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 A103743 a(n) = 2^(2^n-1) - 2^n - 1. 0
 -1, -1, 3, 119, 32751, 2147483615, 9223372036854775743, 170141183460469231731687303715884105599, 57896044618658097711785492504343953926634992332820282019728792003956564819711 (list; graph; refs; listen; history; text; internal format)
 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 rn+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

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Last modified January 19 08:18 EST 2022. Contains 350464 sequences. (Running on oeis4.)