Is number of the following form is always composite:
$$N=2^{2^n}+2^{2^{n-1}}+2^{2^{n-2}}+2^{2^{n-3}} + . . . 2^{2^3}+2^{2^2}+n$$
I found similar question for $2^{2^n}+5$ .
My attempt: We can rewrite N as :
$$N=2^{2^n}+1+2^{2^{n-1}}+1+2^{2^{n-2}}+1+2^{2^{n-3}}+1 + . . . 2^{2^3}+1+2^{2^2}+1$$
So we have a sum of n Fermat numbers. we may write:
$2^{2a}+1=(3-1)^{2a}+1=3k+2$
Therefore the sum each three numbers will be divisible by 3, hence for N being a primes, n must not be divisible by 3. Also n must not be divisible by 2 that means n must be divisible by 6. I checked this up to $n=6$ .
$2^{2^n}≡6 \mod (10)$
So:
$2^{2^n}+1≡7 \mod (10)$:
$2^{2^2}+1+2^{2^3}+1+2^{2^4}+1=165811≡1 \mod(10)$
$2^{2^2}+1+2^{2^3}+1+2^{2^4}+1+2^{2^5}+1≡8 \mod(10)$
$2^{2^2}+1+2^{2^3}+1+2^{2^4}+1+2^{2^5}+1+2^{2^6}+1≡5 \mod(10)$
$2^{2^2}+1+2^{2^3}+1+2^{2^4}+1+2^{2^5}+1+2^{2^6}+1+2^{2^7}+1≡2 \mod(10)$
$2^{2^2}+1+2^{2^3}+1+2^{2^4}+1+2^{2^5}+1+2^{2^6}+1+2^{2^7}+2^{2^8}+1≡9 \mod(10)$
The odd numbers which may appear as last digit are 1, 3, 7 and 9 and the period of repeating of these numbers as the last digit is 10. I could not check for n=8. My answer to this question is no.We can write:
$\displaystyle{\Sigma}^n_{i=1}(2^{2^n}+n)≡2n \mod (3)≡-n\ mod (3)= 3m-n$
Which is a linear equivalent and indicates N may be a prime.