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prime number?
Let f(n) = ((4n-2)/(n+1))*f(n-1), n >1
Supposed f(2) and f(3) are prime numbers. Prove that f(n) is Not prime for any other integer values n>1.
How could I prove that?
1 Answer
- 1 decade agoFavorite Answer
it is a little difficult
notice that f(3)=5/2 * f(2)
and they are prime numbers,
so f(2)=2 and f(3)=5
f(4)=14 is not a prime number
so we can define
f(1)=1 and f(0)=1
therefore
f(n)=(4n-2)/(n+1)*f(n-1)
=(4n-2)(4n-6)/(n+1)n*f(n-2)
=............
=(4n-2)...*6*2/(n+1)...*3*2 *f(0)
=2^n*((2n-1)!!)/((n+1)!)
each odd number in (n+1)!
appear at least once in (2n-1)!!.
when n>4,
the power of 2 in (n+1)!
is not more than n-2
(need a simple proof).
so (n+1)! divides 2^n((2n-1)!!)
and the quotient can be
divided by 4
so f(n) is not prime for n>=4
Source(s): Jessica http://www.tucia.com/