A sequence is defined as follows a 1 = a 2 = a 3 = 1 , and for all positive integers n , a n + 3 = a n + 2 + a n + 1 + a n . Given that a 2 8 = 6 0 9 0 3 0 7 , a 2 9 = 1 1 2 0 1 8 2 1 and a 3 0 = 2 0 6 0 3 3 6 1 .
Find the remainder when k = 1 ∑ 2 8 a k is divided by 1000.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Problem Loading...
Note Loading...
Set Loading...
This is a linear difference equation . This problem is similar a solving the equation A ≡ x 3 − x 2 − x − 1 = 0 adding initial conditions a ( 1 ) = 1 , a ( 2 ) = 1 , a ( 3 ) = 1 . This problem is similar to Tribonacci numbers and here you have the first 28 numbers of this sequence . I added these numbers and collected their last 3 figures... I know it is not a nice solution but what can I say?...