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a(n) = 2^(2^n-1) - 2^n - 1.
0

%I #23 Sep 08 2022 08:45:17

%S -1,-1,3,119,32751,2147483615,9223372036854775743,

%T 170141183460469231731687303715884105599,

%U 57896044618658097711785492504343953926634992332820282019728792003956564819711

%N a(n) = 2^(2^n-1) - 2^n - 1.

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

%C 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

%C The next term -- a(9) -- has 154 digits. - _Harvey P. Dale_, Aug 14 2011

%H Zhi-Wei Sun, <a href="http://pweb.nju.edu.cn/zwsun/">Home page</a>

%H Zhi-Wei Sun, <a href="http://pweb.nju.edu.cn/zwsun/Cover.pdf">Problems and Results on Covering Systems</a>, 2005.

%H Zhi-Wei and M.-H. Le, <a href="http://pweb.nju.edu.cn/zwsun/43l.pdf">Integers not of the form c(2^a+2^b)+p^(alpha)</a>, Acta Arith., 99(2001), no.2, 183--190.

%t Table[2^(2^n-1)-2^n-1,{n,0,9}] (* _Harvey P. Dale_, Aug 14 2011 *)

%o (Magma) [2^(2^n-1)-2^n-1: n in [0..10] ] // _Vincenzo Librandi_, Jan 28 2011

%o (PARI) vector(10,n,2^(2^(n-1)-1)-2^(n-1)-1) \\ _Derek Orr_, Mar 29 2015

%K sign,easy

%O 0,3

%A _N. J. A. Sloane_, Mar 28 2005