A122840 a(n) is the number of 0's at the end of n when n is written in base 10.

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0

Offset: 1

Author

Reinhard Zumkeller, Sep 13 2006

Keywords

• nonn
• base
• easy

Comments

Greatest k such that 10^k divides n.

a(n) = the number of digits in n - A160093(n).

a(A005117(n)) <= 1. - Reinhard Zumkeller, Mar 30 2010

See A054899 for the partial sums. - Hieronymus Fischer, Jun 08 2012

From Amiram Eldar, Mar 10 2021: (Start)

The asymptotic density of the occurrences of k is 9/10^(k+1).

The asymptotic mean of this sequence is 1/9. (End)

Examples

			a(160) = 1 because there is 1 zero at the end of 160 when 160 is written in base 10.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• S. Ikeda and K. Matsuoka, On transcendental numbers generated by certain integer sequences, Siauliai Math. Semin., 8 (16) 2013, 63-69.

Crossrefs

A007814 is the base 2 equivalent of this sequence.

Cf. A160094, A160093, A001511, A070940, A122841, A027868, A054899, A196563, A196564, A004151, A112765.

Programs

Haskell

a122840 n = if n < 10 then 0 ^ n else 0 ^ d * (a122840 n' + 1)
            where (n', d) = divMod n 10
-- Reinhard Zumkeller, Mar 09 2013

Mathematica

a[n_] := IntegerExponent[n, 10]; Array[a, 100] (* Amiram Eldar, Mar 10 2021 *)

PARI

a(n)=valuation(n,10) \\ Charles R Greathouse IV, Feb 26 2014

Python

def a(n): return len(str(n)) - len(str(int(str(n)[::-1]))) # Indranil Ghosh, Jun 09 2017

Python

def A122840(n): return len(s:=str(n))-len(s.rstrip('0')) # Chai Wah Wu, Jul 06 2022

Python

A122840 = lambda n: sympy.multiplicity(10,n) # M. F. Hasler, Apr 05 2024

Formula

a(n) = A160094(n) - 1.

From Hieronymus Fischer, Jun 08 2012: (Start)

With m = floor(log_10(n)), frac(x) = x-floor(x):

a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/10^j))).

a(n) = m + Sum_{j=1..m} (floor(-frac(n/10^j))).

a(n) = A054899(n) - A054899(n-1).

G.f.: g(x) = Sum_{j>0} x^10^j/(1-x^10^j). (End)

a(n) = min(A007814(n), A112765(n)). - Jianing Song, Jul 23 2022