A025616 Numbers of form 3^i*10^j, with i, j >= 0.
1, 3, 9, 10, 27, 30, 81, 90, 100, 243, 270, 300, 729, 810, 900, 1000, 2187, 2430, 2700, 3000, 6561, 7290, 8100, 9000, 10000, 19683, 21870, 24300, 27000, 30000, 59049, 65610, 72900, 81000, 90000, 100000, 177147, 196830, 218700, 243000, 270000
Offset: 1
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Haskell
import Data.Set (singleton, deleteFindMin, insert) a025616 n = a025616_list !! (n-1) a025616_list = f $ singleton (1,0,0) where f s = y : f (insert (3 * y, i + 1, j) $ insert (10 * y, i, j + 1) s') where ((y, i, j), s') = deleteFindMin s -- Reinhard Zumkeller, May 15 2015 -
Mathematica
n = 10^6; Flatten[Table[3^i*10^j, {i, 0, Log[3, n]}, {j, 0, Log10[n/3^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *) -
PARI
list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 10), N=10^n; while(N<=lim, listput(v, N); N*=3)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018
-
Python
from sympy import integer_log def A025616(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(integer_log(x//10**i,3)[0]+1 for i in range(integer_log(x,10)[0]+1)) return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025
Formula
Sum_{n>=1} 1/a(n) = (3*10)/((3-1)*(10-1)) = 5/3. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(3)*log(10)*n)) / sqrt(30). - Vaclav Kotesovec, Sep 25 2020
Comments