A215879 Written in base 3, n ends in a(n) consecutive nonzero digits.

0, 1, 1, 0, 2, 2, 0, 2, 2, 0, 1, 1, 0, 3, 3, 0, 3, 3, 0, 1, 1, 0, 3, 3, 0, 3, 3, 0, 1, 1, 0, 2, 2, 0, 2, 2, 0, 1, 1, 0, 4, 4, 0, 4, 4, 0, 1, 1, 0, 4, 4, 0, 4, 4, 0, 1, 1, 0, 2, 2, 0, 2, 2, 0, 1, 1, 0, 4, 4, 0, 4, 4, 0, 1, 1, 0, 4, 4, 0, 4, 4, 0, 1, 1, 0, 2, 2, 0, 2, 2, 0, 1, 1, 0, 3, 3, 0, 3, 3, 0, 1, 1, 0, 3, 3, 0, 3

Offset: 0

Author

M. F. Hasler, Aug 25 2012

Keywords

• nonn
• base
• nice

Comments

Somehow complementary to A007949, the 3-adic valuation of n.

The base 2 analog of this sequence essentially coincides with the 2-adic valuation A007814 (up to a shift in the index).

One gets back the same sequence by concatenation of the pattern (0,1,1) successively multiplied by a(n)+1 = 1, 2, 2, 1, 3, 3, ... for n = 0, 1, 2, 3, 4, 5, .... This is equivalent to the formula (a(n)+1)*(0, 1, 1) = a(3n, 3n+1, 3n+2). - M. F. Hasler, Aug 26 2012, corrected Aug 23 2022

a(A008585(n)) = 0; a(A001651(n)) > 0. - Reinhard Zumkeller, Dec 28 2012

Examples

			The numbers 0, 1, 2, 3, 4, 5, 6, 7 are written in base 3 as 0, 1, 2, 10, 11, 12, 20, 21 and thus end in a(0..7) = 0, 1, 1, 0, 2, 2, 0, 2 nonzero digits.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

The base-4, base-5 and base-10 analogs of this sequence are given in A215883, A215884 and A215887.

Cf. A007089.

Programs

Haskell

a215879 n = if t == 0 then 0 else a215879 n' + 1
            where (n',t) = divMod n 3
-- Reinhard Zumkeller, Dec 28 2012

Mathematica

cnzd[n_]:=Module[{idn3=IntegerDigits[n,3],len},len=Length[idn3];Which[ idn3[[len]] == 0,0,Position[idn3,0]=={},len,True,len-Position[idn3,0] [[-1,1]]]]; Array[cnzd,110,0] (* Harvey P. Dale, Jun 07 2016 *)

PARI

A215879(n,b=3)=n=divrem(n,b); for(c=0,oo,n[2]||return(c); n=divrem(n[1],b))

PARI

a(n)=my(k);while(n%3,n\=3;k++);k \\ Charles R Greathouse IV, Sep 26 2013

Python

def A215879(n):
    c = 0
    while (a:=divmod(n,3))[1]:
        c += 1
        n = a[0]
    return c # Chai Wah Wu, Oct 15 2022

Formula

a(3^(t+1)*k+m) = t for 3^t > m > 3^(t-1).

a(3n) = 0, a(3n+1) = a(3n+2) = a(n)+1. - M. F. Hasler, Aug 26 2012, corrected thanks to a remark from Jianing Song, Aug 23 2022