A171640 a(n) = 10*a(n-1)-a(n-2)-4 with a(1)=1 and a(2)=3.
1, 3, 25, 243, 2401, 23763, 235225, 2328483, 23049601, 228167523, 2258625625, 22358088723, 221322261601, 2190864527283, 21687323011225, 214682365584963, 2125136332838401, 21036680962799043, 208241673295152025, 2061380051988721203, 20405558846592060001
Offset: 1
Examples
a(2+1) = [5-sqrt(24)+5+sqrt(24)]^2/4 = 100/4 = 25.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
- Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
- Index entries for linear recurrences with constant coefficients, signature (11,-11,1).
Programs
-
Mathematica
RecurrenceTable[{a[n] == 10 a[n - 1] - a[n - 2] - 4, a[1] == 1, a[2] == 3}, a, {n, 1, 21}] (* Michael De Vlieger, Oct 02 2015 *) LinearRecurrence[{11,-11,1},{1,3,25},30] (* Harvey P. Dale, May 05 2018 *) -
PARI
Vec(-x*(3*x^2-8*x+1)/((x-1)*(x^2-10*x+1)) + O(x^30)) \\ Colin Barker, Oct 02 2015
Formula
a(n) = A132596(n-1)+1.
a(n+1) = [(sqrt(3)-sqrt(2))^n +(sqrt(3)+ sqrt(2))^n]^2 / 4.
From Colin Barker, Oct 02 2015: (Start)
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n>3.
G.f.: -x*(3*x^2-8*x+1) / ((x-1)*(x^2-10*x+1)).
(End)
6*a(n)*(a(n)-1) = [A122653(n-1)]^2. - Jean-Luc Manguin, Jun 02 2020