A077267 Number of zeros in base-3 expansion of n.
1, 0, 0, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 3, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 3, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 4, 3, 3, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 1, 1, 2, 1, 1, 3, 2, 2, 2, 1, 1, 2
Offset: 0
Examples
a(8)=0 since 8 written in base 3 is 22 with 0 zeros; a(9)=2 since 9 written in base 3 is 100 with 2 zeros; a(10)=1 since 10 written in base 3 is 101 with 1 zero.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- F. T. Adams-Watters, F. Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS 12 (2009) 09.5.6
- Eric Weisstein's World of Mathematics, Ternary.
- Wikipedia, Ternary numeral system
Crossrefs
Programs
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Haskell
a077267 n = a079978 n + if n < 3 then 0 else a077267 (n `div` 3) -- Reinhard Zumkeller, Feb 21 2013
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Mathematica
Table[Count[IntegerDigits[n,3],0],{n,0,6!}] (* Vladimir Joseph Stephan Orlovsky, Jul 25 2009 *) DigitCount[Range[0,110],3,0] (* Harvey P. Dale, Jul 04 2021 *)
Formula
a(1)=a(2)=0; a(3n)=a(n)+1; a(3n+1)=a(3n+2)=a(n). a(3^n-2)=a(3^n-1)=0; a(3^n)=n. a(n)=A077266(n, 3).
G.f.: (Sum_{k>=0} x^(3^(k+1))/(1 + x^(3^k) + x^(2*3^k)))/(1-x). - Franklin T. Adams-Watters, Nov 03 2005
Extensions
a(0)=1 added, offset changed to 0 and b-file adjusted by Reinhard Zumkeller, Feb 21 2013
Wrong formula deleted by Reinhard Zumkeller, Feb 21 2013