0,3

It appears that this is not prime for n >= 3? Posting to Number Theory List by Georges Z., Mar 25 2005

This is indeed true. Let d_n=2^(2^n-1)-2^n-1. Then 2d_n=2^{2^n}-1-(2^{n+1}+1)=prod_{r=0}^{n-1}F_r-(2^{n+1}+1) where F_r=2^{2^r}+1 is the rth Fermat number. Write n+1=2^{r}*q where q is odd, then r<n (since 2^n>n+1 for n>2) and F_r divides 2^{n+1}+1. Therefore F_r divides 2d_n. On the other hand, F_r<=2^{2^{n-1}}+1<d_n. So d_n is composite. See my recent survey "Problems and results on covering systems" (Lecture 24) and my paper 43 (joint with M. H. Le). - Zhi-Wei Sun (Nanjing Univ., China), Mar 25 2005

Zhi-Wei Sun, Home page

Sequence in context: A143781 A114077 A159522 this_sequence A134230 A141440 A156001

Adjacent sequences: A103740 A103741 A103742 this_sequence A103744 A103745 A103746

sign

N. J. A. Sloane (njas(AT)research.att.com), Mar 28 2005