A103743 a(n) = 2^(2^n-1) - 2^n - 1.

-1, -1, 3, 119, 32751, 2147483615, 9223372036854775743, 170141183460469231731687303715884105599, 57896044618658097711785492504343953926634992332820282019728792003956564819711

Offset

0,3

Comments

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)=Product_{r=0..n-1} F_r-(2^(n+1)+1) where F_r=2^(2^r)+1 is the r-th Fermat number. Write n+1=q*2^r 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, Mar 25 2005

The next term -- a(9) -- has 154 digits. - Harvey P. Dale, Aug 14 2011

Links

Table of n, a(n) for n=0..8.

Zhi-Wei Sun, Home page

Zhi-Wei Sun, Problems and Results on Covering Systems, 2005.

Zhi-Wei and M.-H. Le, Integers not of the form c(2^a+2^b)+p^(alpha), Acta Arith., 99(2001), no.2, 183--190.

Mathematica

Table[2^(2^n-1)-2^n-1, {n, 0, 9}] (* Harvey P. Dale, Aug 14 2011 *)

Prog

(Magma) [2^(2^n-1)-2^n-1: n in [0..10] ] // Vincenzo Librandi, Jan 28 2011
(PARI) vector(10, n, 2^(2^(n-1)-1)-2^(n-1)-1) \\ Derek Orr, Mar 29 2015

Crossrefs

Sequence in context: A114077 A176996 A159522 * A304016 A305364 A304956

Adjacent sequences: A103740 A103741 A103742 * A103744 A103745 A103746

Keyword

sign,easy

Author

N. J. A. Sloane, Mar 28 2005

Status

approved