Problem 5

Determine all sequences

a_0,a_1,a_2,\ldots

of positive integers with

a_0\ge2015

such that for all integers

n\ge1

: (i)

a_{n+2}

is divisible by

a_n

; (ii)

|s_{n+1}-(n+1)a_n|=1

, where

s_{n+1}=a_{n+1}-a_n+a_{n-1}-\cdots+(-1)^{n+1}a_0

Step 4 of 5: Force the one-step ratio

Previous step Next step

a_{n+1}=\frac{(n+1)(n+3)}{n+2}a_n\quad(n\ge1)

Detailed analysis

Applying the previous two-step identity at consecutive indices and comparing with the original recurrence shows that the error term is zero for n at least four, giving the displayed ratio. If the ratio failed at one of the first three indices, take the greatest such index; the divisibility relations would make a_m divide a number at most 14, contradicting the lower bound. Thus the ratio holds for every n at least one.