A141060 Fourth quadrisection of Jacobsthal numbers A001045: a(n)=16a(n-1)-5.

3, 43, 683, 10923, 174763, 2796203, 44739243, 715827883, 11453246123, 183251937963, 2932031007403, 46912496118443, 750599937895083, 12009599006321323, 192153584101141163, 3074457345618258603, 49191317529892137643

Offset: 0

Author

Paul Curtz, Jul 30 2008

Keywords

• nonn
• easy
• less

Comments

Jacobsthal numbers ending with the decimal digit 3. - Jianing Song, Aug 30 2022

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (17,-16).

Crossrefs

The other quadrisections of A001045 are A195156 (first), A139792 (second), and A144864 (third).

Programs

Magma

[(1/3)*(1+8*16^n): n in [0..25]]; // Vincenzo Librandi, May 25 2011

Mathematica

LinearRecurrence[{17,-16},{3,43},30] (* Harvey P. Dale, Mar 16 2015 *)

PARI

a(n)=8*16^n\3+1 \\ Charles R Greathouse IV, May 25 2011

Python

def A141060(n): return ((1<<(n<<2)+3)|1)//3 # Chai Wah Wu, Apr 18 2025

Formula

a(n) = A139792(n) + A013776(n).

a(n+1) - a(n) = 10*A013709(n) = 40*A001025(n).

G.f.: (3-8*x)/((1-x)*(1-16*x)). [Colin Barker, Apr 05 2012]

a(0)=3, a(1)=43, a(n)=17*a(n-1)-16*a(n-2). - Harvey P. Dale, Mar 16 2015

From Jianing Song, Aug 30 2022: (Start)

a(n) = A001045(4*n+3).

a(n) = 10*A141032(n) + 3 = 20*A098704(n+1) + 1 = 40*A131865(n-1) + 1 for n >= 1. (End)