A025635 Numbers of form 9^i*10^j, with i, j >= 0.

1, 9, 10, 81, 90, 100, 729, 810, 900, 1000, 6561, 7290, 8100, 9000, 10000, 59049, 65610, 72900, 81000, 90000, 100000, 531441, 590490, 656100, 729000, 810000, 900000, 1000000, 4782969, 5314410, 5904900, 6561000, 7290000, 8100000, 9000000

Offset: 1

Author

David W. Wilson

Keywords

• easy
• nonn

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A025612, A025616, A025621, A025625, A025629, A025632, A025634, A108761, A003596, A003597, A107988, A003598, A108698, A003599, A107788, A108687, A108779, A108090.

Programs

Haskell

import Data.Set (singleton, deleteFindMin, insert)
a025635 n = a025635_list !! (n-1)
a025635_list = f $ singleton (1,0,0) where
   f s = y : f (insert (9 * y, i + 1, j) $ insert (10 * y, i, j + 1) s')
         where ((y, i, j), s') = deleteFindMin s
-- Reinhard Zumkeller, May 15 2015

Mathematica

With[{max = 10^7}, Flatten[Table[9^i*10^j, {i, 0, Log[9, max]}, {j, 0, Log[10, max/9^i]}]] // Sort] (* Amiram Eldar, Mar 29 2025 *)

PARI

list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 10), N=10^n; while(N<=lim, listput(v, N); N*=9)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018

Python

from sympy import integer_log
def A025635(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,9)[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) = 5/4. - Amiram Eldar, Mar 29 2025

a(n) = 9^A025683(n) * 10^A025691(n). - R. J. Mathar, Jul 06 2025