Recursions in Recursions - 2

Algebra Level 4

a n = 2 a n 1 + 3 b n = b n 1 + b n 2 { \begin{aligned} a_{n} & =& 2a_{n-1} + 3 \\ b_{n} &=& b_{n-1} + b_{n-2} \\ \end{aligned} }

For a positive integer n n , consider the two recurrence relations above subjected to the conditions a 1 = 0 a_1 = 0 and b 1 = b 2 = 1 b_{1} = b_{2} = 1 .

If the value of the expression ( a b 2015 + 3 ) ( a b 2016 + 3 ) \big(a_{b_{2015}} + 3\big)\big(a_{b_{2016}} + 3\big) can be expressed as p b q r s p^{b_{q}} \frac{r}{s} , where b q b_{q} is one of the terms in the recurrence relations sequence above and ( p , r ) (p, r) and ( r , s ) (r, s) are pairwise coprime integers.

Find the value of ( p + q + r + s ) m o d 673 (p+q+r+s) \bmod {673} .


The answer is 13.

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1 solution

Daniel Branscombe
Jul 31, 2015

one can show with induction that

a n = 3 2 n 1 3 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 (a_{b_n}+3)(a_{b_{n+1}}+3)=9*2^{b_n+b_{n+1}-2}

From the definition of b b

2 b n + 2 9 4 2^{b_{n+2}}\frac{9}{4}

Thus for n = 2015 n=2015 we get

2 b 2017 9 4 2^{b_{2017}}\frac{9}{4}

Thus we need 2 + 2017 + 9 + 4 = 2032 = 13 m o d 673 2+2017+9+4=2032=13 mod 673

Sir please prove the induction part too..

Vishal Yadav - 4 years, 2 months ago

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