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.

At the outset, we will prove that:

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

PROOF

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

and, for n>1

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

=> 2^{2^(n-1)} * f(n-1) = 2^{2^(n-1) * 2^(2^(n-1))) + 2^(2^(n-1))

=> 2^(2^(n-1)) * f(n-1) - 2^(2^(n-1)) = 2^(2^n) = f(n) -1

=> f(n) -1 = {f(n-1) - 1} * f(n-1) - f(n-1) + 1

=> f(n)= {f(n-1)}^2 - 2f(n-1) +2

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Let us denote the digital root of a positive integer x by R(x)

We know that:

R(x+y) = R{R(x) + R(y)}

R(xy) = R{R(x) * R(y)}

We will now prove that:

R{f(n)} = 5, if n is even

= 8, if n is odd

PROOF

R{f(n)}

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

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

= R(R(R(f(n-1) * R(f(n-1)) - 2R(f(n-1)) +2

If R(n-1) =5, then:

R(n) = 5^2 -2*5 +2 = 17 == 8 mod 9

If R(n-1) = 8, then:

R(n) = 8^2-2*8+2 = 50 == 5 mod 9

From the above result, it trivially follows that:

{R(f(n))} (- {3,5,8,5,8,5,8,5,8,..........}

Consequently, except the first term - the digital root of the sequence's members are either 5 or 8, depending on the parity of n.

*Edited on ***August 27, 2022, 9:59 pm**