a n b n = = 2 a n − 1 + 3 b n − 1 + b n − 2
For a positive integer n , consider the two recurrence relations above subjected to the conditions a 1 = 0 and b 1 = b 2 = 1 .
If the value of the expression ( a b 2 0 1 5 + 3 ) ( a b 2 0 1 6 + 3 ) can be expressed as p b q s r , where b q is one of the terms in the recurrence relations sequence above and ( p , r ) and ( r , s ) are pairwise coprime integers.
Find the value of ( p + q + r + s ) m o d 6 7 3 .
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one can show with induction that
a n = 3 ∗ 2 n − 1 − 3
Thus
( a b n + 3 ) ( a b n + 1 + 3 ) = 9 ∗ 2 b n + b n + 1 − 2
From the definition of b
2 b n + 2 4 9
Thus for n = 2 0 1 5 we get
2 b 2 0 1 7 4 9
Thus we need 2 + 2 0 1 7 + 9 + 4 = 2 0 3 2 = 1 3 m o d 6 7 3