Modularity and Recursive Sequences

The sequence a n a_n is defined by

{ a 0 = 3 a n + 1 a n = n ( a n 1 ) for n 0 \large \begin{cases} a_0=3 \\ a_{n+1}-a_n=n(a_n-1) & \text{for } n \ge 0 \end{cases}

Find all positive integers m m such that gcd ( m , a n ) = 1 \gcd(m,a_n)=1 for all n 0 n \geq 0 . What can the solution set for m m be best described as?

23, 529, 12167 Powers of any arbitrary integer Powers of 23 Powers of 2

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1 solution

a n + 1 a n = n ( a n 1 ) a n + 1 = ( n + 1 ) a n n a n + 1 1 = ( n + 1 ) a n ( n + 1 ) a n + 1 1 = ( n + 1 ) ( a n 1 ) Let b n = a n 1 b n + 1 = ( n + 1 ) b n b 1 = 1 b 0 b 2 = 2 b 1 = 2 1 b 0 b 3 = 3 ! b 0 . . . = . . . b n = n ! b 0 a n 1 = n ! ( a 0 1 ) a n = 2 n ! + 1 a n 1 (mod 2 ) \begin{aligned} a_{n+1} - a_n & = n(a_n-1) \\ a_{n+1} & = (n+1)a_n - n \\ a_{n+1} - 1 & = (n+1)a_n - (n + 1) \\ a_{n+1} - 1 & = (n+1)(a_n - 1) & \small \color{#3D99F6}{\text{Let }b_n = a_n - 1} \\ b_{n+1} & = (n+1)b_n \\ b_1 & = 1\cdot b_0 \\ b_2 & = 2 \cdot b_1 = 2 \cdot 1 \cdot b_0 \\ b_3 & = 3! b_0 \\ ... & = \ ... \\ \implies b_n & = n! b_0 \\ a_n - 1 & = n!(a_0-1) \\ \implies a_n & = 2n! + 1 \\ \implies a_n & \equiv 1 \text{ (mod }\boxed{2}) \end{aligned}

m \implies m are the powers of 2 \boxed{\text{powers of 2}} .

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