A369375 -id:A369375

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A138290

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

+10
6

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

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

LINKS

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

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

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[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 *)

PROG

(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

(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

(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

CROSSREFS

Many common terms with A092112.

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

KEYWORD

nonn

AUTHOR

T. D. Noe, Mar 13 2008

STATUS

approved

A369378

a(n) is the smallest k such that 2^(2^n+1)-1 + 2^k is prime, or -1 if no such k exists.

+10
1

1, 2, 4, 16, 1968, 64, 64

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Conjecture: if n > 0, then 2^(2^n+1)-1 + 2^k is not prime for every k < 2^n.

This conjecture seems provable, because it is not too strong for large n > 6.

If, for n > 6, a(n) does not exist, then 2^(2^n+1)-1 + 2^k is composite for every natural k. Thus, by the dual Sierpinski conjecture, for n > 6, 2^(2^n+1)-1 is a Sierpinski number, i.e., if n > 6, then (2^(2^n+1)-1)2^k+1 is composite for every natural k. For example, the Mersenne number 2^(2^8+1)-1 may be a dual Sierpinski number.

Similarly, if for n > 5, |2^(2^n-1)-1 - 2^m| is not prime for every m > 0, then by the dual Riesel conjecture, 2^(2^n-1)-1 is a Riesel number, i.e., if n > 5, then (2^(2^n-1)-1)2^m-1 is composite for every integer m > 0. For example, the double Mersenne prime 2^(2^7-1)-1 may be a dual Riesel number. By Crocker's theorem; if n > 2, then positive 2^(2^n-1)-1 - 2^m are composite. Let b(n) be the smallest k such that 2^k - (2^(2^n-1)-1) is prime, for n >= 0: {1, 2, 39, 47, 447, 191, ?}.

a(7) > 65000, a(8) thru a(12) > 25000, if they exist. - Robert G. Wilson v, Jan 22 2024

LINKS

Table of n, a(n) for n=0..6.

EXAMPLE

a(1) = 2 since for 2^(2^1+1)-1 = 7 we have 7 + 2^1 = 9 and 7 + 2^2 = 11 is prime.

MATHEMATICA

a[n_] := Module[{m = 2^(2^n + 1) - 1, k = 1}, While[! PrimeQ[m + 2^k], k++]; k]; Array[a, 7, 0] (* Amiram Eldar, Jan 22 2024 *)

CROSSREFS

Cf. A076336, A101036, A156695, A369375.

KEYWORD

nonn,more

AUTHOR

Thomas Ordowski, Jan 22 2024

STATUS

approved

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