

A006254


Numbers k such that 2k1 is prime.


80



2, 3, 4, 6, 7, 9, 10, 12, 15, 16, 19, 21, 22, 24, 27, 30, 31, 34, 36, 37, 40, 42, 45, 49, 51, 52, 54, 55, 57, 64, 66, 69, 70, 75, 76, 79, 82, 84, 87, 90, 91, 96, 97, 99, 100, 106, 112, 114, 115, 117, 120, 121, 126, 129, 132, 135, 136, 139, 141, 142, 147, 154, 156, 157
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OFFSET

1,1


COMMENTS

The following sequences (allowing offset of first term) all appear to have the same parity: A034953, triangular numbers with prime indices; A054269, length of period of continued fraction for sqrt(p), p prime; A082749, difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254, numbers n such that 2n1 is prime; A067076, 2n+3 is a prime.  Jeremy Gardiner, Sep 10 2004
Solutions of the equation (2*k1)'=1, where k' is the arithmetic derivative of k.  Paolo P. Lava, Nov 15 2012
Positions of prime numbers among odd numbers.  Zak Seidov, Mar 26 2013
Also, the integers remaining after removing every third integer following 2, and, recursively, removing every pth integer following the next remaining entry (where p runs through the primes, beginning with 5).  Pete Klimek, Feb 10 2014
Also, numbers k such that k^2 = m^2 + p, for some integers m and every prime p > 2. Applicable m values are m = k  1 (giving p = 2k  1). Less obvious is: no solution exists if m equals any value in A047845, which is the complement of (A006254  1).  Richard R. Forberg, Apr 26 2014
If you define a different type of multiplication (*) where x (*) y = x * y + (x  1) * (y  1), (which has the commutative property) then this is the set of primes that follows.  Jason Atwood, Jun 16 2019


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KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



