Composite recursion

Probability Level 5

A sequence is formed by the following rules: s 1 = 3 , s 2 = b s_1=3,\textrm{ }s_2=b and s n + 2 = s n + 1 + ( − 1 ) n s n s_{n+2}=s_{n+1}+(-1)^ns_n for all n ≥ 1 n\geq 1 .

What is the largest integer value of b < 1000 b < 1000 for which 2015 is a member of the sequence?


The answer is 253.

Miles Koumouris by Miles Koumouris , 20, Australia

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1 solution

Miles Koumouris
Jan 2, 2017

For m ≥ 1 , we have: \textrm{For }m\geq 1,\textrm{ we have:}

s 2 m + 2 = s 2 m + 1 + s 2 m s_{2m+2}=s_{2m+1}+s_{2m}

s 2 m + 3 = s 2 m + 2 − s 2 m + 1 = s 2 m s_{2m+3}=s_{2m+2}-s_{2m+1}=s_{2m}

s 2 m + 4 = s 2 m + 3 + s 2 m + 2 = s 2 m + 2 + s 2 m s_{2m+4}=s_{2m+3}+s_{2m+2}=s_{2m+2}+s_{2m}

This means that the coefficients of a and b in the even terms \textrm{This means that the coefficients of }a\textrm{ and }b\textrm{ in the even terms}

form a Fibonacci sequence, and, from the 5th term, every odd \textrm{form a Fibonacci sequence, and, from the 5th term, every odd}

term is a repeat of the third term before it. \textrm{term is a repeat of the third term before it.}

So, defining F 1 = 1 , F 2 = 2 , and F n = F n − 1 + F n − 2 for n ≥ 3 , \textrm{So, defining }F_1=1,\textrm{ }F_2=2,\textrm{ and }F_n=F_{n-1}+F_{n-2}\textrm{ for }n\geq 3,

we have s 2 n = b F n − a F n − 2 for n ≥ 3 . \textrm{we have }s_{2n}=bF_n-aF_{n-2}\textrm{ for }n\geq 3\textrm{.}

Since a = 3 and b < 1000 , none of the first five terms of the \textrm{Since }a=3\textrm{ and }b<1000\textrm{, none of the first five terms of the}

given sequence equal 2015. So we are looking for integer \textrm{given sequence equal 2015. So we are looking for integer}

solutions of b F n − 3 F n − 2 = 2015 for n ≥ 3 . \textrm{solutions of }bF_n-3F_{n-2}=2015\textrm{ for }n\geq 3\textrm{.}

s 6 = 3 b − 3 = 2015 , has no solution. s_6=3b-3=2015,\textrm{ has no solution.}

s 8 = 5 b − 6 = 2015 , has no solution. s_8=5b-6=2015,\textrm{ has no solution.}

s 10 = 8 b − 9 = 2015 implies b = 253 . s_{10}=8b-9=2015\textrm{ implies }b=253\textrm{.}

For n ≥ 6 , we have b = 2015 / F n + 3 F n − 2 / F n . \textrm{For }n\geq 6,\textrm{ we have }b=2015/F_n+3F_{n-2}/F_n\textrm{.}

Since F n increases, we have F n ≥ 13 and F n − 2 / F n < 1 for n ≥ 6 . \textrm{Since }F_n\textrm{ increases, we have }F_n\geq 13\textrm{ and }F_{n-2}/F_n<1\textrm{ for }n\geq 6\textrm{.}

Hence b < 2015 / 13 + 3 = 158 . So the largest value of b is 253 . \textrm{Hence }b<2015/13+3=158\textrm{. So the largest value of }b\textrm{ is }\boxed{253}\textrm{.}

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