

A138290


Numbers n such that 2^(n+1)2^k1 is composite for all 0 <= k < n.


4



6, 14, 22, 26, 30, 36, 38, 42, 54, 57, 62, 70, 78, 81, 90, 94, 110, 122, 126, 132, 134, 138, 142, 147, 150, 158, 166, 168, 171, 172, 174, 178, 182, 190, 194, 198, 206, 210, 222, 238, 254, 285, 294, 312, 315, 318, 334, 336, 350, 366, 372, 382, 405, 414, 416, 432
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OFFSET

1,1


COMMENTS

The binary representation of 2^(n+1)2^k1 has n 1bits and one 0bit. Note that prime n are very rare: 577 is the first and 5569 is the second.
A208083(a(n)) = 0 (cf. A081118). [Reinhard Zumkeller, Feb 23 2012]


LINKS

T. D. Noe, Table of n, a(n) for n=1..275


FORMULA

For these n, A095058(n)=0 and A110700(n)>1.


EXAMPLE

6 is here because 95, 111, 119, 123, 125 and 126 are all composite.


MATHEMATICA

t={}; Do[num=2^(n+1)1; k=0; While[k<n && !PrimeQ[num2^k], k++ ]; If[k==n, AppendTo[t, n]], {n, 100}]; t


PROG

(Haskell)
import Data.List (elemIndices)
a138290 n = a138290_list !! (n1)
a138290_list = map (+ 1) $ tail $ elemIndices 0 a208083_list
 Reinhard Zumkeller, Feb 23 2012


CROSSREFS

Sequence in context: A182081 A125086 A195063 * A023057 A197127 A197171
Adjacent sequences: A138287 A138288 A138289 * A138291 A138292 A138293


KEYWORD

nonn


AUTHOR

T. D. Noe, Mar 13 2008


STATUS

approved



