Problem 3

A sequence of real numbers

a_0,a_1,\ldots

is said to be good if: (i)

a_0

is a positive integer; (ii) for each non-negative integer

i

a_{i+1}=2a_i+1

a_{i+1}=\frac{a_i}{a_i+2}

; (iii) there exists a positive integer

k

such that

a_k=2014

. Find the smallest positive integer

n

such that there exists a good sequence with

a_n=2014

Step 2 of 4: Run the sequence backwards

Previous step Next step

a_k=2014\Longrightarrow (m_0,n_0)=(2014,1),\quad a_{k-i}=\frac{m_i}{n_i}

Detailed analysis

Starting from 2014, each predecessor is uniquely determined: if the current value is above one use the inverse of the doubling move, and if it is below one use the inverse of the fractional move. Write each predecessor in lowest terms. The numerator and denominator stay positive, coprime, and have sum 2015.