A085118 Primes together with twice the odd primes.

2, 3, 5, 6, 7, 10, 11, 13, 14, 17, 19, 22, 23, 26, 29, 31, 34, 37, 38, 41, 43, 46, 47, 53, 58, 59, 61, 62, 67, 71, 73, 74, 79, 82, 83, 86, 89, 94, 97, 101, 103, 106, 107, 109, 113, 118, 122, 127, 131, 134, 137, 139, 142, 146, 149, 151, 157, 158, 163, 166, 167, 173, 178

Offset: 1

Author

Leroy Quet, Apr 25 2004

Keywords

• nonn
• easy

Comments

Probably the same sequence as: numbers n such that phi(n)+1 divides n.

Cohen and Segal showed that in case there were other solutions to this problem (which appeared to be posed by Schinzel), then they should have at least 15 distinct prime factors. Moreover, there is a connection with the Lehmer's totient problem which asks whether there is a composite n such that phi(n)|(n-1). If no such composite exists, then p and 2p are the only members for Leroy's sequence. - Francisco Salinas (franciscodesalinas(AT)hotmail.com), Apr 25 2004

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• G. L. Cohen and S. L. Segal, A note concerning those n for which phi(n)+1 divides n, Fibonacci Quarterly, Vol. 27, No. 3 (1989), pp. 285-286.
• Eric Weisstein's World of Mathematics, Lehmer's Totient Problem

Crossrefs

Cf. A000010, A068422.

Equals A001751\{4}.

Programs

Mathematica

With[{nn=40},Take[Sort[Join[Prime[Range[2nn]],2Prime[Range[2,nn]]]],2nn]] (* Harvey P. Dale, Oct 03 2013 *)

Python

from sympy import primepi
def A085118(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+x-primepi(x)-primepi(x>>1)+(x>=4))
    return bisection(f,n,n) # Chai Wah Wu, Oct 17 2024

Extensions

More terms from David Wasserman, Jan 27 2005