login
A025616
Numbers of form 3^i*10^j, with i, j >= 0.
14
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
1,2
LINKS
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
a(n) = 3^A025644(n) *10^A025685(n). - R. J. Mathar, Jul 06 2025
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 *)
PROG
(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
(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
KEYWORD
easy,nonn
STATUS
approved