A003592 Numbers of the form 2^i*5^j with i, j >= 0.

1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200, 250, 256, 320, 400, 500, 512, 625, 640, 800, 1000, 1024, 1250, 1280, 1600, 2000, 2048, 2500, 2560, 3125, 3200, 4000, 4096, 5000, 5120, 6250, 6400, 8000, 8192, 10000, 10240, 12500, 12800

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

These are the natural numbers whose reciprocals are terminating decimals. - David Wasserman, Feb 26 2002

A132726(a(n), k) = 0 for k <= a(n); A051626(a(n)) = 0; A132740(a(n)) = 1; A132741(a(n)) = a(n). - Reinhard Zumkeller, Aug 27 2007

Where record values greater than 1 occur in A165706: A165707(n) = A165706(a(n)). - Reinhard Zumkeller, Sep 26 2009

Also numbers that are divisible by neither 10k - 7, 10k - 3, 10k - 1 nor 10k + 1, for all k > 0. - Robert G. Wilson v, Oct 26 2010

A204455(5*a(n)) = 5, and only for these numbers. - Wolfdieter Lang, Feb 04 2012

Since p = 2 and q = 5 are coprime, sum_{n >= 1} 1/a(n) = sum_{i >= 0} sum_{j >= 0} 1/p^i * 1/q^j = sum_{i >= 0} 1/p^i q/(q - 1) = p*q/((p-1)*(q-1)) = 2*5/(1*4) = 2.5. - Franklin T. Adams-Watters, Jul 07 2014

Conjecture: Each positive integer n not among 1, 4 and 12 can be written as a sum of finitely many numbers of the form 2^a*5^b + 1 (a,b >= 0) with no one dividing another. This has been verified for n <= 3700. - Zhi-Wei Sun, Apr 18 2023

1,2 and 4,5 are the only consecutive terms in the sequence. - Robin Jones, May 03 2025

References

• Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 73.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Vaclav Kotesovec, Graph - the asymptotic ratio (200000 terms)
• Math Stack Exchange, Are (1,2) and (4,5) the only two consecutive pairs in A003592, the integers of the form 2^i5^j?
• Eric Weisstein's World of Mathematics, Regular Number
• Eric Weisstein's World of Mathematics, Decimal Expansion

Crossrefs

Complement of A085837. Cf. A094958, A022333 (list of j), A022332 (list of i).

Cf. A003586, A003591, A003593, A003594, A003595, A257997.

Cf. A164768 (difference between consecutive terms)

Programs

GAP

Filtered([1..10000],n->PowerMod(10,n,n)=0); # Muniru A Asiru, Mar 19 2019

Haskell

import Data.Set (singleton, deleteFindMin, insert)
a003592 n = a003592_list !! (n-1)
a003592_list = f $ singleton 1 where
   f s = y : f (insert (2 * y) $ insert (5 * y) s')
               where (y, s') = deleteFindMin s
-- Reinhard Zumkeller, May 16 2015

Magma

[n: n in [1..10000] | PrimeDivisors(n) subset [2,5]]; // Bruno Berselli, Sep 24 2012

Maple

isA003592 := proc(n)
      if n = 1 then
        true;
    else
        return (numtheory[factorset](n) minus {2,5} = {} );
    end if;
end proc:
A003592 := proc(n)
     option remember;
     if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isA003592(a) then
                return a;
            end if;
        end do:
    end if;
end proc: # R. J. Mathar, Jul 16 2012

Mathematica

twoFiveableQ[n_] := PowerMod[10, n, n] == 0; Select[Range@ 10000, twoFiveableQ] (* Robert G. Wilson v, Jan 12 2012 *)
twoFiveableQ[n_] := Union[ MemberQ[{1, 3, 7, 9}, # ] & /@ Union@ Mod[ Rest@ Divisors@ n, 10]] == {False}; twoFiveableQ[1] = True; Select[Range@ 10000, twoFiveableQ] (* Robert G. Wilson v, Oct 26 2010 *)
maxExpo = 14; Sort@ Flatten@ Table[2^i * 5^j, {i, 0, maxExpo}, {j, 0, Log[5, 2^(maxExpo - i)]}] (* Or *)
Union@ Flatten@ NestList[{2#, 4#, 5#} &, 1, 7] (* Robert G. Wilson v, Apr 16 2011 *)

PARI

list(lim)=my(v=List(),N);for(n=0,log(lim+.5)\log(5),N=5^n;while(N<=lim,listput(v,N);N<<=1));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

Python

# A003592.py
from heapq import heappush, heappop
def A003592():
    pq = [1]
    seen = set(pq)
    while True:
        value = heappop(pq)
        yield value
        seen.remove(value)
        for x in 2*value, 5*value:
            if x not in seen:
                heappush(pq, x)
                seen.add(x)
sequence = A003592()
A003592_list = [next(sequence) for _ in range(100)]

Python

from sympy import integer_log
def A003592(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//5**i).bit_length() for i in range(integer_log(x,5)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Feb 24 2025

Sage

def isA003592(n) :
    return not any(d != 2 and d != 5 for d in prime_divisors(n))
@CachedFunction
def A003592(n) :
    if n == 1 : return 1
    k = A003592(n-1) + 1
    while not isA003592(k) : k += 1
    return k
[A003592(n) for n in (1..48)]  # Peter Luschny, Jul 20 2012

Formula

The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} mu(10*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

a(n) ~ exp(sqrt(2*log(2)*log(5)*n)) / sqrt(10). - Vaclav Kotesovec, Sep 22 2020

a(n) = 2^A022332(n) * 5^A022333(n). - R. J. Mathar, Jul 06 2025

Extensions

Incomplete Python program removed by David Radcliffe, Jun 27 2016