A003596 Numbers of the form 2^i * 11^j.

1, 2, 4, 8, 11, 16, 22, 32, 44, 64, 88, 121, 128, 176, 242, 256, 352, 484, 512, 704, 968, 1024, 1331, 1408, 1936, 2048, 2662, 2816, 3872, 4096, 5324, 5632, 7744, 8192, 10648, 11264, 14641, 15488, 16384, 21296, 22528, 29282, 30976, 32768

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

A204455(11*a(n)) = 11, and only for these numbers. - Wolfdieter Lang, Feb 04 2012

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 100 terms from Vincenzo Librandi)

Crossrefs

Cf. A025612, A025616, A025621, A025625, A025629, A025632, A025634, A025635, A108761, A003597, A107988, A003598, A108698, A003599, A107788, A108687, A108779, A108090.

Programs

GAP

Filtered([1..33000],n->PowerMod(22,n,n)=0); # Muniru A Asiru, Mar 19 2019

Haskell

import Data.Set (singleton, deleteFindMin, insert)
a003596 n = a003596_list !! (n-1)
a003596_list = f $ singleton (1,0,0) where
   f s = y : f (insert (2 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
         where ((y, i, j), s') = deleteFindMin s
-- Reinhard Zumkeller, May 15 2015

Magma

[n: n in [1..2*10^5] | PrimeDivisors(n) subset [2, 11]]; // Vincenzo Librandi, Jun 27 2016

Mathematica

fQ[n_] := PowerMod[22,n,n]==0; Select[Range[40000], fQ] (* Vincenzo Librandi, Feb 04 2012 *)

PARI

list(lim)=my(v=List(),N);for(n=0,log(lim)\log(11),N=11^n;while(N<=lim,listput(v,N);N<<=1));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

Python

from sympy import integer_log
def A003596(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = 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 n+x-sum((x//11**i).bit_length() for i in range(integer_log(x,11)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Formula

The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} mu(22*n)*x^n/(1 - x^n), where mu(n) is the Möbius function A008683. Cf. with the formula of Hanna in A051037. - Peter Bala, Mar 18 2019

Sum_{n>=1} 1/a(n) = (2*11)/((2-1)*(11-1)) = 11/5. - Amiram Eldar, Sep 23 2020

a(n) ~ exp(sqrt(2*log(2)*log(11)*n)) / sqrt(22). - Vaclav Kotesovec, Sep 23 2020