A160093 Number of digits in n, excluding any trailing zeros.
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 3, 3, 3, 3
Offset: 1
Examples
a(1060000) = 3 because discarding the trailing zeros from 1060000 leaves 106, which is a 3-digit number.
Links
- Indranil Ghosh, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
lnzd[n_]:=Module[{spl=Last[Split[IntegerDigits[n]]]},If[!MemberQ[ spl,0], IntegerLength[n], IntegerLength[n]-Length[spl]]]; Array[lnzd,110] (* Harvey P. Dale, Jun 05 2013 *) Table[IntegerLength[n] - IntegerExponent[n, 10], {n, 100}] (* Amiram Eldar, Sep 14 2020 *)
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PARI
a(n)=if(n==0,1,#digits(n/10^valuation(n,10))) \\ Joerg Arndt, Jan 11 2017
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PARI
a(n)=logint(n,10)+1-valuation(n,10) \\ Charles R Greathouse IV, Jan 12 2017
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Python
def A160093(n): return len(str(int(str(n)[::-1]))) # Indranil Ghosh, Jan 11 2017
Formula
From Hieronymus Fischer, Jun 08 2012: (Start)
With m = floor(log_10(n)), frac(x) = x-floor(x):
a(n) = 1 + Sum_{j=0..m} ceiling(frac(n/10^j)).
a(n) = 1 - Sum_{j=1..m} (floor(-frac(n/10^j))).
G.f.: (x/(1-x)) + (1/(1-x))*Sum_{j>0} x^(10^j+1)*(1 - x^(10^j-1))/(1-x^10^j). (End)
Extensions
Simpler definition and changed example from Jon E. Schoenfield, Feb 15 2014