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I believe that the number

$$2^{2^{2t+1}+2t-1}-1$$

is composite for all positive integer $t$. I tested this for many $t$'s, but so far I didn't get a proof. Any idea?

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  • $\begingroup$ Hmm, so you mean $M=2^m-1$ where $m=2^{2t+1}-1+2t=4\cdot 2^w+w $. If $m$ is composite, then $M$ must also be composite. So it might be easier to prove that $m$ is composite? $\endgroup$ Aug 24, 2014 at 7:53
  • $\begingroup$ @GottfriedHelms $m$ can be prime e.g. $t=71$. $\endgroup$
    – joro
    Aug 24, 2014 at 8:09
  • $\begingroup$ $4*2^w+w$ is prime (or at least prp) for $w=141,411,5495,6647,7427,7889,14565,17933,...$ $\endgroup$ Aug 24, 2014 at 8:09
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    $\begingroup$ The first possible prime is for $2^m-1,m=11150372599265311570767859136324180752990349,t=71$. PRP test is not tractable for current hardware unless it is divisible by small primes. $\endgroup$
    – joro
    Aug 24, 2014 at 8:12
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    $\begingroup$ @joro No typo there. Any prime factor is of the form $k*2m+1$. I calculated $2^m \mod p$ with all primes of that form for $k$ up to $2^{28}$. $\endgroup$ Aug 24, 2014 at 10:34

1 Answer 1

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If your expression is composite for all large $t$, then either (a) $4\cdot 2^w+w$ is composite for all large $w$ or (b) $2^p-1$ is composite for infinitely many primes $p$. Now (a) is probably false while (b) is probably (surely!) true --- however, no one has succeeded in showing (b) unconditionally!

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  • $\begingroup$ :-) Nice approach to sort things out... $\endgroup$ Aug 25, 2014 at 7:43

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