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A138290
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Numbers m such that 2^(m+1) - 2^k - 1 is composite for all 0 <= k < m.
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6
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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
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1,1
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COMMENTS
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The binary representation of 2^(m+1) - 2^k - 1 has m 1-bits and one 0-bit. Note that prime m are very rare: 577 is the first and 5569 is the second.
Conjecture: 2^j - 2 are terms for j > 2. - Chai Wah Wu, Sep 07 2021
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LINKS
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FORMULA
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EXAMPLE
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6 is here because 95, 111, 119, 123, 125 and 126 are all composite.
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MATHEMATICA
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t={}; Do[num=2^(n+1)-1; k=0; While[k<n && !PrimeQ[num-2^k], k++ ]; If[k==n, AppendTo[t, n]], {n, 100}]; t
Select[Range[500], AllTrue[2^(#+1)-1-2^Range[0, #-1], CompositeQ]&] (* Harvey P. Dale, Apr 09 2022 *)
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PROG
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(Haskell)
import Data.List (elemIndices)
a138290 n = a138290_list !! (n-1)
a138290_list = map (+ 1) $ tail $ elemIndices 0 a208083_list
(Python)
from sympy import isprime
for n in range(1, 10**3):
k2, n2 = 1, 2**(n+1)
for k in range(n):
if isprime(n2-k2-1):
break
k2 *= 2
else:
(PARI) isok(m) = my(nb=0); for (k=0, m-1, if (!ispseudoprime(2^(m+1) - 2^k - 1), nb++, break)); nb==m; \\ Michel Marcus, Sep 13 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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