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I found some unexpected data for primes. Consider $p(n)$ being the product over all primes smaller than or equal to $n$. When factoring $p(n)^a +1$ for $a=1$ or $a=2$ we get the expected amount of primes.

For instance for $a=2$ , $p(7)^2+1$ is prime and we get primes for $p(T(n))^2+1$ where $T(n)=e^{2n}+O(\sqrt {ln(n)})$ as expected.

However for $a=3$ we expect primes for $p(T_3(n))^2+1$ where $T_3(n)=e^{3n}+O(\sqrt {ln(n)})$.

However for $1<n<61$ , $p(n)^3+1$ is never a prime. And up to $n=101$ I find at best semiprimes and some pseudoprimes.

This contradicts statistics based on the prime number theorem alot !!

I havent seen such an unexpected result since the MERSENNE VS FERMAT primes.

How to explain this ?

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Maybe there's something I don't understand, but I get $p(3)^2 + 1 = 2^2+1 = 5$ is prime, while $p(7)^2 + 1 = 901$ is composite. –  Robert Israel Sep 8 '13 at 22:27
    
I correct sorry. –  mick Sep 8 '13 at 22:28
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Well, now it should be $p(3)^2 + 1 = 6^2 + 1 = 37$. –  Robert Israel Sep 8 '13 at 22:32
    
Yes, well the point is $p(n)^2+1$ has about the expected amount of primes when considering an interval $0<n<k$ whereas $p(n)^3+1$ does not apparantly. Thanks for your helpfull comments. Nice to see you again. @RobertIsrael –  mick Sep 8 '13 at 22:36

1 Answer 1

up vote 8 down vote accepted

Recall that $x^3+1=(x+1)(x^2-x+1)$.

The same non-primes phenomenon will occur whenever $a$ has an odd divisor greater than $1$, that is, whenever $a$ is not a power of $2$.

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I feel so silly now. –  mick Sep 8 '13 at 22:39

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