A033845 Numbers k of the form 2^i*3^j, where i and j >= 1.
6, 12, 18, 24, 36, 48, 54, 72, 96, 108, 144, 162, 192, 216, 288, 324, 384, 432, 486, 576, 648, 768, 864, 972, 1152, 1296, 1458, 1536, 1728, 1944, 2304, 2592, 2916, 3072, 3456, 3888, 4374, 4608, 5184, 5832, 6144, 6912, 7776, 8748, 9216, 10368, 11664
Offset: 1
References
- J-M. de Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 733, page 94.
- J-M. de Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 654, page 85.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Haskell
import Data.Set (singleton, deleteFindMin, insert) a033845 n = a033845_list !! (n-1) a033845_list = f (singleton (2*3)) where f s = m : f (insert (2*m) $ insert (3*m) s') where (m,s') = deleteFindMin s -- Reinhard Zumkeller, Sep 13 2011
-
Mathematica
mx = 12000; Sort@ Flatten@ Table[2^i*3^j, {i, Log[2, mx]}, {j, Log[3, mx/2^i]}] (* Robert G. Wilson v, Aug 17 2012 *)
-
PARI
list(lim)=my(v=List(), N); for(n=0, log(lim\2)\log(3), N=6*3^n; while(N<=lim, listput(v, N); N<<=1)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jan 02 2012
-
Python
from sympy import integer_log def A033845(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 n+x-sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1)) return 6*bisection(f,n,n) # Chai Wah Wu, Sep 15 2024
-
Python
# faster for initial segment of sequence import heapq from itertools import islice def A033845gen(): # generator of terms v, oldv, h, psmooth_primes, = 1, 0, [1], [2, 3] while True: v = heapq.heappop(h) if v != oldv: yield 6*v oldv = v for p in psmooth_primes: heapq.heappush(h, v*p) print(list(islice(A033845gen(), 50))) # Michael S. Branicky, Sep 18 2024
Formula
Six times the 3-smooth numbers (A003586). - Ralf Stephan, Apr 16 2004
A143201(a(n)) = 2. - Reinhard Zumkeller, Sep 13 2011
Sum_{n>=1} 1/a(n) = 1/2. - Amiram Eldar, Oct 13 2020
Extensions
Edited by N. J. A. Sloane, Jan 31 2010 and May 26 2024.
Comments