A138290 Numbers m such that 2^(m+1) - 2^k - 1 is composite for all 0 <= k < m.

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

Offset: 1

Author

T. D. Noe, Mar 13 2008

Keywords

• nonn

Comments

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.

A208083(a(n)+1) = 0 (cf. A081118). - Reinhard Zumkeller, Feb 23 2012 [Corrected by Thomas Ordowski, Feb 19 2024]

Conjecture: 2^j - 2 are terms for j > 2. - Chai Wah Wu, Sep 07 2021

The proof of this conjecture is in A369375. - Thomas Ordowski, Mar 20 2024

Examples

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

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..996 (terms 1..275 from T. D. Noe)

Crossrefs

Many common terms with A092112.

Cf. A081118, A095058, A110700, A208083, A369375.

Programs

Haskell

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

Mathematica

t={}; Do[num=2^(n+1)-1; k=0; While[kHarvey P. Dale, Apr 09 2022 *)

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

Python

from sympy import isprime
A138290_list = []
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:
        A138290_list.append(n) # Chai Wah Wu, Sep 07 2021

Formula

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

For n > 0, a(n) = A369375(n+1) - 1. - Thomas Ordowski, Mar 20 2024