.f.: (1 + 9*x)/(1 - 9*x)^2.
a(n) = 18*a(n-1) - 81*a(n-2) for n>=2.
Sum_{n>=0} 1/a(n) = (3/2)*log(2).
a(n) = A005408(n) * A001019(n).
a(n) = (2*n - 1)*3^(2*n-1)/3 = A060851(n)/3.
Sum_{n>=0} (-1)^n/a(n) = 3*arctan(1/3). - Amiram Eldar, Feb 26 2022
E.g.f.: exp(9*x)*(1 + 18*x). - Stefano Spezia, May 07 2023

A096949 Numerators of partial sums of series for 3*arctanh(1/3) = (3/2)*log(2).

Original entry on oeis.org

1, 28, 421, 26528, 2148803, 7878956, 2765513941, 74668877408, 3808112752813, 651187280816108, 2511722368895123, 173308843453994432, 7798897955430811787, 1895132203169713822916, 54958833891921780540589
Offset: 0

Views

Author

Wolfdieter Lang, Jul 16 2004

Keywords

Comments

Denominators are given in A096950.
From the series of log((1+x)/(1-x)) for x = 1/3, i.e., for log(2) = 2*Sum_{k>=0} (1/3)^(2*k+1)/(2*k+1).

Examples

			n=3: 26528/A096950(3) = 26528/25515 = 1.0397021... approximates 3*arctanh(1/3) = 1.039720771...
n = 3: Sum_{k = 0..3} (1/3)^(2*k + 1)/(2*k + 1) = 1/3 + 1/81 + 1/1215 + 1/15309 = 26528/76545. Hence a(3) = 26528. - _Peter Bala_, Feb 06 2024
		

Crossrefs

Cf. A096950.

Formula

a(n) = numerator(A(n)) with the rational number A(n) := Sum_{k=0..n} (1/3)^(2*k)/(2*k+1) in lowest terms.
(3/2)*log(2) = a(n)/A096950(n) + 3*Integral_{x >= 3} 1/(x^(2*n+4) - x^(2*n+2)) dx. - Peter Bala, Feb 05 2024
Showing 1-2 of 2 results.