A sequence is formed by the following rules:
and for all .
If and , what is the largest integral value of for which 2015 is a member of the sequence?
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If the second term is b , then after writing a few terms of the series, I found that the terms were of the form: b , b − 3 , 2 b − 3 , 3 b − 3 , 5 b − 6 , 8 b − 9 and so on. Equating these values to 2015 gives the first integer answer for 8 b − 9 = 2 0 1 5 ⇒ b = 2 5 3 .
FunFact: This sequence follows a Fibonacci pattern except for its second term, b − 3 .