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Fermat numbers (Posted on 2017-07-13) Difficulty: 1 of 5
For any non-negative integer n define f(n)=2^(2^n)+1.
The above formula produces Fermat numbers i.e. 3,5,17,257 ...etc

Prove: Except the 1st term, the digital root of sequence's members are either 5 or 8, depending on parity of n.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 1 of 2

f(n+1) = (f(n) - 1)^2 + 1

Define r(x) the digit root  of x. We can verify :

r(x+y) = r(r(x)+r(y)), r(x*y) = r(r(x)*r(y))

if r(f(n)) = 5, then  r(f(n+1)) = r((5-1)^2 + 1) = 8

if r(f(n)) = 8, then  r(f(n+1)) = r((8-1)^2 + 1) = 5


  Posted by chun on 2017-07-13 13:26:22
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