A160380 a(0) = 0; for n >= 1, a(n) = number of 0's in base-4 representation of n.

0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1

Offset: 0

Author

Frank Ruskey, Jun 05 2009

Keywords

• nonn
• base
• easy

Comments

The base-4 representation of 0 is 0, and contains a single zero. - N. J. A. Sloane, Apr 26 2021

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

Crossrefs

Cf. A007090, A023416, A077267.

Programs

Haskell

import Data.List (unfoldr)
a160380 = sum . map ((0 ^ ) . (`mod` 4)) .
   unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4))
-- Reinhard Zumkeller, Apr 22 2011

Mathematica

Join[{0},Table[DigitCount[n,4,0],{n,110}]] (* Harvey P. Dale, Oct 18 2015 *)

PARI

a(n) = #select(x->(x==0), digits(n, 4)); \\ Michel Marcus, Apr 26 2021

Formula

Recurrence relation: a(0) = 0, a(4m) = 1+a(m), a(4m+1) = a(4m+2) = a(4m+3) = a(m).

Generating function: (1/(1-z))*Sum_{m>=0} (z^(4^(m+1))*(1 - z^(4^m))/(1 - z^(4^(m+1)))).

Extensions

Definition clarified by Georg Fischer, Apr 26 2021