Consider the sequence set by $ a_1 = 0 $, $ a_2 = 1 $, and for $ n \geq 3 $ $$ a_n = (n-1) (a_ {n-1} + a_ {n-2}) $$ Determine the remainder in the division of $ a_ {2019} $ by 2019
a) 1
b)2
c) 2017
d)2018
Can anyone give me any tips for this problem?
Attemp: a3 = 2 * ( 1 + 0) = 2
a4 = 3 * ( 2 + 1) = 9 , r = 1
a5 = 4 * ( 9 + 2) = 44, r = 4
a6 = 5 * ( 44 + 9) = 265, r = 1
a7 = 6 * ( 265 + 44) = 1854, r = 6
Can you show $a_n-na_{n-1}=a_{n-2}-(n-2)a_{n-3}$?
It follows then that $a_{2019}-2019a_{2018}=a_3-3a_2=-1.$