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There is much to know about the Fermat numbers. Now I propose the numbers of the form $f = 2^{2^n} +2^j +1 $. Of course if $j$ is even, then $ 3 | f$. Also, If $ n = 1, j = 1 $, then $ f = $ 7. If $ n = 2, j = 1 $, then $ f = 19 $. If $ n = 4, j = 1 $, then $ f = $ 65539, i.e. they are twin Fermat primes. The question is, we can say that $ f = 2^{2^n} +2^j +1 $ with $ n> 4 $ and $ 1 \leq j <2^n$ is composite? I have found it is true up to n = 15. Some one know anything else?

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Was there any noticeable pattern in the factors of $f$ for $n>4$? –  Alex Becker Jul 3 '12 at 5:10
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$f=(2^{2^n}-1)+(2)(2^{j-1}+1)$, and the first summand is divisible by $2^{2^r}+1$ for all $r\lt n$, so it suffices to prove that $2^{2^r}+1$ divides $2^{j-1}+1$ for some $r$. –  Gerry Myerson Jul 3 '12 at 5:50
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And that's easy, since $2^{j-1}=2^{q2^r}$ for some $r$ and some odd $q$. –  Gerry Myerson Jul 3 '12 at 5:53
    
@Gerry: I don't believe your reasoning covers the case $j=1$. (It is by no coincidence that all of the OP's examples are when $j=1$) –  Eric Naslund Jul 3 '12 at 15:41
    
If $n$ is odd then $2^{2^n}+3$ is divisible by $7$. –  François Brunault Jul 3 '12 at 19:19
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