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A103743
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2^(2^n-1)-2^n-1.
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0
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-1, -1, 3, 119, 32751, 2147483615, 9223372036854775743, 170141183460469231731687303715884105599, 57896044618658097711785492504343953926634992332820282019728792003956564819711
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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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)=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
The next term -- a(9) -- hast 154 digits. [From Harvey P. Dale, Aug 14 2011]
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LINKS
| Zhi-Wei Sun, Home page
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MATHEMATICA
| Table[2^(2^n-1)-2^n-1, {n, 0, 9}] (* From Harvey P. Dale, Aug 14 2011 *)
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PROG
| (MAGMA) [2^(2^n-1)-2^n-1: n in [0..10] ] [From Vincenzo Librandi, Jan 28 2011]
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CROSSREFS
| Sequence in context: A114077 A176996 A159522 * A185554 A134230 A202768
Adjacent sequences: A103740 A103741 A103742 * A103744 A103745 A103746
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KEYWORD
| sign
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Mar 28 2005
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